TITLE: Closed- form of $\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$
QUESTION [14 upvotes]: Is there a possibility to find a closed-form for
 $$\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$$

REPLY [13 votes]: $\def\Li{{\rm Li}}$
Repeatedly integrate by parts.
\begin{align}
&\int^1_0\frac{\Li_3^2(-x)}{x^2}{\rm d}x\\
=&-\frac{\Li_3^2(-x)}{x}\Bigg{|}^1_0+\int^1_0\frac{2\Li_2(-x)\Li_3(-x)}{x^2}{\rm d}x\\
=&-\frac{9}{16}\zeta^2(3)-\frac{2\Li_2(-x)\Li_3(-x)}{x}\Bigg{|}^1_0+\int^1_0\frac{2\Li_2^2(-x)}{x^2}{\rm d}x-\int^1_0\frac{2\Li_3(-x)\ln(1+x)}{x^2}{\rm d}x\\
=&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{2\Li_2^2(-x)}{x}\Bigg{|}^1_0-\int^1_0\frac{4\Li_2(-x)\ln(1+x)}{x^2}{\rm d}x\\
&+\frac{2\Li_3(-x)\ln(1+x)}{x}\Bigg{|}^1_0-\int^1_0\frac{2\Li_3(-x)}{x(1+x)}{\rm d}x-\int^1_0\frac{2\Li_2(-x)\ln(1+x)}{x^2}{\rm d}x\\
=&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\int^1_0\frac{2\Li_3(-x)}{1+x}{\rm d}x\\
&+\frac{6\Li_2(-x)\ln(1+x)}{x}\Bigg{|}^1_0-\int^1_0\frac{6\Li_2(-x)}{x(1+x)}{\rm d}x+\int^1_0\frac{6\ln^2(1+x)}{x^2}{\rm d}x\\
=&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\frac{9}{2}\zeta(3)-3\zeta(2)\ln{2}\\
&+\int^1_0\frac{6\Li_2(-x)}{1+x}{\rm d}x+\int^1_0\frac{2\Li_3(-x)}{1+x}{\rm d}x-\frac{6\ln^2(1+x)}{x}\Bigg{|}^1_0+\int^1_0\frac{12\ln(1+x)}{x(1+x)}{\rm d}x\\
=&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\frac{9}{2}\zeta(3)-3\zeta(2)\ln{2}\\
&+6\zeta(2)-12\ln^2{2}+\int^1_0\frac{6\Li_2(-x)}{1+x}{\rm d}x+\int^1_0\frac{2\Li_3(-x)}{1+x}{\rm d}x
\end{align}
Since
\begin{align}
\int^1_0\frac{6\Li_2(-x)}{1+x}{\rm d}x
=&6\Li_2(-x)\ln(1+x)\Bigg{|}^1_0+\int^1_0\frac{6\ln^2(1+x)}{x}{\rm d}x\\
=&\frac{3}{2}\zeta(3)-3\zeta(2)\ln{2}
\end{align}
(see Log Integrals I)
and
\begin{align}
\int^1_0\frac{2\Li_3(-x)}{1+x}{\rm d}x
=&2\Li_3(-x)\ln(1+x)\Bigg{|}^1_0-\int^1_0\underbrace{\frac{2\Li_2(-x)\ln(1+x)}{x}{\rm d}x}_{\displaystyle\small{2\Li_2(-x)\ {\rm d}\Li_2(-x)}}\\
=&\frac{1}{4}\zeta^2(2)-\frac{3}{2}\zeta(3)\ln{2}
\end{align}
The closed form is
\begin{align}
\int^1_0\frac{\Li_3^2(-x)}{x^2}{\rm d}x
=&-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{4}\zeta^2(2)+\frac{7}{4}\zeta(4)-3\zeta(3)\ln{2}\\&+6\zeta(3)-6\zeta(2)\ln{2}+6\zeta(2)-12\ln^2{2}\\
\end{align}