This is going to be strange,
$$I=\int_0^1 \frac{\ln(1-x) \ln^2 (x) \ln^2(1+x)}{1-x} \, dx$$
Performing $\frac{1-x}{1+x}=u$,
$$I = \int_{0}^{1} \frac{\log\left(\frac{2u}{1+u}\right) \log^2\left(\frac{1-u}{1+u}\right) \log^2\left(\frac{2}{1+u}\right)}{u(1+u)} \, du$$
Using,
$$\log\left(\frac{2u}{1+u}\right) = \log 2 - \log u - \log(1+u)\tag1$$
$$\log^2\left(\frac{1-u}{1+u}\right) = \log^2(1-u) + \log^2(1+u)-2\log(1-u)\log(1+u)\tag2$$
$$\log^2\left(\frac{2}{1+u}\right) = \log^2 2 + \log^2(1+u)-2\log(2)\log(1+u)\tag3$$
Also,
$$\frac{1}{u(1+u)}=\frac{1}{u}-\frac{1}{1+u}$$
$$I = \int_{0}^{1} \frac{\log\left(\frac{2u}{1+u}\right) \log^2\left(\frac{1-u}{1+u}\right) \log^2\left(\frac{2}{1+u}\right)}{u} \, du-\int_{0}^{1} \frac{\log\left(\frac{2u}{1+u}\right) \log^2\left(\frac{1-u}{1+u}\right) \log^2\left(\frac{2}{1+u}\right)}{1+u} \, du$$
$$I = I_1-I_2$$
Section - 1
Working with our first integral,
$$I_1= \int_{0}^{1} \frac{\log\left(\frac{2u}{1+u}\right) \log^2\left(\frac{1-u}{1+u}\right) \log^2\left(\frac{2}{1+u}\right)}{u} \, du$$
Now after multiplying $(1),(2),(3)$,
$$\log^3(2)\log^2(1-u)+\log^3(2)\log^2(1+u)-2\log^3(2)\log(1-u)\log(1+u)-\log^2(2)\log(u)\log^2(1-u)-\log^2(2)\log(u)\log^2(1+u)+2\log^2(2)\log(u)\log(1-u)\log(1+u)-\log^2(2)\log(1+u)\log^2(1-u)-\log^2(2)\log^3(1+u)-2\log^2(2)\log(1-u)\log^2(1+u)+\log(2)\log^2(1-u)\log^2(1+u)+\log(2)\log^4(1+u)-2\log(2)\log(1-u)\log^3(1+u)-\log(u)\log^2(1-u)\log^2(1+u)-\log(u)\log^4(1+u)+2\log(u)\log(1-u)\log^3(1+u)-\log^3(1+u)\log^2(1-u)-\log^5(1+u)-2\log(1-u)\log^4(1+u)-2\log^2(2)\log(1+u)\log^2(1-u)-2\log^2(2)\log^2(1+u)+4\log^2(2)\log(1-u)\log^2(1+u)+2\log(2)\log(u)\log(1+u)\log^2(1-u)+2\log(2)\log(u)\log^3(1+u)-4\log(2)\log(u)\log(1-u)\log^2(1+u)+2\log(2)\log^2(1+u)\log^2(1-u)+2\log(2)\log^4(1+u)+4\log(2)\log(1-u)\log^3(1+u)$$
Dividing by $u$,
$$\frac{\log^3(2)\log^2(1-u)}{u}+\frac{\log^3(2)\log^2(1+u)}{u}-\frac{2\log^3(2)\log(1-u)\log(1+u)}{u}-\frac{\log^2(2)\log(u)\log^2(1-u)}{u}-\frac{\log^2(2)\log(u)\log^2(1+u)}{u}+\frac{2\log^2(2)\log(u)\log(1-u)\log(1+u)}{u}-\frac{\log^2(2)\log(1+u)\log^2(1-u)}{u}-\frac{\log^2(2)\log^3(1+u)}{u}-\frac{2\log^2(2)\log(1-u)\log^2(1+u)}{u}+\frac{\log(2)\log^2(1-u)\log^2(1+u)}{u}+\frac{\log(2)\log^4(1+u)}{u}-\frac{2\log(2)\log(1-u)\log^3(1+u)}{u}-\frac{\log(u)\log^2(1-u)\log^2(1+u)}{u}-\frac{\log(u)\log^4(1+u)}{u}+\frac{2\log(u)\log(1-u)\log^3(1+u)}{u}-\frac{\log^3(1+u)\log^2(1-u)}{u}-\frac{\log^5(1+u)}{u}-\frac{2\log(1-u)\log^4(1+u)}{u}-\frac{2\log^2(2)\log(1+u)\log^2(1-u)}{u}-\frac{2\log^2(2)\log^2(1+u)}{u}+\frac{4\log^2(2)\log(1-u)\log^2(1+u)}{u}+\frac{2\log(2)\log(u)\log(1+u)\log^2(1-u)}{u}+\frac{2\log(2)\log(u)\log^3(1+u)}{u}-\frac{4\log(2)\log(u)\log(1-u)\log^2(1+u)}{u}+\frac{2\log(2)\log^2(1+u)\log^2(1-u)}{u}+\frac{2\log(2)\log^4(1+u)}{u}+\frac{4\log(2)\log(1-u)\log^3(1+u)}{u}$$
We have,
$$I_1=\log^3(2)A+\log^3(2)B-2\log^3(2)C-D-\log^2(2)E+2\log^2(2)F-\log^2(2)G-H-2\log^2(2)K+\log(2)L+\log(2)M-2\log(2)N-P-Q+2R-S-T-2W-2\log^2(2)X-2\log^2(2)Y+4\log^2(2)Z+2\log(2)\alpha+2\log(2)\beta-4\log(2)\gamma+2\log(2)\delta+2\log(2)\Omega+4\log(2)\tau$$
Where,
