\begin{align}
I &= \int_0^1 \frac{2(x-1) \ln^3(x) \left[2 \ln(x+1) - \ln \left(2(x^2+1)\right)\right]}{(x+1)^3} \, dx \\
&= 4 \underbrace{\int_0^1 \frac{x \ln^3(x) \ln(1+x)}{(1+x)^3} \, dx}_{I_1} - 2 \underbrace{\int_0^1 \frac{x \ln^3(x) \ln \left(2(1+x^2)\right)}{(1+x)^3} \, dx}_{I_2} \\
&\quad - 4 \underbrace{\int_0^1 \frac{\ln^3(x) \ln(1+x)}{(1+x)^3} \, dx}_{I_3} + 2 \underbrace{\int_0^1 \frac{\ln^3(x) \ln \left(2(1+x^2)\right)}{(1+x)^3} \, dx}_{I_4}
\end{align}
In this first part, the integral $I$ is split into four integrals labeled $I_1, I_2, I_3,$ and $I_4$, based on the terms in the integrand.
\begin{align}
I_1 &= \text{Integration by parts} - \left.\frac{1}{2} \ln^3(x) \ln(1+x) \frac{2x+1}{(1+x)^2}\right|_0^1 \\
&\quad + \frac{1}{2} \int_0^1 \frac{2x+1}{(1+x)^2} \left(\frac{\ln^3(x)}{1+x} + \frac{3 \ln^2(x) \ln(1+x)}{x}\right) \, dx \\
&= \underbrace{\int_0^1 \frac{x \ln^3(x)}{(1+x)^3} \, dx}_{I_5} + 3 \underbrace{\int_0^1 \frac{\ln^2(x) \ln(1+x)}{(1+x)^2} \, dx}_{I_6} \\
&\quad + \frac{1}{2} \underbrace{\int_0^1 \frac{\ln^3(x)}{(1+x)^3} \, dx}_{I_7} + \frac{3}{2} \underbrace{\int_0^1 \frac{\ln^2(x) \ln(1+x)}{x(1+x)^2} \, dx}_{I_8}
\end{align}
In this second part, the integral $I_1$ is computed using integration by parts, yielding expressions involving new integrals $I_5, I_6, I_7,$ and $I_8$.
\begin{align}
I_5 &= \frac{1}{2} \sum_{n=2}^{\infty} (-1)^n n(n-1) \int_0^1 x^{n-1} \ln^3(x) \, dx \\
&= -3 \sum_{n=2}^{\infty} \frac{(-1)^n (n-1)}{n^3} \\
&= -3 \left(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} - \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}\right) \\
&= 3 (\eta(2) - \eta(3)) \\
&= \frac{3}{2} \zeta(2) - \frac{9}{4}
\end{align}
Here, the integral $I_5$ is evaluated using a series representation, resulting in terms involving the Dirichlet eta function $\eta(s)$ and the Riemann zeta function $\zeta(s)$.
\begin{align}
I &= \text{Integration by parts} - \left.\frac{\ln^2(x) \ln(1+x)}{1+x}\right|_0^1 \\
&\quad + \int_0^1 \frac{\ln(x)}{1+x} \left(\frac{\ln(x)}{1+x} + \frac{2 \ln(1+x)}{x}\right) \, dx \\
&= \underbrace{\int_0^1 \frac{\ln^2(x)}{(1+x)^2} \, dx}_{I_9} + 2 \underbrace{\int_0^1 \frac{\ln(x) \ln(1+x)}{x(1+x)} \, dx}_{I_{10}} \\
I_9 &= -\sum_{n=1}^{\infty} (-1)^n n \int_0^1 x^{n-1} \ln^2(x) \, dx \\
&= 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \\
&= 2 \eta(2) \\
&= \zeta(2) \\
I_{10} &= -\sum_{n=1}^{\infty} (-1)^n H_n \int_0^1 x^{n-1} \ln(x) \, dx \\
&= \sum_{n=1}^{\infty} \frac{(-1)^n H_n}{n^2} \\
&= -\frac{5}{8} \zeta(3)
\end{align}
In this final part, the integrals $I_9$ and $I_{10}£ are computed using series representations involving harmonic numbers and the Riemann zeta function.
