How to Evaluate $\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$? How to find $$\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$$
In fact,
\begin{align*}
&\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du}  \\
=& \int_0^1 {\frac{{u\arcsin u}}{{\left( {{u^2} - \sqrt {2\sqrt {13}  - 2} u + \sqrt {13} } \right)\left( {{u^2} + \sqrt {2\sqrt {13}  - 2} u + \sqrt {13} } \right)}}du} \\
 = &\frac{1}{{2\sqrt {2\sqrt {13}  - 2} }}\left[ {\int_0^1 {\frac{{\arcsin u}}{{{u^2} - \sqrt {2\sqrt {13}  - 2} u + \sqrt {13} }}du}  - \int_0^1 {\frac{{\arcsin u}}{{{u^2} + \sqrt {2\sqrt {13}  - 2} u + \sqrt {13} }}du} } \right].
\end{align*}
But how can we continue?

This is probably not an answer
Being too lazy to attack such monsters, I should use Taylor series $$\sin^{-1}(u)=\sum_ 0^\infty\frac{(2n)!}{4^n(n!)^2(2n+1)}u^{2n+1}\qquad (|u|\leq 1)$$ and perform the long division by the denominator. This would lead to $$\frac{{u\sin^{-1} u}}{{{u^4} + 2{u^2} + 13}}=\frac{u^2}{13}+\frac{u^4}{1014}-\frac{79 u^6}{263640}+O\left(u^7\right)$$ If you are brave enough, take more terms !