$$\boxed{A=\int_0^1 \frac{\log^2(1-u)}{u} \, du=2\zeta(3)}$$
solution of A
$$\boxed{B=\int_0^1 \frac{\log^2(1+u)}{u} \, du=\frac{1}{4}\zeta(3)}$$
solution of B
$$\boxed{C=\int_0^1 \frac{\log(1-u)\log(1+u)}{u} \, du=-\frac{5}{8}\zeta(3)}$$
solution of C
$$\boxed{D=\int_0^1 \frac{\log(u)\log^2(1-u)}{u} \, du=-\frac12 \zeta(4)}$$
solution of D
$$\boxed{E=\int_0^1 \frac{\log(u)\log^2(1+u)}{u} \, du=\frac{\pi^4}{24}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}}$$
solution of E
$$\boxed{F=\int_0^1 \frac{\log(u)\log(1-u)\log(1+u)}{u} \, du=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12}\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right)}$$
solution of F
$$\boxed{G=\int_0^1 \frac{\log^2(1-u)\log(1+u)}{u} \, du=2\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac58\zeta(4)+\frac72\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)+\frac{1}{12}\ln^4(2)}$$
G was self-computed
$$\boxed{H=\int_0^1 \frac{\log^3(1+u)}{u} \, du=6\zeta(4)-\frac{21}{4}\ln(2)\zeta(3)+\frac{3}{2}\ln^{2}(2)\zeta(2)-\frac{1}{4}\ln^{4}(2)-6\operatorname{Li}_4\left(\frac{1}{2}\right)}$$
H was self-computed
$$\boxed{K=\int_0^1 \frac{\log(1-u)\log^2(1+u)}{u} \, du=-\frac {\pi^4}{240}}$$
solution of K
$$\boxed{L=\int_0^1 \frac{\log^2(1-u)\log^2(1+u)}{u} \, du=\frac{21}{24}\zeta(5)-\frac16\left(24\zeta(5)-\frac{21}{2}\ln^2(2)\zeta(3)+4\ln^3(2)\zeta(3)-\frac45\ln^5(2)-24\ln(2)\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)\right)}$$
L was self-computed
$$\boxed{M=\int_0^1 \frac{\log^4(1+u)}{u}du=24\zeta(5)-\frac{21}{2}\ln^{2}(2)\zeta(3)+4\ln^{3}(2)\zeta(2)-\frac{4}{5}\ln^{5}(2)-24\ln(2)\operatorname{Li}_4\left(\frac{1}{2}\right)-24\operatorname{Li}_5\left(\frac{1}{2}\right)}$$
M was self-computed
$$\boxed{N=\int_0^1 \frac{\log(1-u)\log^3(1+u)}{u} \, du=-6\operatorname{Li}_5\left(\frac12\right)-6\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{3}{4}\zeta(5)+\frac{21}{8}\zeta(2)\zeta(3)\\\quad-\frac{21}8\ln^22\zeta(3)+\ln^32\zeta(2)-\frac15\ln^52}$$
Solution of N
$$\boxed{P=\int_0^1 \frac{\log(u)\log^2(1+u)\log^2(1-u)}{u} \, du=-2 \zeta(\bar5,1)+8 \text{Li}_6\left(\frac{1}{2}\right)+4 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+8 \text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{13 \zeta (3)^2}{16}+\frac{7}{6} \zeta (3) \log ^3(2)-\frac{221 \pi ^6}{30240}+\frac{\log ^6(2)}{9}-\frac{1}{12} \pi ^2 \log ^4(2)}$$
Solution of P
$$Q=\int_0^1 \frac{\log(u)\log^4(1+u)}{u} \, du$$
from here,
$$\mathcal{J}_{2}^{(4,1)} = - \frac{[\log(\tfrac{1}{2})]^{6}}{6}=-\frac16\log^6(2)$$
$$\mathcal{J}_{1}^{(4,1)}=-24\left[\zeta(6) - \operatorname{Li}_6\left(\frac12\right)-\log(2)\operatorname{Li}_5\left(\frac12\right)-\frac12\log^2(2)\operatorname{Li}_4\left(\frac12\right)-\frac16\log^3(2)\operatorname{Li}_3\left(\frac12\right)-\frac1{24}\log^4(2)\operatorname{Li}_2\left(\frac12\right) \right]$$
$$\mathcal{J}_{0}^{(4,1)} =-12 \left[ \zeta(5,1) - \zeta_1(2; 5, 1) - \log(2) \zeta_1(2; 4, 1) - \frac12\log^2(2) \zeta_1(2; 3, 1) - \frac16 \log^3(2) \zeta_1(2; 2, 1) \right]$$
Substituting these we can (possibly) obtain a closed form!
