Interesting Logarithmic Integral: $\int_{0}^{1} \frac{\ln^2 x \ln^2(1+x)}{x} \;dx $ Other than numerical approximation, how can we calculate the closed form of this integral?
$$\int_{0}^{1} \frac{\ln^2 x \ln^2(1+x)}{x} \;dx $$
 A: OK as I suggested in a comment here is an approach.Let us begin with the Taylor series of $\ln^2(1+x)$. It is known to be:
$$\ln^2 (1+x) = 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^n$$
Then we have successively:
\begin{align*}
\int_{0}^{1} \frac{\ln^2 x \ln^2 (1+x)}{x} \, {\rm d}x &= 2 \int_{0}^{1} \frac{\ln^2 x}{x} \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^n \, {\rm d}x \\ 
 &=2 \int_{0}^{1} \ln^2 x \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^{n-1} \, {\rm d}x\\ 
 &= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} \int_{0}^{1} x^{n-1} \ln^2 x \, {\rm d}x\\ 
 &= 2 \sum_{k=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n^4}\\ 
 &=2 \sum_{k=2}^{\infty} \frac{(-1)^n \left [ \mathcal{H}_n - \frac{1}{n} \right ]}{n^4} \\ 
 &= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} - 2 \sum_{n=2}^{\infty} \frac{(-1)^n}{n^5} \\
 &= 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} - 2 - \frac{15 \zeta(5)}{16}
\end{align*}
Well for the Euler sum first of all we have the generating function:
\begin{align} 
\sum^\infty_{n=1}\frac{H_n}{n^3}z^n 
=&2{\rm Li}_4(z)+{\rm Li}_4\left(\tfrac{z}{z-1}\right)-{\rm Li}_4(1-z)-{\rm Li}_3(z)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\tfrac{z}{z-1}\right)\\ 
&+\frac{1}{2}{\rm Li}_2(z)\ln^2(1-z)+\frac{1}{2}{\rm Li}_2^2(z)+\frac{1}{6}\ln^4(1-z)-\frac{1}{6}\ln{z}\ln^3(1-z)\\ 
&+\frac{\pi^2}{12}\ln^2(1-z)+\zeta(3)\ln(1-z)+\frac{\pi^4}{90} 
\end{align}
Integrating once and plugging $z=-1$ we get the value of the sum. I am not presenting the full calculatios but it suffices to say that:
$$\sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^4} = \frac{\pi^2 \zeta(3)}{4} -\frac{43 \zeta(5)}{32}+1$$
Thus:
$$\int_{0}^{1} \frac{\ln^2 x \ln^2 (1+x)}{x} \, {\rm d}x = \frac{\pi^2 \zeta(3)}{3} - \frac{29 \zeta(5)}{8}$$
