Evaluate $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx$ 
How to prove $$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2 \ ?$$
  where $\operatorname{Li}_3(x)=\sum\limits_{n=1}^\infty\frac{x^n}{n^3}$ is the trilogarithm and $G=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}$ is Catalan's constant

Trying the algebraic identity $\ 4ab=(a+b)^2-(a-b)^2\ $ where $\ a=\ln(1-x)$ and $b=\ln(1+x)\ $is not helpful here and the integral will be more complicated.
Also, applying IBP or substituting $x=\frac{1-y}{1+y}$ is not that useful either.
All approaches are appreciated. 

lets start with $\displaystyle\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\operatorname{Li}_3(1+i)\quad$ (proved here)
\begin{align}
\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx&=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ d+\underbrace{\int_1^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx}_{\small\displaystyle x\mapsto1/x}\\
2\Im\operatorname{Li}_3(1+i)&=2\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx-2\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}\ dx+\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}\ dx}_{2\beta(3)}
\end{align}
then 

$$\int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}\ dx=\Im\operatorname{Li}_3(1+i)-\beta(3)\tag{1}$$

now lets start with $\ I=\displaystyle\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx$ and by setting $x=\frac{1-y}{1+y}$, we get
$$I=\displaystyle\int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}-\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx+\ln2\underbrace{\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}\ dx}_{x=(1-y)/(1+y)}+I$$
then

\begin{align}
\int_0^1\frac{\ln^2(1+x)-\ln x\ln(1+x)}{1+x^2}=\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx-\ln2\underbrace{\int_0^1\frac{\ln x}{1+x^2}}_{-G}\tag{2}
\end{align}

from $(1)$ and $(2)$ and substituting $\displaystyle\beta(3)=\frac{\pi^3}{32}\ $, the result follows.
