<s>[INST] Closed form for $\sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2} \,dx$ I am trying to get this to equal $\displaystyle-\frac {\pi^a}{b}$ for some positive integers $a$ and $b$ . My efforts so far give:
$\displaystyle \sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2} \,dx=A-B+C$
where   $A=\displaystyle \sum_{n=1}^\infty\frac{2\zeta(3)}{n^3}=2\zeta(3)^2$
and  $C=\displaystyle \sum_{n=1}^\infty\frac{\pi^2}{3n^4}=\frac{\pi^6}{270}$  
and  $B=\displaystyle \sum_{n=1}^\infty\frac{f(n)}{n^6}$ with f(1)=6, f(2)=34, f(3)=393/4, f(4)=5750/27, f(5)=339059/864, f(6)=325493/500, f(7)=36107863/36000, ... ,f(11)=29071954407257/8001504000, ...
I can not figure out what $f(n)$ might be. The sequences of numerators and denominators are unrecognized by  OEIS. Maybe I am off the track.
Perhaps a different approach from the beginning would be better. The results for  A and C were interesting anyway.
 [/INST] Hint. One may use some MZVs algebra.
For $n\geq1$, set
$$
 I_n:=\frac1{n^2}\int_0^1 x^{n-1}\ln^2(x)\ln(1-x)\:dx,
$$ then, differentiating the Euler beta integral three times, one gets
$$
I_n=-\frac4{n^6}-2\frac{H_{n-1}}{n^5}+2\frac{\frac{\pi^2}6-H_{n-1,2}}{n^4}+2\frac{\zeta(3)-H_{n-1,3}}{n^3}
$$ summing with respect to $n$ gives
$$
\sum_{n=1}^\infty I_n=-4\zeta(6)-2\zeta(5,1)+\frac{\pi^2}3\zeta(4)-2\zeta(4,2)+2\zeta(3)^2-2\zeta(3,3),
$$ using
$$
\begin{align}
\zeta(5,1)&=-\frac12\zeta(3)^2 +\frac{\pi^6}{1260}
\\\\ \zeta(4,2)&= \zeta(3)^2 -\frac{4\pi^6}{2835}
\\\\ \zeta(3,3)&=\frac12\zeta(3)^2 -\frac{\pi^6}{1890}
\end{align}
$$ leads to

$$
\sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2}=-\frac{\pi^6}{2835}
$$ 

or

$$
\int_0^1 \frac{\text{Li}_2(x)\ln^2(x)\ln(1-x)}x=-\frac{\pi^6}{2835}.
$$

Probably there is  a direct path using $\rm{Li}_2(\cdot)$ properties.
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