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: $$\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i)$$
Different approach using the harmonic series:
I am going to use the following identities:
$$\int_0^1 x^{n}\ln(1-x)=-\frac{H_{n+1}}{n+1}$$ $$\int_0^1\frac{x^n\ln x}{1-x}dx=-\zeta(2)+H_n^{(2)}$$ $$\sum_{n=0}^\infty(-1)^nf(2n+1)=\text{Im}\sum_{n=1}^\infty(i)^nf(n)$$
Start with applying integration by parts then writing $\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}$ we get
$$I=\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}dx=\int_0^1\arctan x\left(\frac{\ln x}{1-x}-\frac{\ln(1-x)}{x}\right)dx$$
$$=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\left[\int_0^1\frac{x^{2n+1}\ln x}{1-x}dx-\int_0^1x^{2n}\ln(1-x)dx\right]$$
$$=-\zeta(2)\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}+\sum_{n=0}^\infty(-1)^n\left(\frac{H_{2n+1}^{(2)}}{2n+1}+\frac{H_{2n+1}}{(2n+1)^2}\right)$$
$$=-\zeta(2)\frac{\pi}{4}+\text{Im}\sum_{n=1}^\infty(i)^n\left(\frac{H_{n}^{(2)}}{n}+\frac{H_{n}}{n^2}\right)\tag1$$
From here we have
$$\small{\sum_{n=1}^\infty x^n\left(\frac{H_{n}^{(2)}}{n}+\frac{H_{n}}{n^2}\right)=2\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\frac12\ln x\ln^2(1-x)-\zeta(2)\ln(1-x)-\zeta(3)}$$
By setting $x=i$ and considering the imaginary part we get
$$\text{Im}\sum_{n=1}^\infty(i)^n\left(\frac{H_{n}^{(2)}}{n}+\frac{H_{n}}{n^2}\right)=\frac{17\pi^3}{192}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i)\tag2$$
Plug (2) in (1), the closed form of $I$ follows.