How to find the value of this integral? This integral to the value 

\begin{align}
\int_0^1\frac{\ln^2(1+x)\ln^2 x}{1-x}\ dx=&\ \color{blue}{-\frac{13\pi^2}{24}\zeta(3)+\frac{47}{2}\zeta(5)-\frac15\ln^52+\frac{\pi^2}9\ln^32-\frac{49\pi^4}{360}\ln2+\frac{7}2\zeta(3)\ln^22}\\&\color{blue}{-8\operatorname{Li}_4\left(\frac12\right)\ln2-16\operatorname{Li}_5\left(\frac12\right)},
\end{align}
   How to find this result?
   In fact,I find that $$\int_0^1\frac{\ln(1+x)\ln{x}}{1-x}dx=\zeta(3)-\frac{\pi^2}{4}\ln2$$
   $$\int_0^1\frac{\ln^2(1+x)\ln{x}}{1-x}dx=\frac{21}{4}\zeta(3)\ln{2}-\frac{5\pi^2}{12}\ln^2{2}+\frac{1}{6}\ln^4{2}-\frac{7\pi^4}{144}+4\operatorname{Li}_4\left(\frac12\right)$$

 A: First, this problem can be done in many creative ways. In the following, I'll present a solution using more advanced machinery, but still very nice.
Combining integration by parts and Dilogarithm reflection formula, we immediately arrive at 
$$\mathcal{I}=2\underbrace{\int_0^1 \frac{\log (1-x)\log ^2(x)\log (1+x)}{x}\textrm{d}x}_{\displaystyle I_1}-2\underbrace{\int_0^1 \frac{\log (1-x)\log ^2(x)\log (1+x) }{x(1+x)}\textrm{d}x}_{\displaystyle I_2}$$
$$+ 2\underbrace{\int_0^1\frac{\log (1-x) \log (x)\log ^2(1+x)}{x}\textrm{d}x}_{\displaystyle I_3}-\zeta(2)\underbrace{\int_0^1\frac{\log ^2(1+x)}{x}\textrm{d}x}_{\displaystyle \text{Trivial}}$$
$$-2\zeta(2)\underbrace{\int_0^1\frac{\log (1+x) \log (x)}{1+x} \textrm{d}x}_{\displaystyle \text{Trivial}}.$$
Now, the first integral, in a generalized form, is calculated in the book, (Almost) Impossible Integrals, Sums, and Series (see page $6$) 

$$I_1=\int_0^1 \frac{\log(1-x)\log^{2}(x)\log(1+x)}{x} \textrm{d}x=\frac{3}{4} \zeta (2) \zeta (3)-\frac{27}{16}\zeta (5),$$

the integral $I_3$ is already calculated in this post A Challenging Logarithmic Integral $\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx$, and finally, with respect to $I_2$, we have 

$$I_2=\underbrace{\int_0^1 \frac{\log (1-x)\log ^2(x)\log (1+x) }{x(1+x)}\textrm{d}x}_{\text{Note that} \sum _{n=1}^{\infty } (-1)^{n-1} H_n x^n=\log(1+x)/(1+x)  \ \text{and} \int_0^1 x^{n-1} \log(1-x)\textrm{d}x=-H_n/n} $$
$$=2\sum_{n=1}^{\infty}(-1)^n\frac{  H_n H_n^{(2)}}{n^2}+2\sum_{n=1}^{\infty}(-1)^n\frac{ H_n H_n^{(3)}}{n}+2\sum_{n=1}^{\infty}(-1)^n\frac{H_n^2}{n^3}$$
$$-2\zeta (3) \sum_{n=1}^{\infty}(-1)^n \frac{ H_n}{n}-2\zeta(2) \sum_{n=1}^{\infty}(-1)^n\frac{H_n}{n^2}$$
$$=\frac{1}{10}\log ^5(2)+\frac{13 }{4}\zeta (2) \zeta (3)-\frac{1}{3}  \log ^3(2)\zeta (2)-\frac{7}{4}\log ^2(2) \zeta (3) +\frac{49}{8} \log (2) \zeta (4)-15 \zeta (5)$$
$$+4 \operatorname{Li}_4\left(\frac{1}{2}\right) \log (2)+8 \operatorname{Li}_5\left(\frac{1}{2}\right),$$

which is easily calculated if we use the values of the first three series which may be found in the mention book, pages $311-312$ and $529$. I skipped the last two series which are trivial.
Alternatively, we may cleverly use algebraic identities and Cornel's Master Theorem of Series presented in the paper http://jca.ele-math.com/10-10/A-master-theorem-of-series-and-an-evaluation-of-a-cubic-harmonic-series, or in the mentioned book to get simpler ways that circumvent the necessity of calculating advanced alternating harmonic series of weight $5$.
This solution has been suggested by Cornel I. Valean.
