How to evaluate $\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}dx$ in an elegant way? How to prove, in an elegant way that

$$I=\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}dx=\frac{11}{4}\zeta(4)-\frac14\ln^42-6\operatorname{Li}_4\left(\frac12\right)\ ?$$


First, let me show you how I did it
\begin{align}
I&=\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}\ dx\overset{1-x\ \mapsto x}{=}\int_0^1\frac{\ln(1-x)\ln^2x}{2-x}\ dx\\
&=\sum_{n=1}^\infty\frac1{2^n}\int_0^1x^{n-1}\ln^2x\ln(1-x)\ dx\\
&=\sum_{n=1}^\infty\frac1{2^n}\frac{\partial^2}{\partial n^2}\int_0^1x^{n-1}\ln(1-x)\ dx\\
&=\sum_{n=1}^\infty\frac1{2^n}\frac{\partial^2}{\partial n^2}\left(-\frac{H_n}{n}\right)\\
&=\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2\zeta(2)}{n^2}+\frac{2\zeta(3)}{n}-\frac{2H_n}{n^32^n}-\frac{2H_n^{(2)}}{n^22^n}-\frac{2H_n^{(3)}}{n2^n}\right)\\
&=2\zeta(2)\operatorname{Li}_2\left(\frac12\right)+2\ln2\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}
\end{align}
By substituting 
$$S_1=\sum_{n=1}^\infty \frac{H_n}{n^32^n}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42$$
$$
S_2=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42$$
$$S_3=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$
along with $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$ we get the closed form on $I$. Note that $S_1$, $S_2$ and $S_3$ can be found here, here and here respectively.

Now we can see how boring and tedious our calculations are as we used results of three harmonic series with powers of 2 in the denominator. A friend ( who proposed this problem ) suggested that the integral can be done without using harmonic series, so any idea how to do it that way?
Thanks
 A: I'll just show an idea that avoids those type of sum, but skip the calculations. You might also have better ideas to solve them.
For start we will denote $a=\ln(1-x)$ and $b=\ln(1+x)$ and use the following identity:
$$a^2=\frac12 (a+b)^2+\frac12(a-b)^2-b^2$$
$$\Rightarrow I=\frac12 \underbrace{\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1+x}dx}_{I_1}+\frac12\underbrace{ \int_0^1 \frac{\ln x\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx}_{I_2}-\underbrace{\int_0^1 \frac{\ln x\ln^2(1+x)}{1+x}dx}_{I_3}$$

For the first integral we will write the denominator as:
$$\frac{1}{1+x}=\frac{1}{1-x^2}-\frac{x}{1-x^2}$$
$$\Rightarrow I_1=\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1-x^2}dx-{\int_0^1 \frac{x\ln x\ln^2(1-x^2)}{1-x^2}dx}$$
$$\overset{x^2\to x}=\frac14 \int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}\frac{dx}{\sqrt x}-\frac14\int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}dx$$
Those two integral can be found using the second identity from here.

Let's also take $I_2$ and substitute $\frac{1-x}{1+x}\to x$.
$$\Rightarrow I_2=\underbrace{\int_0^1 \frac{\ln(1-x)\ln^2 x}{1+x}dx}_{P}-\underbrace{\int_0^1 \frac{\ln(1+x)\ln^2 x}{1+x}dx}_{Q}$$
$$P-Q=I_2;\quad P+Q=\int_0^1 \frac{\ln(1-x^2)\ln^2 x}{1+x}dx$$
And again with the same trick done for $I_1$, we have:
$$P+Q=\int_0^1 \frac{\ln(1-x^2)\ln^2 x}{1-x^2}dx-\int_0^1 \frac{x\ln(1-x^2)\ln^2 x}{1-x^2}dx$$
$$\overset{x^2\to x}=\frac18\int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}\frac{dx}{\sqrt x}-\frac18 \int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}dx$$
Henceforth we can extract our second integral, $I_2$ as:
$$I_2=P-Q=(P+Q)-2Q$$
Note that $P+Q$ can again be found using the second identity from here.
Finally, we only need to find $Q$.
$$Q=\int_0^1 \frac{\ln(1+x)\ln^2 x}{1+x}dx=\sum_{n=1}^\infty (-1)^{n+1} H_n\int_0^1 x^{n}\ln^2 x=2\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{(n+1)^3}$$
So $Q$ is actually an Euler sum in disguise, but you nicely found it here.