\begin{align}
I_8 &= \text{Integration by parts} \, \ln^2(x) \ln(1+x) \left[\frac{1}{1+x} + \ln \left(\frac{x}{1+x}\right)\right] \bigg|_0^1 \\
&\quad - \int_0^1 \left(\frac{1}{1+x} + \ln(x) - \ln(1+x)\right) \left(\frac{\ln^2(x)}{1+x} + \frac{2 \ln(x) \ln(1+x)}{x}\right) \, dx \\
&= - \underbrace{\int_0^1 \frac{\ln^2(x)}{(1+x)^2} \, dx}_{I_9} - 2 \underbrace{\int_0^1 \frac{\ln(x) \ln(1+x)}{x(1+x)} \, dx}_{I_{10}} - \underbrace{\int_0^1 \frac{\ln^3(x)}{1+x} \, dx}_{I_{11}}
\end{align}
In this first part, $I_8$ is expressed as an integration by parts term minus a complex integral. The integral is split into terms labeled $I_9, I_{10},$ and $I_{11}$.
\begin{align}
& - 2 \underbrace{\int_0^1 \frac{\ln^2(x) \ln(1+x)}{x} \, dx}_{I_{12}} + \underbrace{\int_0^1 \frac{\ln^2(x) \ln(1+x)}{1+x} \, dx}_{I_{13}} + 2 \underbrace{\int_0^1 \frac{\ln(x) \ln^2(1+x)}{x} \, dx}_{I_{14}} \\
I_{11} &= \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln^3(x) \, dx \\
&= -6 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4} \\
&= -6 \eta(4) \\
&= -\frac{21}{4} \zeta(4) \quad \text{(using the relation between eta and zeta functions)}
\end{align}
In this second part, integrals $I_{12}, I_{13},$ and $I_{14}$ are introduced. The integral $I_{11}$ is evaluated using series representations involving the eta function and zeta function.
\begin{align}
I_{12} &= \text{Integration by parts} \, \ln^3(x) \ln(1+x) \bigg|_0^1 \\
&\quad - \int_0^1 \ln^2(x) \left(\frac{\ln(x)}{1+x} + \frac{2 \ln(1+x)}{x}\right) \, dx \\
&= -I_{11} - 2 I_{12} \\
&\Rightarrow I_{12} = -\frac{1}{3} I_{11} \\
&= \frac{7}{4} \zeta \quad \text{(substituting the value of } I_{11})
\end{align}
This third part evaluates $I_{12}$ using integration by parts and solves for $I_{12}$ in terms of $I_{11}$.
\begin{align}
I_{13} &= \int_0^1 \frac{\ln^2(x) \ln(1+x)}{1+x} \, dx \\
&= -\sum_{n=1}^{\infty} (-1)^n H_n \int_0^1 x^n \ln^2(x) \, dx \\
&= 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \left(H_n - \frac{1}{n+1}\right)}{(n+1)^3} \\
&= 2 \left(\sum_{n=1}^{\infty} \frac{(-1)^n H_n}{n^3} + \eta(4)\right) \\
&= 4 \operatorname{Li}_4\left(\frac{1}{2}\right) - \ln^2(2) \zeta(2) \\
&\quad + \frac{7}{2} \ln(2) \zeta(3) + \frac{1}{6} \ln^4(2) - \frac{15}{4} \zeta(4) \quad \text{(evaluating using known series results)}
\end{align}
In the final part, $I_{13}$ is computed using series representations and special functions like polylogarithms and logarithms.
\begin{align}
& \text{Combining all results:} \\
& \Rightarrow I_8 = -4 \operatorname{Li}_4\left(\frac{1}{2}\right) + \ln^2(2) \zeta(2) - \frac{7}{2} \ln(2) \zeta(3) \\
& \quad - \frac{1}{6} \ln^4(2) + \frac{11}{2} \zeta(4) + \frac{5}{4} \zeta(3) - \zeta(2) \\
& \Rightarrow I_1 = -6 \operatorname{Li}_4\left(\frac{1}{2}\right) + \frac{3}{2} \ln^2(2) \zeta(2) \\
& \quad - \frac{21}{4} \ln(2) \zeta(3) - \frac{1}{4} \ln^4(2) \\
& \quad + \frac{33}{4} \zeta(4) - \frac{21}{4} \zeta(3) + \frac{9}{4} \zeta(2)
\end{align}
In this concluding part, the final values of $I_8$ and $I_1$ are obtained by combining the results from the previous steps.