Alternatively,
Here's a partial proof to finding it's closed form;
Substitute $x=\frac{1-x}{x}$,
$$Q=\int_{\frac12}^1\frac{\ln(1-x)\ln^4(x)}{x(1-x)}\,dx-\int_{\frac12}^1\frac{\ln^5(x)}{x(1-x)}\,dx$$
We apply partial fractions to obtain 4 more integrals,
$$Q=\int_{\frac12}^1\frac{\ln(1-x)\ln^4(x)}{x}\,dx+\int_{\frac12}^1\frac{\ln(1-x)\ln^4(x)}{1-x}\,dx-\int_{\frac12}^1\frac{\ln^5(x)}{x}\,dx-\int_{\frac12}^1\frac{\ln^5(x)}{1-x}\,dx$$
While I'm not able to move on with the first two integrals, the last two are quite straightforward;
$$\int_{\frac12}^1\frac{\ln^5(x)}{x}\,dx=-\frac{\ln^6(2)}{6}$$
$$\int_{\frac12}^1\frac{\ln^5(x)}{1-x}\,dx=\sum_{n\ge 1} \int_{\frac12}^1 x^{n-1}\ln^5(x)\,dx$$
$$=\sum_{n\ge 1} \frac{\ln^5(2)}{n2^n}+\frac{5\ln^4(2)}{n^22^n}+\frac{20\ln^3(2)}{n^32^n}+\frac{60\ln^2(2)}{n^42^n}+\frac{120\ln(2)}{n^52^n}+\frac{120}{n^62^n}-\frac{120}{n^6}=120\,\operatorname{Li}_6\left(\tfrac{1}{2}\right) + 60\,\operatorname{Li}_4\left(\tfrac{1}{2}\right)\log^2(2) + 120\,\operatorname{Li}_5\left(\tfrac{1}{2}\right)\log(2) + \tfrac{35}{2}\zeta(3)\log^3(2) - \tfrac{8\pi^6}{63} + \tfrac{11\log^6(2)}{6} - \tfrac{5}{4}\pi^2\log^4(2)$$
All these sums are well known, I am not going to provide the proof for each,
You can compare the results integral and sum
$$R=\int_0^1 \frac{\log(u)\log(1-u)\log^3(1+u)}{u} \, du$$
$$S=\int_0^1 \frac{\log^3(1+u)\log^2(1-u)}{u} \, du$$
$$\boxed{T=\int_0^1 \frac{\log^5(1+u)}{u} \, du=\frac{1}{6}\ln^6(2)+\frac{8\pi^6}{63}+\sum_{n=0}^5 \binom{5}{n}\operatorname{Li}_{n+1}\left(\frac12\right)\ln^{5-n}(2)n!}$$
T was self-computed
$$W=\int_0^1 \frac{\log(1-u)\log^4(1+u)}{u} \, du$$
$$\boxed{X=\int_0^1 \frac{\log(1+u)\log^2(1-u)}{u} \, du=2\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac58\zeta(4)+\frac72\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)+\frac{1}{12}\ln^4(2)}$$
Computed same as $G$
$$\boxed{Y=\int_0^1 \frac{\log^2(1+u)}{u} \, du=\frac{1}{4}\zeta(3)}$$
Above integral computed same as $B$
$$\boxed{Z=\int_0^1 \frac{\log^2(1+u)\log(1-u)}{u} \, du=-\frac {\pi^4}{240}}$$
Above integral computed same as $K$
$$\alpha=\int_0^1 \frac{\log(u)\log(1+u)\log^2(1-u)}{u} \, du$$
$$\boxed{\beta=\int_0^1 \frac{\log(u)\log^3(1+u)}{u} \, du=\frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac{21}4\zeta(3)\ln^22\\-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right)}$$
Solution of the above integral
$$\boxed{\gamma=\int_0^1 \frac{\log(u)\log(1-u)\log^2(1+u)}{u} \, du=\frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)}$$
Solution of the above integral
$$\boxed{\delta=\int_0^1 \frac{\log^2(1-u)\log^2(1+u)}{u} \, du=\frac{21}{24}\zeta(5)-\frac16\left(24\zeta(5)-\frac{21}{2}\ln^2(2)\zeta(3)+4\ln^3(2)\zeta(3)-\frac45\ln^5(2)-24\ln(2)\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)\right)}$$
Above integral computed same as $L$
$$\boxed{\Omega=\int_0^1 \frac{\log^4(1+u)}{u} \, du=24\zeta(5)-\frac{21}{2}\ln^{2}(2)\zeta(3)+4\ln^{3}(2)\zeta(2)-\frac{4}{5}\ln^{5}(2)-24\ln(2)\operatorname{Li}_4\left(\frac{1}{2}\right)-24\operatorname{Li}_5\left(\frac{1}{2}\right)}
$$
$\Omega$ was self-computed
