Integral $\int_0^1 dx \frac{\ln x \ln^2(1-x)\ln(1+x)}{x}$ I am trying to calculate
$$
I:=\int_0^1 dx \frac{\ln x \ln^2(1-x)\ln(1+x)}{x}$$
Note, the closed form is beautiful (yes beautiful) and is given by
$$
I=−\frac{3}{8}\zeta_2\zeta_3 -\frac{2}{3}\zeta_2\ln^3 2  +\frac{7}{4}\zeta_3\ln^2 2-\frac{7}{2}\zeta_5+4\ln 2 \operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{2}{15}\ln^5 2+4\operatorname{Li}_5\left(\frac{1}{2}\right)
$$
where
$$
\zeta_s=\sum_{n=1}^\infty \frac{1}{n^{s}},\qquad \operatorname{Li}_s(z)=\sum_{n=1}^\infty \frac{z^n}{n^s},\qquad\text{for}\ |z|<1.
$$
I succeeded in writing the integral as
$$
I=-\sum_{i=0}^\infty \int_0^1  x^i\ln x\ln(1+x)\ln(1-x)\ dx,
$$
but I am confused as to where to go from here.  Possibly I was thinking of trying to use Mellin transforms or residues.
A reference to aid us is here. (Since somebody has asked for reference)
We can also write I as
$$
I=\sum_{i=0}^\infty \sum_{j=1}^\infty \frac{1}{j}\sum_{k=1}^\infty \frac{1}{k} \int_0^1  x^{i+j+k} \ln x\ dx
$$
using
$$
\int_0^1 x^n \ln x\ dx= -\frac{1}{(n+1)^2},
$$
we can simplify this, but I am not sure then how to compute the triple sum.  Thank you again.
 A: using the following identity proved by Cornel and can be found in his book, (Almost) Impossible Integrals, Sums and Series. $\quad\displaystyle\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n}$.
multiply both sides by $\frac{\ln(1-x)}{x}$ then integrate from $x=0$ to $1$, we get
\begin{align}
I&=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln(1-x)\ dx\\
&=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(-\frac{H_{2n}}{2n}\right)\\
&=2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n)^2}-\frac12\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}+\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}\\
&=\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{n^2}-\frac12\color{blue}{\sum_{n=1}^\infty\frac{H_{n}H_{2n}}{n^2}}+\sum_{n=1}^\infty\frac{H_{n}}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^3}
\end{align}
I managed here to prove $$\color{blue}{\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}}=4\sum_{n=1}^\infty\frac{H_n^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-4\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}-6\zeta(4)$$
which follows that
$$\boxed{I=\sum_{n=1}^\infty\frac{H_n}{n^3}-\sum_{n=1}^\infty\frac{H_n^2}{n^2}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}+3\zeta(4)}$$
Plugging the following results: 
$$\displaystyle\sum_{n=1}^{\infty}\frac{H_n^2}{n^2}=6\zeta(4)-\frac74\zeta(4)=\frac{17}4\zeta(4)$$
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$
The first sum can be found here, and the second sum can be found using the easy-to-prove generating function $\sum_{n=1}^\infty\frac{x^nH_n^{(3)}}{n}=\operatorname{Li_4}(x)-\ln(1-x)\operatorname{Li_3}(x)-\frac12\operatorname{Li_2}^2(x)$,substituting these two sums along with the well known value of $\sum_{n=1}^\infty\frac{H_n}{n^3}=3\zeta(5)-\zeta(2)\zeta(3)$, we get the desired closed form of $I$.
NOTE . Sorry guys I just noticed that my solution is for $\int_0^1\frac{\ln^2(1-x)\ln(1+x)}{x}\ dx$ without $\ln x$ in the numerator as in the original problem. I am keeping the solution as it was voted as useful.