\begin{align}
I_2 &= \ln(2) \underbrace{\int_0^1 \frac{x \ln^3(x)}{(1+x)^3} \, dx}_{I_5} + \underbrace{\int_0^1 \frac{x \ln^3(x) \ln \left(1+x^2\right)}{(1+x)^3} \, dx}_{I_{15}} \\
I_{15} &= \text{Integration by parts} \\
&\quad - \left. \frac{1}{2} \ln^3(x) \ln \left(1+x^2\right) \frac{2x+1}{(1+x)^2} \right|_0^1 \\
&\quad + \frac{1}{2} \int_0^1 \frac{2x+1}{(1+x)^2} \left(\frac{2x \ln^3(x)}{1+x^2} + \frac{3 \ln^2(x) \ln \left(1+x^2\right)}{x(1+x)^2}\right) \, dx \\
&= 2 \underbrace{\int_0^1 \frac{x^2 \ln^3(x)}{(1+x)^2 (1+x^2)} \, dx}_{I_{16}} \\
&\quad + 3 \underbrace{\int_0^1 \frac{\ln^2(x) \ln \left(1+x^2\right)}{(1+x)^2} \, dx}_{I_{17}} \\
&\quad + \underbrace{\int_0^1 \frac{x \ln^3(x)}{(1+x)^2 (1+x^2)} \, dx}_{I_{18}} \\
&\quad + \frac{3}{2} \underbrace{\int_0^1 \frac{\ln^2(x) \ln \left(1+x^2\right)}{x(1+x)^2} \, dx}_{I_{19}}
\end{align}
In this first part, $I_2$ is broken into two integrals: $I_5$ and $I_{15}$. The integral $I_{15}$ is evaluated using integration by parts and separated into simpler integrals $I_{16}$ through $I_{19}$.
\begin{align}
I_{16} &= \int_0^1 \frac{x^2 (1-x)^2 (1+x^2) \ln^3(x)}{(1-x^4)^2} \, dx \\
&= \int_0^1 \frac{\left(x^2 - 2x^3 + 2x^4 - 2x^5 + x^6\right) \ln^3(x)}{(1-x^4)^2} \, dx \\
&= \underbrace{\int_0^1 \frac{x^2 \ln^3(x)}{(1-x^4)^2} \, dx}_{I_{20}} \\
&\quad - 2 \underbrace{\int_0^1 \frac{x^3 \ln^3(x)}{(1-x^4)^2} \, dx}_{I_{21}} \\
&\quad + 2 \underbrace{\int_0^1 \frac{x^4 \ln^3(x)}{(1-x^4)^2} \, dx}_{I_{22}} \\
&\quad - 2 \underbrace{\int_0^1 \frac{x^5 \ln^3(x)}{(1-x^4)^2} \, dx}_{I_{23}} \\
&\quad + \underbrace{\int_0^1 \frac{x^6 \ln^3(x)}{(1-x^4)^2} \, dx}_{I_{24}}
\end{align}
Here, $I_{16}$ is decomposed into simpler integrals $I_{20}$ through $I_{24}$.
\begin{align}
I_{20} &= \sum_{n=1}^{\infty} n \int_0^1 x^{4n-2} \ln^3(x) \, dx \\
&= -\frac{3}{2} \sum_{n=1}^{\infty} \frac{(4n-1) + 1}{(4n-1)^4} \\
&= -\frac{3}{2} \left(\frac{1}{4^3} \sum_{n=1}^{\infty} \frac{1}{\left(n - \frac{1}{4}\right)^3} + \frac{1}{4^4} \sum_{n=1}^{\infty} \frac{1}{\left(n - \frac{1}{4}\right)^4}\right) \\
&= \frac{1}{1024} \psi_3 \left(\frac{1}{4}\right) - \frac{45}{32} \zeta(4) - \frac{21}{32} \zeta(3) + \frac{3 \pi^3}{128}
\end{align}
In this part, $I_{20}$ is computed using series and special functions.
