Evaluating $\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx$ I've been trying to evaluate
$$\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx$$
With no success, i tried to consider the following integrals
$$I=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx,J=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx$$
$$I+J=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^2\right)}{1+x^2}\:dx=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^4\right)}{1+x^2}\:dx-\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$$
I managed to express that $1$st integral into somewhat known euler sums but that $2$nd integral arrived at a sum i didnt know how to evaluate which was
$$2\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{\left(2k+1\right)^3}$$
And it seems this approach wont go smooth, could i tackle the main integral differently? maybe with an easier approach?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[10px,#ffd]{\int_{0}^{1}
{\ln^{2}\pars{x}\ln\pars{1 - x} \over 1 + x^{2}}\,\dd x}:\ {\Large ?}}$.

\begin{align}
&\mbox{Lets consider}
\\[1mm] &\
\mathcal{I}\pars{a} \equiv
\left.\int_{0}^{1}
{\ln^{2}\pars{x}\ln\pars{1 - ax} \over 1 + x^{2}}\,\dd x\,
\right\vert_{\ a\ >\ 1}\,,\
\mathcal{I}\pars{0} = 0
\label{1}\tag{1}
\end{align}

\begin{align}
\mathcal{I}'\pars{a} & \equiv
\Im\int_{0}^{1}{x\ln^{2}\pars{x} \over \pars{\ic - x}\pars{1 - ax}}\,\dd x
\\[5mm] & =
-\,\Im\bracks{{1 \over a + \ic}\int_{0}^{1}{\ln^{2}\pars{x} \over
\ic - x}\,\dd x} -
\Im\bracks{{\ic/a \over a + \ic}\int_{0}^{1}{\ln^{2}\pars{x} \over
1/a - x}\,\dd x}
\end{align}
However,
$\ds{\int_{0}^{1}{\ln^{2}\pars{x} \over \xi - x}\,\dd x = 2\,\mrm{Li}_{3}\pars{1 \over \xi}}$. Then,
\begin{align}
\mathcal{I}'\pars{a} & =
-2\,\Im\bracks{{\mrm{Li}_{3}\pars{-\ic} \over a + \ic}} -
2\,\Re\bracks{\mrm{Li}_{3}\pars{a} \over a\pars{a + \ic}}
\\[5mm]
\mathcal{I}\pars{1} & =
-2\,\Im\bracks{\mrm{Li}_{3}\pars{-\ic}\int_{0}^{1}{\dd a \over a + \ic}} +
2\,\Im\int_{0}^{1}{\mrm{Li}_{3}\pars{a} \over \ic + a}\,\dd a
\end{align}
A: Using the same approach as @Felix Marin, making the story short, we have
$$I'(a)=\frac{\pi ^3 a-32 \text{Li}_3(a)-3 \zeta (3)}{16( a^2+1)}$$
$$\int I'(a)\,da=\frac{\pi ^3}{32}  \log \left(a^2+1\right)-\frac{3\zeta (3)}{16}  \tan ^{-1}(a)-2\int\frac{ \text{Li}_3(a)}{a^2+1}\,da$$
$$\int_0^1 I'(a)\,da=\frac{\pi}{64} \left(2\pi ^2 \log (2)-3   \zeta (3)\right)-2\int_0^1\frac{ \text{Li}_3(a)}{a^2+1}\,da$$
As said in comments, the last integral is given here and then the final result.