Also, $I_3$ is pretty easy, one just needs to use the same approach as done for $I_1$ in your following post.
$$I_3=\int_0^1 \frac{\ln x \ln^2(1+x)}{1+x}dx\overset{IBP}=-\frac12\int_0^1 \frac{\ln^3(1+x)}{x}dx$$
A: A fancy way of calculating the integral

Here is a fancy way proposed by Cornel (it's pretty amazing for the mathematical connections involved). Let's briefly start with recalling and using Dilogarithm reflection formula,
  $$\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x),$$
  where if we multiply both sides by $\displaystyle\frac{\log(1-x)}{1+x}$ and then consider to integrate from $x=0$ to $x=1$, we may express our integral as follows
  $$\int_0^1\frac{\log(x)\log^2(1-x)}{1+x}\textrm{d}x$$
$$=\zeta(2)\underbrace{\int_0^1\frac{\log(1-x)}{1+x}\textrm{d}x}_{\displaystyle \text{Trivial}}-\underbrace{\int_0^1\frac{\log(1-x)\operatorname{Li}_2(x)}{1+x}\textrm{d}x}_{\displaystyle I}-\underbrace{\int_0^1\frac{\log(1-x)\operatorname{Li}_2(1-x)}{1+x}\textrm{d}x}_{\displaystyle J}.$$
Let the party begin ...
By Landen's Identity, the integral $I$ may be connected to the integral 
  $$\int_0^1 \frac{\displaystyle \log(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x=\frac{29}{16} \zeta (4)+\frac{1}{4}\log ^2(2) \zeta (2) -\frac{1}{8} \log ^4(2),$$
  which appears in (Almost) Impossible Integrals, Sums, and Serie, page $17$, with a nice solution, and therefore we immediately obtain the desired value of $I$. A different solution than the one presented in the book may be found here.
The last integral (the integral $J$) is also a very pleasant and unexpected game! The magic will happen by letting the variable change $x\mapsto 1-x$, and then connecting the form of the integral to the generalization 
  $$ \int_0^1 \frac{\log (x)\operatorname{Li}_2(x) }{1-a x} \textrm{d}x=\frac{(\operatorname{Li}_2(a))^2}{2 a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a},$$
  which is given in the article A simple idea to calculate a class of polylogarithmic integrals by using the Cauchy product of squared Polylogarithm function by C. I. Valean, and the solution is straightforward if we expand the integral in series and then use the Cauchy product of $(\operatorname{Li_2}(x))^2$.
  In other words, we have that 
  $$J=\int_0^1\frac{\log(1-x)\operatorname{Li}_2(1-x)}{1+x}\textrm{d}x=\frac{1}{2}\int_0^1\frac{\log(x)\operatorname{Li}_2(x)}{1-x/2}\textrm{d}x$$
$$=\frac{1}{2}\left(\frac{(\operatorname{Li}_2(a))^2}{2 a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a}\right) \biggr|_{a=1/2}.$$
End of party (story)

An important note: the necessity of calculating advanced alternating harmonic series or advanced harmonic series with powers of $2$ in the denominator is completely removed by the actual procedure. In fact, by carefully checking the development of the solution to the integral $\int_0^1 \frac{\log(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x$ in the book (Almost) Impossible Integrals, Sums, and Series, one may observe that reaching the point with harmonic series may be completely avoided if necessary and the calculation can be accomplished with the use of integrals only (to be clearer, I'm talking about the famous Au-Yeung series).