\begin{align}
I_{21} &= \sum_{n=1}^{\infty} n \int_0^1 x^{4n-1} \ln^3(x) \, dx \\
&= -\frac{3}{128} \sum_{n=1}^{\infty} \frac{1}{n^3} \\
&= -\frac{3}{128} \zeta
\end{align}
\begin{align}
I_{22} &= \sum_{n=1}^{\infty} n \int_0^1 x^{4n} \ln^3(x) \, dx \\
&= -\frac{3}{2} \sum_{n=1}^{\infty} \frac{(4n+1) - 1}{(4n+1)^4} \\
&= -\frac{3}{2} \left(\frac{1}{4^3} \sum_{n=1}^{\infty} \frac{1}{\left(n + \frac{1}{4}\right)^3} - \frac{1}{4^4} \sum_{n=1}^{\infty} \frac{1}{\left(n + \frac{1}{4}\right)^4}\right) \\
&= \frac{1}{1024} \psi_3 \left(\frac{1}{4}\right) - \frac{21}{32} \zeta(3) - \frac{3 \pi^3}{128}
\end{align}
In this final part, $I_{22}$ is also evaluated using series and special functions, leading to the same results as before.
\begin{align*}
I_{23} &= \sum_{n=1}^{\infty} n \int_0^1 x^{4n+1} \ln^3(x) \, dx = -\frac{3}{16} \sum_{n=1}^{\infty} \frac{(2n+1) - 1}{(2n+1)^4} \\
&= -\frac{3}{16} \left( \frac{1}{2^3} \sum_{n=1}^{\infty} \frac{1}{\left(n + \frac{1}{2}\right)^3} - \frac{1}{2^4} \sum_{n=1}^{\infty} \frac{1}{\left(n + \frac{1}{2}\right)^4} \right) \\
&= -\frac{3}{16} \left( -\frac{1}{2^3} \frac{\psi_2\left(\frac{3}{2}\right)}{2!} - \frac{1}{2^4} \frac{\psi_3\left(\frac{3}{2}\right)}{3!} \right) \\
&= \frac{3}{256} \left[\psi_2\left(\frac{1}{2}\right) + 16 \right] + \frac{1}{512} \left[\psi_3\left(\frac{1}{2}\right) - 96 \right] \\
&= \frac{3}{256} (-14 \zeta(3)) + \frac{1}{512} (90 \zeta(4)) \\
&= \frac{45}{256} \zeta(4) - \frac{21}{128} \zeta(3) \ldots(\vartheta)
\end{align*}
\begin{align*}
I_{24} &= \sum_{n=1}^{\infty} n \int_0^1 x^{4n+2} \ln^3(x) \, dx = -\frac{3}{2} \sum_{n=1}^{\infty} \frac{(4n+3) - 3}{(4n+3)^4} \\
&= -\frac{3}{2} \left( \frac{1}{4^3} \sum_{n=1}^{\infty} \frac{1}{\left(n + \frac{3}{4}\right)^3} - \frac{3}{4^4} \sum_{n=1}^{\infty} \frac{1}{\left(n + \frac{3}{4}\right)^4} \right) \\
&= -\frac{3}{2} \left( -\frac{1}{4^3} \frac{\psi_2\left(\frac{7}{4}\right)}{2!} - \frac{3}{4^4} \frac{\psi_3\left(\frac{7}{4}\right)}{3!} \right) \\
&= \frac{3}{256} \left[\psi_2\left(\frac{3}{4}\right) + \frac{128}{27} \right] + \frac{3}{1024} \left[\psi_3\left(\frac{3}{4}\right) - \frac{512}{27} \right] \\
&= \frac{3}{256} \left(2 \pi^3 - 56 \zeta(3)\right) + \frac{3}{1024} \left[1440 \zeta(4) - \psi_3\left(\frac{1}{4}\right)\right] \\
&= -\frac{3}{1024} \psi_3\left(\frac{1}{4}\right) + \frac{135}{32} \zeta(4) - \frac{21}{32} \zeta(3) + \frac{3 \pi^3}{128} \\
