A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$ While playing around with Mathematica, I found that
$$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$
Please help me prove this result.

We can take $\ 2ab= a^2+b^2-(a-b)^2$ to get:
$$I= \frac12 \int_0^1 \frac{\ln^2 (1-x)}{1+x}dx+\frac12\int_0^1\frac{\ln^2(1+x)}{1+x}dx-\frac12 \int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx$$
By letting $\frac{1-x}{1+x}=t$ and expanding into power series  the last one we get:
$$I=\frac12 J +\frac{\ln^3(1+x)}{6}\bigg|_0^1 -\frac12 \int_0^1 \frac{\ln^2 t}{1+t}dt=\frac12 J +\frac{\ln^3 2}{6}-\frac34\zeta(3) $$
$$J=\int_0^1 \frac{\ln^2(1-x)}{1+x}dx\overset{1-x\to x}=\frac12\int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx=\frac12 \sum_{n=0}^\infty \frac{1}{2^n} \int_0^1 x^{n}\ln^2 xdx$$
$$=\sum_{n=0}^\infty \frac{1}{2^n}\frac{1}{(n+1)^3}=2\operatorname{Li}_3 \left(\frac12\right)\Rightarrow \boxed{I=\operatorname{Li}_3 \left(\frac12\right)+\frac{\ln^3 2}{6}-\frac34\zeta(3)}$$
Of course one can rewrite the trilogarithm's value as seen from $(17)$ in this link, but this form is also valid.
