Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$ In this thread 
a friend posted the following integral 
$$I=\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$$
The best we could do is expressing it in terms of Euler sums 
$$I=-\frac{\zeta^2(2)}{2}+ \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^2} H_{n}^{(2)}+\sum_{n\geq 1}\frac{(-1)^{n-1}}{n^3}H_{n}$$
I am wondering if the approach I followed made the integral complicated ?
What approach would you follow to solve the integral?, can we find a better solution ?
 A: using an identity devoloped by Cornel Ioan Valean and it can be found in his book " Almost impossible integrals, sums, and series":
$$\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n} $$
we get:
\begin{align}
I&=\int_0^1\frac{\ln(1-x)\ln(1+x)\ln x}{x}\ dx=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln x\ dx\\
&=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(-\frac1{(2n)^2}\right)=2\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}-\frac14\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)\\
&=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)\\
&=2\operatorname{Li}_4\left(\frac12\right)-\frac12\ln^22\zeta(2)+\frac74\ln2\zeta(3)-\frac{27}{16}\zeta(4)+\frac1{12}\ln^42
\end{align}
where we used the value of the first sum proved here and the common value of the second sum which is $\frac54\zeta(4)$.
A: Related problems: (I). You can have the following solution

$$ \frac{3\gamma}{4}\,\zeta( 3 )+{\frac {7\pi^4}{360}}+\sum _{m=1}^{\infty }{\frac { \left( -1 \right) ^{m-1}\psi \left( m
 \right) }{{m}^{3}}}+\sum _{m=1}^{\infty 
}-{\frac { \left( -1 \right) ^{m-1}\psi' \left( m \right) }{{m}^{2}}}\sim 0.2907212779,$$

which you might be able to simplify it further.
Note: If you use the identity 

$$ \frac{\pi^4}{90}=\zeta(4), $$

in the above expression, then you will have the form

$$ \frac{3\gamma}{4}\,\zeta( 3 )+{\frac {7}{4}}\zeta(4)+\sum _{m=1}^{\infty }{\frac { \left( -1 \right) ^{m-1}\psi \left( m
 \right) }{{m}^{3}}}+\sum _{m=1}^{\infty 
}-{\frac { \left( -1 \right) ^{m-1}\psi' \left( m \right) }{{m}^{2}}}.$$ 

A: The values of the two Euler Sums are
$$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n}{n^{3}} = \frac{11\pi^4}{360}-2\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7}{4}\log(2) \zeta(3)+\frac{\pi^2}{12}\log^2(2)-\frac{1}{12}\log^4(2)$$
$$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n^{(2)}}{n^{2}}  =-\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$
Therefore the integral evaluates to

$$\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \sim 0.290721 \end{align*}$$

Refer to this page for the evaluation of Euler Sums.
A: I would do the following variable change.  
$$x=e^{-t}$$ Then we can represent the integral as follows:  
$$I=-\int_{0}^{\infty}t\ln(1+e^{-t})\ln(1-e^{-t})\;dt$$ Now, apply the Taylor expansion of the logarithm:  
$$\ln(1+x)=\sum_{i=1}^{\infty}(-1)^{i-1}\frac{x^i}{i}$$  
$$I=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i-1}}{ij}\int_{0}^{\infty}te^{-(i+j)t}dt=$$  
$$=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i-1}}{ij(i+j)^2}$$
A: $$-\frac{\partial^2}{\partial s\partial t}\left[B(s+1,t+1)\;_3 F_2(1,1,s+1;2,s+t+2;-1)\right]_{s=t=0}$$
It may be that the Hypergeometric function is summable. In this case, the differentiation is trivial. (B denotes Euler's beta function.)