&*(\chi) \wedge (\omega) \wedge (\wp) \wedge (\vartheta) \wedge (\varkappa) \rightarrow (\phi) \Rightarrow I_{16} = \frac{315}{128} \zeta(4) - \frac{9}{4} \zeta(3) \ldots(\Delta)
\end{align*}
\begin{align*}
I_{17} &\stackrel{I \underline{\underline{B}} P}{ }-\left.\frac{\ln^2(x) \ln\left(1 + x^2\right)}{1 + x}\right|_0^1 + 2 \int_0^1 \frac{\ln(x)}{x(1 + x)}\left(\frac{x^2 \ln(x)}{1 + x^2} + \ln\left(1 + x^2\right)\right) \, dx \\
&= 2 \left( \underbrace{\int_0^1 \frac{x \ln^2(x)}{\left(1 + x^2\right)(1 + x)} \, dx}_{I_{25}} + \underbrace{\int_0^1 \frac{\ln(x) \ln\left(1 + x^2\right)}{x(1 + x)} \, dx}_{I_{26}} \right) \ldots(\varpi)
\end{align*}
\begin{align*}
I_{25} &= \int_0^1 \frac{x(1 - x) \ln^2(x)}{1 - x^4} \, dx = \sum_{n=0}^{\infty} \int_0^1\left(x^{4n+1} - x^{4n+2}\right) \ln^2(x) \, dx \\
&= 2 \sum_{n=0}^{\infty}\left(\frac{1}{(4n+2)^3} - \frac{1}{(4n+3)^3}\right) \\
&= \frac{1}{32} \left( \sum_{n=0}^{\infty} \frac{1}{\left(n + \frac{1}{2}\right)^3} - \sum_{n=0}^{\infty} \frac{1}{\left(n + \frac{3}{4}\right)^3} \right) \\
&= \frac{1}{32} \left( -\frac{\psi_2\left(\frac{1}{2}\right)}{2!} + \frac{\psi_2\left(\frac{3}{4}\right)}{2!} \right) \\
&= \frac{1}{64} \left[14 \zeta(3) + \left(2 \pi^3 - 56 \zeta(3)\right) \right] \\
&= -\frac{21}{32} \zeta(3) + \frac{\pi^3}{32} \quad \ldots(\varrho)
\end{align*}
\begin{align*}
I_{26} &= \int_0^1 \frac{(1 - x) \ln(x) \ln\left(1 + x^2\right)}{x\left(1 - x^2\right)} \, dx = \sum_{n=1}^{\infty} \bar{H}_n \int_0^1 x^{2n-1}(1 - x) \ln(x) \, dx \\
&= \sum_{n=1}^{\infty} \bar{H}_n\left(-\frac{1}{(2n)^2} + \frac{1}{(2n+1)^2}\right) \\
&= \sum_{n=1}^{\infty} \frac{\bar{H}_n}{(2n+1)^2} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{\bar{H}_n}{n^2} \\
&= \frac{\pi}{2} G + \frac{3}{4} \ln(2) \zeta(2) - \frac{7}{4} \zeta(3) - \frac{1}{4}\left(\frac{3}{2} \ln(2) \zeta(2) - \frac{1}{4} \zeta(3)\right) \\
&= \frac{3}{8} \ln(2) \zeta(2) - \frac{27}{16} \zeta(3) + \frac{\pi}{2} G \quad \ldots(\varsigma)
\end{align*}
\begin{align*}
&*(\varrho) \wedge (\varsigma) \rightarrow (\varpi) \Rightarrow I_{17} = \frac{3}{4} \ln(2) \zeta(2) - \frac{75}{16} \zeta(3) + \frac{\pi^3 }{16} + \pi G \ldots(\varphi)
\end{align*}
\begin{align*}
I_{18} &= \int_0^1 \frac{x(1-x)^2\left(1+x^2\right) \ln^3(x)}{\left(1-x^4\right)^2} \, dx \\
