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? $Li_{5}\left(\frac{1}{2}\right)$

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A very nice question (+1) –  Chris's sis Dec 28 '12 at 14:01

$$\begin{eqnarray*} \int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx} &=& \left. \left(\frac{\partial^2}{\partial s^2}\frac{\partial^2}{\partial t^2} \int_0^1 dx\, (1-x)^s(1+x)^{t-1} \right) \right|_{s=t=0} \\ &=& \left. \left(\frac{\partial^2}{\partial s^2}\frac{\partial^2}{\partial t^2} \,\frac{{}_2F_1(1-t,1;s+2;-1)}{s+1} \right) \right|_{s=t=0} \end{eqnarray*}$$ 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.