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. 

Different approach:
Start with subbing $x\mapsto \frac{1-x}{1+x}$ 
$$\small{\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx=\ln2\underbrace{\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx}_{-G}-\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx+\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx}\tag1$$
where 
$$\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx=\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\ln^2(1+x)}{1+x^2}dx}_{x\mapsto 1/x}$$
$$=\underbrace{\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx}_{2\ \text{Im}\operatorname{Li}_3(1+i)}-\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx+2\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}dx}_{\pi^3/16}$$
$$\Longrightarrow \int_0^1\frac{\ln^2(1+x)}{1+x^2}dx=\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx+\text{Im}\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}\tag2$$
Plug $(2)$ in $(1)$ we obtain
$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2$$