$$\boxed{\tau=\int_0^1 \frac{\log(1-u)\log^3(1+u)}{u} \, du=-6\operatorname{Li}_5\left(\frac12\right)-6\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{3}{4}\zeta(5)+\frac{21}{8}\zeta(2)\zeta(3)\\\quad-\frac{21}8\ln^22\zeta(3)+\ln^32\zeta(2)-\frac15\ln^52}$$
Above integral is computed same as $N$
Section - 2
Working with our second integral,
$$I_2= \int_{0}^{1} \frac{\log\left(\frac{2u}{1+u}\right) \log^2\left(\frac{1-u}{1+u}\right) \log^2\left(\frac{2}{1+u}\right)}{1+u} \, du$$
Now after multiplying $(1),(2),(3)$,
$$\log^3(2)\log^2(1-u)+\log^3(2)\log^2(1+u)-2\log^3(2)\log(1-u)\log(1+u)-\log^2(2)\log(u)\log^2(1-u)-\log^2(2)\log(u)\log^2(1+u)+2\log^2(2)\log(u)\log(1-u)\log(1+u)-\log^2(2)\log(1+u)\log^2(1-u)-\log^2(2)\log^3(1+u)-2\log^2(2)\log(1-u)\log^2(1+u)+\log(2)\log^2(1-u)\log^2(1+u)+\log(2)\log^4(1+u)-2\log(2)\log(1-u)\log^3(1+u)-\log(u)\log^2(1-u)\log^2(1+u)-\log(u)\log^4(1+u)+2\log(u)\log(1-u)\log^3(1+u)-\log^3(1+u)\log^2(1-u)-\log^5(1+u)-2\log(1-u)\log^4(1+u)-2\log^2(2)\log(1+u)\log^2(1-u)-2\log^2(2)\log^2(1+u)+4\log^2(2)\log(1-u)\log^2(1+u)+2\log(2)\log(u)\log(1+u)\log^2(1-u)+2\log(2)\log(u)\log^3(1+u)-4\log(2)\log(u)\log(1-u)\log^2(1+u)+2\log(2)\log^2(1+u)\log^2(1-u)+2\log(2)\log^4(1+u)+4\log(2)\log(1-u)\log^3(1+u)$$
Dividing by $1+u$,
$$\frac{\log^3(2)\log^2(1-u)}{1+u}+\frac{\log^3(2)\log^2(1+u)}{1+u}-\frac{2\log^3(2)\log(1-u)\log(1+u)}{1+u}-\frac{\log^2(2)\log(u)\log^2(1-u)}{1+u}-\frac{\log^2(2)\log(u)\log^2(1+u)}{1+u}+\frac{2\log^2(2)\log(u)\log(1-u)\log(1+u)}{1+u}-\frac{\log^2(2)\log(1+u)\log^2(1-u)}{1+u}-\frac{\log^2(2)\log^3(1+u)}{1+u}-\frac{2\log^2(2)\log(1-u)\log^2(1+u)}{1+u}+\frac{\log(2)\log^2(1-u)\log^2(1+u)}{1+u}+\frac{\log(2)\log^4(1+u)}{1+u}-\frac{2\log(2)\log(1-u)\log^3(1+u)}{1+u}-\frac{\log(u)\log^2(1-u)\log^2(1+u)}{1+u}-\frac{\log(u)\log^4(1+u)}{1+u}+\frac{2\log(u)\log(1-u)\log^3(1+u)}{1+u}-\frac{\log^3(1+u)\log^2(1-u)}{1+u}-\frac{\log^5(1+u)}{1+u}-\frac{2\log(1-u)\log^4(1+u)}{1+u}-\frac{2\log^2(2)\log(1+u)\log^2(1-u)}{1+u}-\frac{2\log^2(2)\log^2(1+u)}{1+u}+\frac{4\log^2(2)\log(1-u)\log^2(1+u)}{1+u}+\frac{2\log(2)\log(u)\log(1+u)\log^2(1-u)}{1+u}+\frac{2\log(2)\log(u)\log^3(1+u)}{1+u}-\frac{4\log(2)\log(u)\log(1-u)\log^2(1+u)}{1+u}+\frac{2\log(2)\log^2(1+u)\log^2(1-u)}{1+u}+\frac{2\log(2)\log^4(1+u)}{1+u}+\frac{4\log(2)\log(1-u)\log^3(1+u)}{1+u}$$
We have,
$$I_1=\log^3(2)A+\log^3(2)B-2\log^3(2)C-D-\log^2(2)E+2\log^2(2)F-\log^2(2)G-H-2\log^2(2)K+\log(2)L+\log(2)M-2\log(2)N-P-Q+2R-S-T-2W-2\log^2(2)X-2\log^2(2)Y+4\log^2(2)Z+2\log(2)\alpha+2\log(2)\beta-4\log(2)\gamma+2\log(2)\delta+2\log(2)\Omega+4\log(2)\tau$$
Where,
Please note I am out of notations to use so Ill be reusing the old ones,
$$\boxed{A=\int_0^1 \frac{\log^2(1-u)}{1+u} \, du=\frac{7}{4}\zeta(3)+\frac13\ln^3(2)-\frac{\pi^2}{6}\ln(2)}$$
A was self-computed
$$\boxed{B=\int_0^1 \frac{\log^2(1+u)}{1+u} \, du=\frac13 \ln^3(2)}$$
B was self-computed
$$\boxed{C=\int_0^1 \frac{\log(1-u)\log(1+u)}{1+u} \, du = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}}$$
Solution to C
$$\boxed{D=\int_0^1 \frac{\log(u)\log^2(1-u)}{1+u} \, du=\frac{11}{4}\zeta(4)-\frac14\ln^42-6\operatorname{Li}_4\left(\frac12\right)}$$
Solution to D
