How to integrate $\int_0^\frac{1}{2}\frac{\ln(1+x)}{x}\ln\left(\frac{1}{x}-1\right)\mathrm{d}x$

Question

Question; how to integrate $$\int_0^\frac{1}{2}\frac{\ln(1+x)}{x}\ln\left(\frac{1}{x}-1\right)\mathrm{d}x$$

here is my attempt to solve the integral

\begin{align} I&=\int_0^\frac{1}{2}\frac{\ln(1+x)}{x}\ln\left(\frac{1}{x}-1\right)\mathrm{d}x\\ &=\frac{1}{2}\int_0^\frac{1}{2}\frac{2\ln(1-x)\ln(1+x)}{x}\mathrm{~d}x\\ &\quad-\int_0^\frac{1}{2}\frac{\ln x\ln(1+x)}{x}\mathrm{~d}x\\ &=\frac{1}{2}\int_0^\frac{1}{2}\frac{\ln^2\left(1-x^2\right)-\ln^2(1-x)-\ln^2(1+x)}{x}\mathrm{~d}x\\ &\quad-\int_0^\frac{1}{2}\frac{\ln x\ln(1+x)}{x}\mathrm{~d}x\\ \end{align}

At first glance, I didn't know how to approach this integral. I spent a lot of time on it, and this is as far as I got. (I was planning to post this integral yesterday, but then I remembered I don't have enough reputation to get the answer without making an attempt.)

Edit: according to mathematica the closed form the integral is $$\frac{1}{24} \left(4 i \pi ^3+4 \log (2)^3+4 \log (3)^3-6 \log (2)^2 \log (4)+12 \log (2) \log (3) \log (27)-6 \log (3)^2 \log (36)-12 \log (4) \text{Li}_2\left(-\frac{1}{2}\right)-12 \log (9) \text{Li}_2\left(-\frac{1}{3}\right)+12 \log (9) \text{Li}_2\left(\frac{1}{3}\right)+12 \log (4) \text{Li}_2\left(\frac{2}{3}\right)-12 \log (9) \text{Li}_2\left(\frac{2}{3}\right)-12 i \pi \left(\log (2)^2+3 \log (3)^2-\log (2) \log (4)-\log (3) \log (9)-2 \text{Li}_2\left(-\frac{1}{2}\right)+2 \text{Li}_2\left(\frac{2}{3}\right)\right)-24 \text{Li}_3\left(-\frac{1}{2}\right)-24 \text{Li}_3\left(-\frac{1}{3}\right)+24 \text{Li}_3\left(\frac{1}{3}\right)-24 \text{Li}_3\left(\frac{2}{3}\right)-15 \zeta (3)\right) $$

After using FULLSIMPLIFY I GOT

$$\frac{1}{12} \left(i \pi ^3+\pi ^2 \log (81)+6 i \pi \left(2 \log (2)^2-\log (3)^2+\text{Li}_2\left(\frac{1}{4}\right)-2 \text{Li}_2\left(\frac{2}{3}\right)\right)-2 \left(4 \log (2)^3+2 \log (3)^2 \left(6 \text{ArcCoth}[17]+\log (3)\right)+\log (27) \text{Li}_2\left(\frac{1}{9}\right)+\log (8) \text{Li}_2\left(\frac{1}{4}\right)+6 \log \left(\frac{27}{2}\right) \text{Li}_2\left(\frac{2}{3}\right)\right)-3 \left(\text{Li}_3\left(\frac{1}{9}\right)+\text{Li}_3\left(\frac{1}{4}\right)-8 \text{Li}_3\left(\frac{1}{3}\right)+4 \text{Li}_3\left(\frac{2}{3}\right)-\zeta (3)\right)\right)$$