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.
 A: Set  $x=2t-1$
$$\begin{align}
  & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)}{t}\ln \left( 2-2t \right)dt} \\ 
 & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)}{t}\left( \ln \left( 2 \right)-\sum\nolimits_{n=1}^{\infty }{\frac{{{t}^{n}}}{n}} \right)dt} \\ 
 & =\int_{\frac{1}{2}}^{1}{\left\{ \frac{\ln \left( t \right)\ln \left( 2t \right)\ln \left( 2 \right)}{t}-\sum\nolimits_{n=1}^{\infty }{\frac{{{t}^{n-1}}\ln \left( t \right)\ln \left( 2t \right)}{n}} \right\}dt} \\ 
 & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)\ln \left( 2 \right)}{t}dt-}\sum\nolimits_{n=1}^{\infty }{\frac{1}{n}\int_{\frac{1}{2}}^{1}{{{t}^{n-1}}\ln \left( t \right)\ln \left( 2t \right)}dt} \\ 
 & =-\frac{1}{6}{{\ln }^{4}}\left( 2 \right)-\sum\nolimits_{n=1}^{\infty }{\left( \frac{2}{{{n}^{4}}}-\frac{2}{{{2}^{n}}{{n}^{4}}}-\frac{\ln \left( 2 \right)}{{{n}^{3}}}-\frac{\ln \left( 2 \right)}{{{2}^{n}}{{n}^{3}}} \right)} \\ 
 & \vdots  \\ 
 & \vdots  \\ 
\end{align}$$
A: I proved here
$$\small{\int_0^a\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\text{Li}_3\left(\frac{1+a}{2}\right)-\text{Li}_3\left(\frac{1}{2}\right)-\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)+\frac12\ln2\ln^2(1+a)}$$
Divide both sides by $1+a$ the integrate from $a=0$ to $a=1$ we get
$$\int_0^1\int_0^a\frac{\ln(1-x)\ln(1+x)}{(1+x)(1+a)} \ dxda=\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da-\text{Li}_3\left(\frac{1}{2}\right)\underbrace{\int_0^1\frac{da}{1+a}}_{\ln(2)}$$
$$-\int_0^1\frac{\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)}{1+a}\ da+\frac12\ln2\underbrace{\int_0^1\frac{\ln^2(1+a)}{1+a}\ da}_{1/3 \ln^3(2)}$$
where
$$\int_0^1\int_0^a\frac{\ln(1-x)\ln(1+x)}{(1+x)(1+a)} \ dxda=\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x} \left(\int_x^1\frac{da}{1+a}\right)\ dx$$
$$=-\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx=-I$$
$$\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da=\text{Li}_4\left(\frac{1+a}{2}\right)\bigg|_0^1=\zeta(4)-\text{Li}_4\left(\frac{1}{2}\right)$$
$$\int_0^1\frac{\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)}{1+a}\ da\overset{IBP}{=}\ln(1+a)\text{Li}_3\left(\frac{1+a}{2}\right)\bigg|_0^1-\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da$$
$$=\ln(2)\zeta(3)-\zeta(4)+\text{Li}_4\left(\frac{1}{2}\right)$$
Combine all results and use $\text{Li}_3\left(\frac{1}{2}\right)=\frac78\zeta(3)-\frac12\ln(2)\zeta(2)+\frac16\ln^3(2)$, we get the closed form of $I$.
A: Substitute $t=\frac{1+x}2$
\begin{align}
I=&\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\frac{1+x}{2}\ dx\\
=&\int_{\frac12}^1 \frac{\ln(2(1-t))(\ln^2t+\ln2\ln t)}t \>dt
\overset{ibp}=\frac13 \int_{\frac12}^1 \frac{\ln^3t}{1-t}dt+ 
 \frac12\ln2 \int_{\frac12}^1 \frac{\ln^2t}{1-t}dt
\end{align}
where $\int_{0}^1 \frac{\ln^3t}{1-t}dt=-6Li_4(1)$, $\int_{0}^1 \frac{\ln^2t}{1-t}dt=2Li_3(1)$
\begin{align}
&\int^{\frac12}_0\frac{\ln^2t}{1-t}dt
=2Li_3(\frac12)+2\ln2 Li_2(\frac12)+\ln^3 2\\
&\int^{\frac12}_0\frac{\ln^3t}{1-t}dt
=-6Li_4(\frac12)-6\ln2 Li_3(\frac12)-3\ln^22Li_2(\frac12)-\ln^42
\end{align}
Thus
\begin{align}
I=&\>2\>[ Li_4(\frac12)-Li_4(1) ]+\ln2 \>[Li_3(\frac12)+Li_3(1)] -\frac16\ln^42\\
\end{align}