$$\boxed{E=\int_0^1 \frac{\log(u)\log^2(1+u)}{1+u} \, du=2 \, \text{Li}_4\left(\frac{1}{2}\right) + \frac{7}{4} \, \zeta(3) \ln(2) - \frac{\pi^4}{45} + \frac{\ln^4(2)}{12} - \frac{1}{12} \pi^2 \ln^2(2)}
$$
E was self-computed
$$\boxed{F=\int_0^1 \frac{\log(u)\log(1-u)\log(1+u)}{1+u} \, du= 2\operatorname{Li}_4\left(\dfrac{1}{2}\right)+\dfrac{21}{8}\zeta(3)\ln 2 - \dfrac{1}{45}\pi^4 + \dfrac{1}{12}\ln^4 2 - \dfrac{5}{24}\pi^2\ln^2 2}$$
F was self-computed
$$\boxed{G=\int_0^1 \frac{\log^2(1-u)\log(1+u)}{1+u} \, du=-\frac{1}{4}\zeta \left(4\right)+2\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{4}\ln ^4\left(2\right)}$$
Solution to G
$$\boxed{H=\int_0^1 \frac{\log^3(1+u)}{1+u} \, du=\frac14 \ln^4(2)}$$
H was self-computed
$$\boxed{K=\int_0^1 \frac{\log(1-u)\log^2(1+u)}{1+u} \, du=2\operatorname{Li}_4\left(\frac{1}{2}\right)+2\zeta(3)\log(2)+\frac13\log^4(2)-\frac{\pi^4}{45}-\frac{\pi^2}{6}\log^2(2)}$$
K was self-computed
$$\boxed{L=\int_0^1 \frac{\log^2(1-u)\log^2(1+u)}{1+u} \, du=-4\operatorname{Li}_5\left(\frac12\right)+4\zeta(3)\log^2(2)-\frac{2\pi^2}9 \log^3(2)-\frac{\pi^2}{3}\zeta(3)-\frac{\pi^4}{20}\log2+\frac7{30}\log^5(2)+ \frac{63}8\zeta(5) }$$
Solution to above integral
$$\boxed{M=\int_0^1 \frac{\log^4(1+u)}{1+u} \, du=\frac15 \ln^5(2)}$$
M was self-computed
$$\boxed{N=\int_0^1 \frac{\log(1-u)\log^3(1+u)}{1+u} \, du=6\zeta(5)-6\operatorname{Li}_5\left(\frac12\right)-\frac{\pi^4}{15}\ln(2)+3\ln^2(2)\zeta(3)-\frac{\pi^2}{6}\ln^3(2)+\frac14 \ln^5(2)}$$
Solution to N
$$P=\int_0^1 \frac{\log(u)\log^2(1+u)\log^2(1-u)}{1+u} \, du$$
$$\boxed{Q=\int_0^1 \frac{\log(u)\log^4(1+u)}{1+u} \, du=24\operatorname{Li}_6\left(\frac12\right)-\frac{8}{315}\pi^6-\frac{\pi^2}{4}\ln^4(2)+24\ln(2)\operatorname{Li}_5\left(\frac12\right)+12\ln^2(2)\operatorname{Li}_4\left(\frac12\right)+\frac{7}{2}\zeta(3)\ln^3(2)+\frac13\ln^6(2)}$$
Q was self-computed
$$R=\int_0^1 \frac{\log(u)\log(1-u)\log^3(1+u)}{1+u} \, du$$
Here's a partial proof :
$$ab^3=\frac{1}{8}(a+b)^4-\frac{1}{8}(a-b)^4-a^3b$$
$$R=\int _0^1\frac{\log (u)\log (1-u)\log ^3(1+u)}{1+u}du$$
$$a=\log(1-u), b=\log(1+u)$$
$$\implies\log(1-u)\log^3(1+u)=\frac{1}{8}\log^4(1-u^2)-\frac{1}{8}\log^4\left(\frac{1-u}{1+u}\right)-\log^3(1-u)\log(1+u)$$
$$R=\frac{1}{8}\int_0^1\frac{\log(u)\log^4(1-u^2)}{1+u}\,du-\frac{1}{8}\int_0^1\frac{\log(u)\log^4\left(\frac{1-u}{1+u}\right)}{1+u}\,du-\int_0^1\frac{\log(u)\log^3(1-u)\log(1+u)}{1+u}\,du$$
$$\int_0^1\frac{\log(u)\log^4(1-u^2)}{1+u}\,du=\int_0^1\frac{\log(u)\log^4(1-u^2)}{1-u^2}(1-u)\,du$$
$$u^2=t\implies \,du=\frac{1}{2\sqrt{t}}\,dt$$
$$\int_0^1\frac{\log(\sqrt{t})\log^4(1-t)}{1-t}\left(1-\sqrt{t}\right)\frac{1}{2\sqrt{t}}\,dt\implies\frac14\int_0^1\frac{\log(u)\log^4(1-u)}{\sqrt{t}(1-t)}-\frac14\int_0^1\frac{\log(u)\log^4(1-u)}{(1-t)}$$
The above two integrals can be evaluated using the Beta function (routine method)
$$\int_0^1\frac{\log(u)\log^4\left(\frac{1-u}{1+u}\right)}{1+u}\,du$$
$$\underset{\frac{1-u}{1+u}=t}{=}\int_0^1 \frac{\log(1 - t) \log^4(t)}{1 + t} \, dt-\int_0^1 \frac{\log(1 + t) \log^4(t)}{1 + t} \, dt$$
Need to work on the above two integrals....
$$\int_0^1 \frac{\log(u)\log(1+u)\log^3(1-u)}{1+u} \, du$$
We can follow a similar path as followed in calculating $\alpha$ from section 2.
