Closed form of $\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$ I remember that some time ago I was asking this question Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$ ,
and now, while I was making a review, I asked myself if we can get the closed form of
$$
\int_{0}^{1}\left[\vphantom{\Large A}\ln\left(1 - x\right)\ln\left(1 + x\right)\ln\left(x\right)\right]^{\,2}\,\mathrm{d}x
$$
by using the similar tools as in that proof. The problem is that we have to cope with some
series that seem far more complicated. In case you have some fruitful ideas here ...
 A: Process 1:

It is known that
\begin{align}
\int_{0}^{1} x^{\nu -1} (1+x)^{\lambda} (1-x)^{\mu -1} \, dx = B(\mu, \nu) \, {}_{2}F_{1}(- \lambda, \nu; \mu+\nu; -1)
\end{align}
for which 
\begin{align}
\int_{0}^{1} \left[ \ln(x) \, \ln(1-x) \, \ln(1+x) \right]^{2} \, dx = \partial_{\nu}^{2} \partial_{\mu}^{2} \partial_{\lambda}^{2} \left[ B(\mu, \nu) \, {}_{2}F_{1}(- \lambda, \nu; \mu+\nu; -1) \right]_{\mu=\nu=1}^{\lambda = 0} 
\end{align}
The resulting value will then lead to the calculation of series involving the polygamma function of up to order three.

Process 2

In the view of using the series
\begin{align}
\frac{1}{2} \, \ln^{2}(1-x) = \sum_{n=1}^{\infty} \frac{H_{n} \, x^{n+1}}{n+1}
\end{align}
then
\begin{align}
\frac{1}{4} \, \ln^{2}(1-x) \, \ln^{2}(1+x) = \sum_{n=1}^{\infty} A_{n} \, x^{n+2}  
\end{align}
where 
\begin{align}
A_{n} = \sum_{s=1}^{n} \frac{(-1)^{s-1} \, H_{n-s} H_{s} }{(s+1) (n-s+1)}.
\end{align}
Now consider the integral
\begin{align}
J = \int_{0}^{1} x^{\mu + \nu} \, dx = \frac{1}{\mu+\nu+1}
\end{align}
for which
\begin{align}
\partial_{\mu}^{2} J = \int_{0}^{1} x^{\mu+\nu} \, \ln^{2}(x) \, dx = \frac{2}{(\mu+\nu+1)^{3}}.
\end{align}
Now, for the integral
\begin{align}
I = \int_{0}^{1} \left[ \ln(1-x) \, \ln(1+x) \, \ln(x) \right]^{2} \, dx,
\end{align}
it is seen that
\begin{align}
I &= 4 \sum_{n=1}^{\infty} A_{n} \, \int_{0}^{1} x^{n+2} \, \ln^{2}(x) \, dx \\
&= 8 \, \sum_{n=1}^{\infty} \frac{A_{n}}{(n+3)^{3}} \\
&= 8 \, \sum_{n=1}^{\infty} \sum_{s=1}^{\infty} \frac{(-1)^{s-1} \, H_{s} H_{n}}{(n+1) (s+1) (n+s+3)^{3}}.
\end{align}