&= \int_0^1 \frac{\left(x - 2x^2 + 2x^3 - 2x^4 + x^5\right) \ln^3(x)}{\left(1-x^4\right)^2} \, dx \\
&= I_{27} - 2I_{20} + 2I_{21} - 2I_{22} + I_{23} \quad \ldots(\Lambda)
\end{align*}
\begin{align*}
I_{27} &= \int_0^1 \frac{x \ln^3(x)}{\left(1-x^4\right)^2} \, dx \\
&= \sum_{n=1}^{\infty} n \int_0^1 x^{4n-3} \ln^3(x) \, dx \\
&= -\frac{3}{16} \sum_{n=1}^{\infty} \frac{(2n-1)+1}{(2n-1)^4} \\
&= -\frac{3}{16} \left( \frac{1}{2^3} \sum_{n=1}^{\infty} \frac{1}{\left(n-\frac{1}{2}\right)^3}
+ \frac{1}{2^4} \sum_{n=1}^{\infty} \frac{1}{\left(n-\frac{1}{2}\right)^4} \right) \\
&= -\frac{3}{16} \left( -\frac{1}{2^3} \frac{\psi_2\left(\frac{1}{2}\right)}{2!}
+ \frac{1}{2^4} \frac{\psi_3\left(\frac{1}{2}\right)}{3!} \right) \\
&= \frac{3}{256} (-14 \zeta(3)) - \frac{1}{512} (90 \zeta(4)) \\
&= -\frac{45}{256} \zeta(4) - \frac{21}{128} \zeta(3) \quad \ldots(\Xi)
\end{align*}
\begin{align*}
I_{19} &\stackrel{IBP}{=} \left.\ln^2(x) \ln\left(1+x^2\right) \left[\frac{1}{1+x} + \ln\left(\frac{x}{1+x}\right)\right]\right|_0^1 \\
&\quad - 2 \int_0^1 \left(\frac{1}{1+x} + \ln(x) - \ln(1+x)\right) \frac{\ln(x)}{x}
\left(\frac{x^2 \ln(x)}{1+x^2} + \ln\left(1+x^2\right)\right) \, dx \\
&= -2 \Bigg(\underbrace{\int_0^1 \frac{x \ln^2(x)}{\left(1+x^2\right)(1+x)} \, dx}_{I_{25}}
+ \underbrace{\int_0^1 \frac{x \ln^3(x)}{1+x^2} \, dx}_{I_{28}} \\
&\quad - \underbrace{\int_0^1 \frac{x \ln^2(x) \ln(1+x)}{1+x^2} \, dx}_{I_{29}}
+ \underbrace{\int_0^1 \frac{\ln(x) \ln\left(1+x^2\right)}{x(1+x)} \, dx}_{I_{26}} \\
&\quad + \underbrace{\int_0^1 \frac{\ln^2(x) \ln\left(1+x^2\right)}{x} \, dx}_{I_{30}}
- \underbrace{\int_0^1 \frac{\ln(x) \ln(1+x) \ln\left(1+x^2\right)}{x} \, dx}_{I_{31}} \Bigg)
\end{align*}
\begin{align*}
I_{28} &= \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^{2n+1} \ln^3(x) \, dx \\
&= -6 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)^4} \\
&= -\frac{3}{8} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4} = -\frac{3}{8} \eta(4)
\end{align*}
\begin{align*}
I_{29} &\stackrel{IBP}{=} \frac{1}{2} \left.\ln^2(x) \ln(1+x) \ln\left(1+x^2\right)\right|_0^1 \\
&\quad - \frac{1}{2} \int_0^1 \ln\left(1+x^2\right) \left(\frac{\ln^2(x)}{1+x}
+ \frac{2 \ln(x) \ln(1+x)}{x}\right) \, dx \\
&= -\frac{1}{2} \underbrace{\int_0^1 \frac{\ln^2(x) \ln\left(1+x^2\right)}{1+x} \, dx}_{I_{32}} - I_{31} \quad \ldots(\Theta)