$$\boxed{S=\int_0^1 \frac{\log^3(1+u)\log^2(1-u)}{1+u} \, du=-12 \text{Li}_6\left(\frac{1}{2}\right)-12 \text{Li}_5\left(\frac{1}{2}\right) \ln (2)+6 \zeta^2 (3)+8 \zeta (3) \ln ^3(2)-12 \zeta(2) \zeta (3) \ln (2)+36 \zeta (5) \ln (2)-9\zeta(6)+\frac{1}{4}\ln ^6(2)-2\zeta(2) \ln ^4(2)-\frac{27}{2} \zeta(4) \ln ^2(2)+12 \sum_{n=1}^\infty\frac{H_n}{n^52^n}}$$
S was self-computed, Also no closed form for that sum
$$\boxed{T=\int_0^1 \frac{\log^5(1+u)}{1+u} \, du=\frac16 \ln^6(2)}$$
T was self-computed
$$\boxed{W=\int_0^1 \frac{\log(1-u)\log^4(1+u)}{1+u} \, du=24\operatorname{Li}_6\left(\frac12\right)-\frac{8}{315}\pi^6+24\ln(2)\zeta(5)-\frac{2}{15}\pi^4\ln^2(2)+4\ln^3(2)\zeta(3)-\frac{\pi^2}{6}\ln^4(2)+\frac15\ln^5(2)}$$
Solution to W
$$\boxed{X=\int_0^1 \frac{\log(1+u)\log^2(1-u)}{1+u} \, du= 2 \zeta (3) \log (2)-\frac{\pi ^4}{360}+\frac{\log ^4(2)}{4}-\frac{1}{6} \pi ^2 \log ^2(2)}$$
Solution to X
$$\boxed{Y=\int_0^1 \frac{\log^2(1+u)}{1+u} \, du=\frac13 \ln^3(2)}$$
Y was self-computed
$$\boxed{Z=\int_0^1 \frac{\log^2(1+u)\log(1-u)}{1+u} \, du=2\operatorname{Li}_4\left(\frac{1}{2}\right)+2\zeta(3)\log(2)+\frac13\log^4(2)-\frac{\pi^4}{45}-\frac{\pi^2}{6}\log^2(2)}$$
Z was self-computed
$$\boxed{\alpha=\int_0^1 \frac{\log(u)\log(1+u)\log^2(1-u)}{1+u} \, du = \frac{21}{8}\log^2(2)\zeta(3)-\frac{15}{8}\log(2)\zeta(4)+\frac{121}{16}\zeta(5)-\frac23\log^3(2)\zeta(2)-\frac{1}{15}\log^5(2)-2\log(2)\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right)-3\zeta(2)\zeta(3)}$$
Proof : Here, we can re-write our integral as follows,
$$\int_0^1 \frac{\log(u)\log(1+u)\log^2(1-u)}{1+u} \, du=\frac{1}{6}\int_0^1\frac{(1-u)\log(u)\log^3(1-u^2)}{1-u^2}\,du-\frac{1}{6}\int_0^1\frac{(1-u)\log^3(u)\log(1-u^2)}{1-u^2}\,du+\frac13\int_0^1\frac{\log^3(u)\log(1+u)}{1+u}\,du-\frac13\int_0^1\frac{\log(u)\log^3(1+u)}{1+u}\,du$$
Apply IBP to the last two integrals and for first two make the substitution of $u^2\to u$, we can obtain known integrals in text.
$$\boxed{\beta=\int_0^1 \frac{\log(u)\log^3(1+u)}{1+u} \, du=6 \, \text{Li}_5\left(\frac{1}{2}\right) + 6\text{Li}_4\left(\frac{1}{2}\right) \ln(2) - 6 \, \zeta(5) + \frac{21}{8} \, \zeta(3) \ln^2(2) + \frac{\ln^5(2)}{5} - \frac{1}{6} \pi^2 \ln^3(2)}$$
