Evaluate $\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx$ 
How to prove $\ \displaystyle\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i)$

Where $\ \displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}\ $ 
is the trilogarithm.
I managed to prove the above equality using integral manipulation, but I would like to see different approaches. 

Using   $\displaystyle\int_0^1\frac{u\ln^nx}{1-ux}\ dx=(-1)^nn!\operatorname{Li}_{n+1}(u)$ .
which can be found in Cornel's book, (Almost) impossible integrals, sums and series.
and with $n=1$, we get  $\displaystyle\int_0^1\frac{u\ln x}{1-ux}\ dx=-\operatorname{Li}_2(u)$ 
Divide both sides by $1+u^2$ then integrate from $u=0$ to $u=1$, we get
\begin{align}
\int_0^1\int_0^1\frac{u\ln x}{(1-ux)(1+u^2)}\ dx\ du=-\int_0^1\frac{\operatorname{Li}_2(u)}{1+u^2}\ du=I
\end{align}
Lets evaluate the double integral:
\begin{align}
I&=\int_0^1\ln x\left(\int_0^1\frac{u}{(1-ux)(1+u^2)}\ du\right)\ dx\\
&=\int_0^1\ln x\left(\frac{\ln2}{2}\frac{1}{1+x^2}-\frac{\ln(1-x)}{1+x^2}-\frac{\pi}{4}\frac{x}{1+x^2}\right)\ dx\\
&=\frac{\ln2}{2}\int_0^1\frac{\ln x}{1+x^2}\ dx-\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx-\frac{\pi}{4}\int_0^1\frac{x\ln x}{1+x^2}\ dx\\
&=-\frac12G\ln2-\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx-\frac{\pi}{4}\left(\frac{\pi^2}{48}\right)\tag{1}
\end{align}
And by applying IBP to $\displaystyle\int_0^1\frac{\operatorname{Li}_2(u)}{1+u^2}\ du\ $, we get
$$I=-\int_0^1\frac{\operatorname{Li}_2(u)}{1+u^2}\ du=-\frac{\pi^3}{24}-\int_0^1\frac{\arctan u\ln(1-u)}{u}\ du\tag{2}$$
From $(1)$ and $(2)$, we get
$$\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}-\frac12G\ln2+\int_0^1\frac{\arctan u\ln(1-u)}{u}\ du$$
substituting $\ \displaystyle\int_0^1 \frac{\arctan u\ln(1-u)}{u}\ du=\frac{\pi}{16}\ln^22+\frac12G\ln2-\text{Im}\operatorname{Li}_3(1+i)\ $ ( proved here)
gives: $$\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i)$$
