Other challenging logarithmic integral $\int_0^1 \frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx$ How can we prove that:

$$\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx=\frac{\pi^2}{8}\zeta(3)-\frac{27}{16}\zeta(5)  $$

 A: A generalization of the present integral is given in the book (Almost) Impossible Integrals, Sums, and Series (see page 6),
$$\int_0^1\frac{\log(1-x)\log^{2n}(x)\log(1+x)}{x} \textrm{d}x =\frac{1}{2}(2n)!\left(1-\frac{1}{2^{2n+1}}\right)\sum_{k=1}^{2n} \zeta(k+1)\zeta(2n-k+2)$$
$$-(2n)!\sum_{k=1}^{n}\left(1-\frac{1}{2^{2k-1}}\right)\zeta(2k)\zeta(2n-2k+3) + \frac{1}{2^{2n+3}} (2n+3-2^{2n+3})(2n)!\zeta(2n+3),$$
where in his solution the author exploits a classical series representation of $\log(1-x)\log(1+x)$.
A: Let $I$ denotes our integral $\displaystyle\int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx$
By using the generating function: $$\displaystyle\ln(1+x)\ln(1-x)=\sum_{n=1}^\infty \frac{H_n-H_{2n}}{n}x^{2n}-\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2}$$
we can write 
\begin{align}
I&=\sum_{n=1}^\infty \frac{H_n-H_{2n}}{n}\int_0^1x^{2n-1}\ln^2x\ dx-\frac12\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{2n-1}\ln^2x\ dx\\
&=\frac14\sum_{n=1}^\infty \frac{H_n-H_{2n}}{n^4}-\frac1{8}\sum_{n=1}^\infty\frac1{n^5}\\
&=-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-\frac74\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)
\end{align}
By plugging $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$ we get 

$$I=\frac34\zeta(2)\zeta(3)-\frac{27}{16}\zeta(5)$$

A: Different approach:
Let $I$ denotes our integral $\displaystyle\int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx$
By using the identity $4ab=(a+b)^2-(a-b)^2$ and setting $a=\ln(1-x)$ and $b=\ln(1+x)$ we have
$$4I=\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{x}\ dx-\int_0^1\frac{\ln^2x\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx$$

The first integral:
$$\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{x}\ dx=\frac18\int_0^1\frac{\ln^2x\ln^2(1-x)}{x}\ dx=\frac14\sum_{n=1}^\infty\frac{H_n}{n+1}\int_0^1 x^n\ln^2x\ dx\\=\frac12\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac12\zeta(5)=\boxed{\zeta(5)-\frac12\zeta(2)\zeta(3)}$$

The second integral:
We proved here  that 
$$\ln^2\left(\frac{1-x}{1+x}\right)=-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}x^{2n}$$
Then we can write 
$$\int_0^1\frac{\ln^2x\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx=-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}\int_0^1x^{2n-1}\ln^2x\ dx=16\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}-\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}\\
=8\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{15}2\sum_{n=1}^\infty\frac{H_n}{n^4}=\boxed{\frac{31}{4}\zeta(5)-\frac72\zeta(2)\zeta(3)}$$
and the answer follows by substituting these boxed results.
A: \begin{align}J&=\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx\\
U&=\int_0^1\frac{\log(1+x)\log^3 x}{1-x}dx,V=\int_0^1\frac{\log(1-x)\log^3 x}{1+x}dx\\
J&\overset{\text{IBP}}=\frac{1}{3}\Big[\ln^3 x\ln(1-x)\ln(1+x)\Big]_0^1-\frac{1}{3}\int_0^1 \left(\frac{1}{1+x}-\frac{1}{1-x}\right)\ln^3 x\,dx\\
&=\frac{1}{3}\Big(U-V\Big)\\
\end{align}
From Integrating $\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx$ with restricted techniques
One obtains:
\begin{align*}U&=-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)\\
V&=\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)\\
J&=\boxed{\frac{3}{4}\zeta(2)\zeta(3)-\frac{27}{16}\zeta(5)}
\end{align*}