$\beta$ was self-computed
$$\gamma=\int_0^1 \frac{\log(u)\log(1-u)\log^2(1+u)}{1+u} \, du$$
You can try the following for $a=1,b=1,c=2$,
$$\sin \pi (b-a) \int_0^1 x^{b-1}(1-x)^{c-b-1}(1+x)^{-a} dx = \sin \pi (c-a) \int_0^1 x^{a-c} \left[ (1-x)^{c-a-1}(1+x)^{-b} - 1 \right] dx - \sin \pi (b-a) \frac{\Gamma(c-b)\Gamma(b)}{\Gamma(a)\Gamma(c-a)} \int_0^1 x^{b-c} \left[ (1-x)^{c-a-1}(1+x)^{-b} - 1 \right] dx + \frac{\sin \pi (c-a)}{a-c+1} - \frac{\pi \Gamma(b)}{\Gamma(2-c+b)\Gamma(a)\Gamma(c-a)}$$
You will have to perform the below,
$$\frac{\partial}{\partial a^m \partial b^n \partial c^r}$$
Picking appropriate $m,n,r$
$$\boxed{\delta=\int_0^1 \frac{\log^2(1-u)\log^2(1+u)}{1+u} \, du=-4\operatorname{Li}_5\left(\frac12\right)+4\zeta(3)\log^2(2)-\frac{2\pi^2}9 \log^3(2)-\frac{\pi^2}{3}\zeta(3)-\frac{\pi^4}{20}\log2+\frac7{30}\log^5(2)+ \frac{63}8\zeta(5) }$$
Computed same as $L$
$$\boxed{\Omega=\int_0^1 \frac{\log^4(1+u)}{1+u} \, du=\frac15\ln^5(2)}$$
$\Omega$ was self-computed
$$\boxed{\tau=\int_0^1 \frac{\log(1-u)\log^3(1+u)}{1+u} \, du=6\zeta(5)-6\operatorname{Li}_5\left(\frac12\right)-\frac{\pi^4}{15}\ln(2)+3\ln^2(2)\zeta(3)-\frac{\pi^2}{6}\ln^3(2)+\frac14 \ln^5(2)}$$
Solution to above integral
All the above integrals are known and trivial, derived from the beta
function, Not to mention all the "self-computed" and other related/similar integrals are well-documented in text
and on/off this site as well
$$\int_0^1 t^{a-1}(1-t)^{b-1}\,dt=\beta(a,b)$$
$$\int_{-1}^1(1-t)^{a-1}(1+t)^{b-1}\,dt=2^{a+b-1}\beta(a,b)$$
$$\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} \textrm{d}x = \beta(a,b)$$
Recap : We are still missing the closed form/proofs for the following integrals :
From section-1,
$R=\int_0^1 \frac{\log(u)\log(1-u)\log^3(1+u)}{u} \, du$
$S=\int_0^1 \frac{\log^3(1+u)\log^2(1-u)}{u} \, du$
$W=\int_0^1 \frac{\log(1-u)\log^4(1+u)}{u} \, du$
$Q=\int_0^1 \frac{\log(u)\log^4(1+u)}{u} \, du$, which also gave us two more $\int_{\frac12}^1\frac{\ln(1-x)\ln^4(x)}{x}\,dx$ and $\int_{\frac12}^1\frac{\ln(1-x)\ln^4(x)}{1-x}\,dx$, the other refernce does involve a generalized solution.