\end{align*}
\begin{align*}
I_{32} &= \int_0^1 \frac{(1-x) \ln^2(x) \ln\left(1+x^2\right)}{1-x^2} \, dx \\
&= \sum_{n=1}^{\infty} \bar{H}_n \int_0^1 x^{2n}(1-x) \ln^2(x) \, dx \\
&= 2 \sum_{n=1}^{\infty} \bar{H}_n \left(\frac{1}{(2n+1)^3} - \frac{1}{(2n+2)^3}\right) \\
&= 2 \sum_{n=1}^{\infty} \frac{\bar{H}_n}{(2n+1)^3}
- \frac{1}{4} \sum_{n=1}^{\infty} \frac{\bar{H}_{n+1} - \frac{(-1)^n}{n+1}}{(n+1)^3} \\
&= 2 \sum_{n=1}^{\infty} \frac{\bar{H}_n}{(2n+1)^3}
- \frac{1}{4} \left(\sum_{n=1}^{\infty} \frac{\bar{H}_n}{n^3}
- \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4}\right) \\
&= 2 \left(\frac{7}{8} \ln(2) \zeta(3) - \frac{45}{32} \zeta(4) + G^2\right)
- \frac{1}{4} \left(\frac{7}{4} \ln(2) \zeta(3) - \frac{5}{16} \zeta(4) - \eta(4)\right) \\
&= \frac{21}{16} \ln(2) \zeta(3) - \frac{161}{64} \zeta(4) + 2G^2
\end{align*}
\begin{align*}
&*(\Xi) \rightarrow (\Theta) \Rightarrow I_{29} = -\frac{21}{32} \ln(2) \zeta(3)
+ \frac{161}{128} \zeta(4) - G^2 - I_{31} \quad \ldots(\Phi)
\end{align*}
\begin{align*}
& \left. I_{30} \stackrel{IBP}{=} \ln^3(x) \ln\left(1+x^2\right) \right|_0^1
- 2 \int_0^1 \frac{\ln^2(x)}{x} \left(\frac{x^2 \ln(x)}{1+x^2} + \ln\left(1+x^2\right)\right) \, dx \\
& = -2 \left(\underbrace{\int_0^1 \frac{x \ln^3(x)}{1+x^2} \, dx}_{I_{28}}
+ \underbrace{\int_0^1 \frac{\ln^2(x) \ln\left(1+x^2\right)}{x} \, dx}_{I_{30}}\right) \\
& \Rightarrow I_{30} = -\frac{2}{3} I_{28} = \frac{7}{32} \zeta(4)
\end{align*}
\begin{align*}
& *(\varrho) \wedge (\Psi) \wedge (\Phi) \wedge (\varsigma) \wedge (\Psi) \rightarrow (\Phi): \\
& \Rightarrow I_{19} = -\frac{21}{16} \ln(2) \zeta(3) - \frac{3}{4} \ln(2) \zeta(2)
+ \frac{175}{64} \zeta(4) + \frac{75}{16} \zeta(3) - \frac{\pi^3}{16} - 2G^2 - \pi G \ldots
\end{align*}
\begin{align*}
& *(\Delta) \wedge (\varphi) \wedge (\Upsilon) \wedge (\Lambda) \rightarrow (v): \\
& \Rightarrow I_{15} = -\frac{1}{256} \psi_3\left(\frac{1}{4}\right) - \frac{63}{32} \ln(2) \zeta(3)
+ \frac{9}{8} \ln(2) \zeta(2) + \frac{1515}{128} \zeta(4) - \frac{297}{32} \zeta(3) \\
& \quad + \frac{3\pi^3}{32} - 3G^2 + \frac{3\pi}{2} G
\end{align*}
\begin{align*}
& *(\gamma) \wedge (\Gamma) \rightarrow (\tau): \\
& \Rightarrow I_2 = -\frac{1}{256} \psi_3\left(\frac{1}{4}\right) - \frac{135}{32} \ln(2) \zeta(3)
+ \frac{21}{8} \ln(2) \zeta(2) + \frac{1515}{128} \zeta(4) - \frac{297}{32} \zeta(3) \\
