How to evaluate $\int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx}$ I want to evaluate $$\int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx}$$
I run this integral on Maple, It does converge. How we get a closed form?
Is that related to polylogs? $\operatorname{Li}_{5}\left(\frac{1}{2}\right)$
 A: $$\begin{eqnarray*}
\int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx} 
    &=& \left. \left(\frac{\partial^2}{\partial s^2}\frac{\partial^2}{\partial t^2}
\int_0^1 dx\, (1-x)^s(1+x)^{t-1} \right) \right|_{s=t=0} \\
    &=& \left. \left(\frac{\partial^2}{\partial s^2}\frac{\partial^2}{\partial t^2}
\,\frac{{}_2F_1(1-t,1;s+2;-1)}{s+1} \right) \right|_{s=t=0}
\end{eqnarray*}$$
Here we've used Euler's integral representation for the hypergeometric function. 
Addendum: 
Using the series representation for the hypergeometric function we can take the derivatives. After a little work we find the integral can be written as
$$\sum_{k=2}^\infty  \frac{(-1)^k}{k+1} 
\left(H_k^2 - H_k^{(2)}\right)
\left(H_{k+1}^2 + H_{k+1}^{(2)}\right),$$
where $H_k$ and $H_k^{(n)}$ are the harmonic and generalized harmonic numbers, respectively. 
This sum has bad convergence behavior, the terms go like $(-1)^k(\log k)^4/k$ $(k\to\infty)$.
Since the sum is alternating we could accelerate it using the Euler transform, for example. 
A: The result is
-4〖Li〗_5 (1/2)+4ζ(3) 〖Log〗^2 2-(2π^2)/9 〖Log〗^3 2- π^2/3 ζ(3)-π^4/20 Log2+ 7/30 〖Log〗^5 2+ 63/8 ζ(5)
 =0,418709830998418751408037243448…
