Integral $\ 4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx$ How to prove that

$$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx=\frac{29}4\zeta(2)\zeta(3)-\frac{91}8\zeta(5)$$

Where $\chi_2(x)=\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}$ is the Legendre Chi function and $ \operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$ is the Dilogarithm function.
This integral was proposed by Cornel.
 A: Using the relation between the Chi function and Dilogarithm we can rewrite the first integral as:
$$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}dx=2\int_0^1\frac{\operatorname{Li}^2_2(x)}{x} dx-2\int_0^1\frac{\operatorname{Li}_2(x)\operatorname{Li}_2(-x)}{x} dx$$
You solved the first part here.
$$\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}dx=2\zeta(2)\zeta(3)-3\zeta(5)$$
And the second one is found here:
$$\int_0^1\frac{\operatorname{Li}_2(x){\operatorname{Li}_2(-x)}}{x}dx =-\frac54\zeta(2)\zeta(3)+\frac{59}{32}\zeta(5)$$
Combinging the two results from above yields:
$$\boxed{4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}dx=\frac{13}{2}\zeta(2)\zeta(3)-\frac{155}{16}\zeta(5)}$$
The second integral is solved here.
$$\boxed{\int_0^1\frac{\ln(1-x)\ln^2 x\ln(1+x)}{x}dx=\frac34 \zeta(2)\zeta(3)-\frac{27}{16}\zeta(5)}$$
Combining the two boxed results gives:
$$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x} dx+\int_0^1\frac{\ln(1-x)\ln^2(x)\ln(1+x)}{x} dx=\frac{29}4\zeta(2)\zeta(3)-\frac{91}8\zeta(5)$$

Remark.
We know from above that:
$$\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}dx=\frac{13}{8}\zeta(2)\zeta(3)-\frac{155}{64}\zeta(5)$$
But integating by parts also gives us:
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}\int_0^1 x^{2n}\operatorname{Li}_2 (x)dx$$$$\overset{IBP}=\sum_{n=0}^\infty \frac{\operatorname{Li}_2(1)}{(2n+1)^3}+\sum_{n=0}^\infty \frac{1}{(2n+1)^3}\int_0^1 x^{2n}\ln(1-x)dx$$
$$=\frac{7}{8}\zeta(2)\zeta(3) +\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^4}$$
Which results in:
$$\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^4}=\frac34\zeta(2)\zeta(3)-\frac{155}{64}\zeta(5)$$
Alteratively one can compute that sum in a different way to find the value of the first integral.
A: This approach is pretty identical to Cornel's solution posted on his FB page.
using the fact that $\quad\displaystyle \sum_{n=1}^\infty a_{2n}=\frac12\left(\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty (-1)^na_n\right),\ $ we have 
\begin{align}
\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}&=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}=\frac12\left(\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)^2}+\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{(n+1)^2}\right)\\
&=\frac12\left(\sum_{n=1}^\infty\frac{x^n}{n^2}-\sum_{n=1}^\infty(-1)^n\frac{x^n}{n^2}\right)=\frac12\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)\right)
\end{align}
then, the first integral:
\begin{align}
I_1&=4\int_0^1\left(\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}\right)\frac{\operatorname{Li}_2(x)}{x}\ dx\\
&=2\sum_{n=1}^\infty\left(\frac1{n^2}-\frac{(-1)^n}{n^2}\right)\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\
&=2\sum_{n=1}^\infty\left(\frac1{n^2}-\frac{(-1)^n}{n^2}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\
&=\zeta(2)\zeta(3)-2\zeta(2)\operatorname{Li}_3(-1)-2\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}\\
&\boxed{=\frac72\zeta(2)\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}}
\end{align}
and the second integral: 
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 $\displaystyle\frac{\ln^2x}{x}$ then integrate from $0$ to $1$, we get
\begin{align}
I_2&=\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln^2x\ dx\\
&=\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(\frac{2}{(2n)^3}\right)\\
&=-4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}+\frac14\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)\\
&=-2\sum_{n=1}^\infty\frac{H_n}{n^4}-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}+\frac14\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)\\
&\boxed{=-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}-\frac74\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)}
\end{align}
Finally 
\begin{align}
I&=I_1+I_2\\
&=\frac72\zeta(2)\zeta(3)-\frac18\zeta(5)-\frac{15}4\sum_{n=1}^\infty\frac{H_n}{n^4}\\
&=\frac72\zeta(2)\zeta(3)-\frac18\zeta(5)-\frac{15}4\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\
&\boxed{=\frac{29}{4}\zeta(2)\zeta(3)-\frac{91}{8}\zeta(5)}
\end{align}