& \quad + \frac{3\pi^3}{32} - 3G^2 + \frac{3\pi}{2} G
\end{align*}
\begin{align*}
& I_3^I \underline{BP} - \left.\frac{1}{2} \frac{\ln^3(x) \ln(1+x)}{(1+x)^2}\right|_0^1
+ \frac{1}{2} \int_0^1 \frac{\ln^2(x)}{(1+x)^2} \left(\frac{\ln(x)}{1+x} + \frac{3 \ln(1+x)}{x}\right) \, dx \\
& = \frac{1}{2} \underbrace{\int_0^1 \frac{\ln^3(x)}{(1+x)^3} \, dx}_{I_7}
+ \frac{3}{2} \underbrace{\int_0^1 \frac{\ln^2(x) \ln(1+x)}{x(1+x)^2} \, dx}_{I_8}
\end{align*}
\begin{align*}
& *(\kappa) \wedge (\rho) \rightarrow (\Gamma): \\
& \Rightarrow I_3 = -6 \text{Li}_4\left(\frac{1}{2}\right) + \frac{3}{2} \ln^2(2) \zeta(2)
- \frac{21}{4} \ln(2) \zeta(3) - \frac{1}{4} \ln^4(2) + \frac{33}{4} \zeta(4) \\
& \quad + \frac{3}{4} \zeta(3) - \frac{9}{4} \zeta(2) \ldots (\aleph)
\end{align*}
\begin{align*}
& I_4 = \ln(2) \underbrace{\int_0^1 \frac{\ln^3(x)}{(1+x)^3} \, dx}_{I_7}
+ \underbrace{\int_0^1 \frac{\ln^3(x) \ln\left(1+x^2\right)}{(1+x)^3} \, dx}_{I_{33}}
\end{align*}
\begin{align*}
& \left. I_{33} \underline{IP} \frac{\ln^3(x) \ln\left(1+x^2\right)}{2(1+x)^2} \right|_0^1
+ \frac{1}{2} \int_0^1 \frac{1}{(1+x)^2} \left(\frac{2x \ln^3(x)}{1+x^2}
+ \frac{3 \ln^2(x) \ln\left(1+x^2\right)}{x}\right) \, dx \\
& = \underbrace{\int_0^1 \frac{x \ln^3(x)}{(1+x)^2\left(1+x^2\right)} \, dx}_{I_{18}}
+ \frac{3}{2} \underbrace{\int_0^1 \frac{\ln^2(x) \ln\left(1+x^2\right)}{x(1+x)^2} \, dx}_{I_{19}}
\end{align*}
\begin{align*}
& *(\Upsilon) \wedge (\Lambda) \rightarrow (\mathrm{O}): \\
& \Rightarrow I_{33} = -\frac{1}{256} \psi_3\left(\frac{1}{4}\right) - \frac{63}{32} \ln(2) \zeta(3)
- \frac{9}{8} \ln(2) \zeta(2) + \frac{885}{128} \zeta(4) + \frac{297}{32} \zeta(3) \\
& \quad - \frac{3\pi^3}{32} - 3G^2 - \frac{3\pi}{2} G \ldots
\end{align*}
\begin{align*}
& *(\kappa) \wedge (\Omega) \rightarrow (\Sigma): \\
& \Rightarrow I_4 = -\frac{1}{256} \psi_3\left(\frac{1}{4}\right) - \frac{135}{32} \ln(2) \zeta(3)
- \frac{21}{8} \ln(2) \zeta(2) + \frac{885}{128} \zeta(4) + \frac{297}{32} \zeta(3) \\
& \quad - \frac{3\pi^3}{32} - 3G^2 - \frac{3\pi}{2} G
\end{align*}
\begin{align*}
& * \text{Finally} (\sigma) \wedge (\Pi) \wedge (\aleph) \wedge (\Omega) \rightarrow (\alpha): \\
& \therefore I = -\frac{21}{2} \ln(2) \zeta(2) - \frac{315}{32} \zeta(4)
+ \frac{105}{8} \zeta(3) + 18 \zeta(2) - \frac{3\pi^3}{8} - 6\pi G
\end{align*}
