Is a closed form possible for $\int\frac{\text{Li}_2(x)^2}{x}dx$?

Question

Can $\,\displaystyle\int\frac{\text{Li}_2(x)^2}{x}dx\,$ be calculated

by a sum/term of polylogarithm functions and the natural logarithm and polynomials (“closed form”) ?

For the special case $\,\displaystyle\int\limits_0^1\frac{\text{Li}_2(x)^2}{x}dx\,$ see here .

Note:

Closed forms are possible for $\,\displaystyle\int\frac{\text{Li}_n(x)^2}{x}dx\,$ where $n\in\mathbb{Z}\,$ and $\,n<2\,$ , e.g. $\,\displaystyle\int\frac{\text{Li}_1(x)^2}{x}dx = -2 \text{Li}_3(1-x) - 2 \text{Li}_2(1-x) \text{Li}_1(x) - \text{Li}_1(1-x) \text{Li}_1(x)^2 + C\,$

and $\,\displaystyle\int\frac{\text{Li}_0(x)^2}{x}dx = \frac{\text{Li}_0(x)}{x} - \text{Li}_1(2-x) + C\,$ .

Answer 1

\begin{align} \int\frac{\operatorname{Li}_2(x)^2}{x}\ dx&\overset{IBP}{=}\operatorname{Li}_3(x)\operatorname{Li}_2(x)+\int\frac{\operatorname{Li}_3(x)\ln(1-x)}{x}\ dx\\ &=\operatorname{Li}_3(x)\operatorname{Li}_2(x)+\sum_{n=1}^\infty\frac1{n^3}\int x^{n-1}\ln(1-x)\ dx \end{align}

This is the simplest form I can get.