Compute $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$ How to prove that
$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$$
$$=2\text{Li}_4\left(\frac12\right)-2\zeta(4)+\frac{15}8\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)$$
where $\text{Li}_r$ is the polylogarithm function and $\zeta$ is Riemann zeta function.
I managed to prove the equality above using the following harmonic series, 
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n}, \ \sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}H_n}{n},\ \sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}\ \text{and }\ \sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$
so definitely this approach is pretty boring. Is it possible to solve it in a different way? Thank you.

Set  $x=2t-1$
$$\begin{align}
  & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)}{t}\ln \left( 2-2t \right)dt} \\ 
 & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)}{t}\left( \ln \left( 2 \right)-\sum\nolimits_{n=1}^{\infty }{\frac{{{t}^{n}}}{n}} \right)dt} \\ 
 & =\int_{\frac{1}{2}}^{1}{\left\{ \frac{\ln \left( t \right)\ln \left( 2t \right)\ln \left( 2 \right)}{t}-\sum\nolimits_{n=1}^{\infty }{\frac{{{t}^{n-1}}\ln \left( t \right)\ln \left( 2t \right)}{n}} \right\}dt} \\ 
 & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)\ln \left( 2 \right)}{t}dt-}\sum\nolimits_{n=1}^{\infty }{\frac{1}{n}\int_{\frac{1}{2}}^{1}{{{t}^{n-1}}\ln \left( t \right)\ln \left( 2t \right)}dt} \\ 
 & =-\frac{1}{6}{{\ln }^{4}}\left( 2 \right)-\sum\nolimits_{n=1}^{\infty }{\left( \frac{2}{{{n}^{4}}}-\frac{2}{{{2}^{n}}{{n}^{4}}}-\frac{\ln \left( 2 \right)}{{{n}^{3}}}-\frac{\ln \left( 2 \right)}{{{2}^{n}}{{n}^{3}}} \right)} \\ 
 & \vdots  \\ 
 & \vdots  \\ 
\end{align}$$