$\alpha=\int_0^1 \frac{\log(u)\log(1+u)\log^2(1-u)}{u} \, du$
From Section -2,
$P=\int_0^1 \frac{\log(u)\log^2(1+u)\log^2(1-u)}{1+u} \, du$
$R=\int_0^1 \frac{\log(u)\log(1-u)\log^3(1+u)}{1+u} \, du$, which also gave us three more $\int_0^1 \frac{\log(1 - t) \log^4(t)}{1 + t} \, dt$ and $\int_0^1 \frac{\log(1 + t) \log^4(t)}{1 + t} \, dt$ and $\int_0^1 \frac{\log(u)\log(1+u)\log^3(1-u)}{1+u} \, du$
$\gamma=\int_0^1 \frac{\log(u)\log(1-u)\log^2(1+u)}{1+u} \, du$
Another approach may go this way,
$$I=\int_0^1 \frac{\ln(1-x) \ln^2(1+x)\ln^2 x}{1-x} \, dx$$
Using the Identity,
$$ab^2=\frac{1}{6}(a+b)^3+\frac{1}{6}(a-b)^3-\frac{1}{3}a^3$$
Where,
$$a=\ln(1-x)\,||||\, b=\ln(1+x)$$
$$\ln(1-x)\ln^2(1+x)=\frac{1}{6}(\ln(1-x)+\ln(1+x))^3+\frac{1}{6}(\ln(1-x)-\ln(1+x))^3-\frac{1}{3}(\ln(1-x))^3$$
$$\ln(1-x)\ln^2(1+x)=\frac{1}{6}\ln^3(1-x^2)+\frac{1}{6}\ln^3\left(\frac{1-x}{1+x}\right)-\frac{1}{3}\ln^3(1-x)$$
$$\ln(1-x)\ln^2(1+x)\ln^2 x=\frac{1}{6}\ln^3(1-x^2)\ln^2 x+\frac{1}{6}\ln^3\left(\frac{1-x}{1+x}\right)\ln^2 x-\frac{1}{3}\ln^3(1-x)\ln^2 x$$
$$\frac{\ln(1-x)\ln^2(1+x)\ln^2 x}{1-x}=\frac{1}{6}\frac{\ln^3(1-x^2)\ln^2 x}{1-x}+\frac{1}{6}\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln^2 x}{1-x}-\frac{1}{3}\frac{\ln^3(1-x)\ln^2 x}{1-x}$$
$$I=\frac{1}{6}\int_0^1\frac{\ln^3(1-x^2)\ln^2 (x)}{1-x}\,dx+\frac{1}{6}\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln^2(x)}{1-x}\,dx-\frac{1}{3}\int_0^1\frac{\ln^3(1-x)\ln^2 (x)}{1-x}\,dx$$
You can then refer here and here and here for further.
MultipleZetaValuesgives: $$-\frac{4}{3} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{131 \zeta (3)^2}{16}+\frac{7}{3} \zeta (3) \ln ^3(2)-\frac{77}{24} \pi ^2 \zeta (3) \ln (2)+\frac{217}{8} \zeta (5) \ln (2)+\frac{41 \pi ^6}{30240}-\frac{1}{18} \pi ^2 \log ^4(2)-\frac{1}{144} \pi ^4 \log ^2(2)$$ – Mariusz Iwaniuk Oct 06 '24 at 09:58