Evaluating $I=\int_{-\infty}^\infty \frac{u^2}{5u^2\left(u^2+1\right)+2}\,\mathrm{d}u$? This integral popped up when I was trying to solve this integral : $$\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{4(\cos^4x+2\sin^4x)+2\sin^2 2x}\,dx $$
I simplified it a little bit, substituted $\tan(x)=u$ and came up with, 
$$I= \int_{-\infty}^\infty \frac{u^2}{5u^2\left(u^2+1\right)+2}\,\mathrm{d}u$$
any suggestions for how I can take it from here ?

This is more a comment than an answer.
Considering
$$\frac{\sin^2(x)}{4(\cos^4(x)+2\sin^4(x))+2\sin^2 (2x)}$$ the denominator can be simplified as $2\cos^2(2x)-4\cos(2x)+10$ and the numerator as $\frac{1-\cos(2x)} 2$.
So, for the time being, let $X=\cos(2x)$ and the integrand is then $$\frac{1-X}{4 (X-a) (X-b)}=\frac 1{4(a-b)}\left(\frac{1-a}{X-a} -\frac{1-b}{X-b}\right)$$ where $a=1-2i$, $b=1+2i$ are the roots of $2X^2-4X+10=0$.
So, we are basically facing simpler integrals looking like $$I=\int \frac{dx}{\cos(2x)-c}=\frac 1{\sqrt{1-c^2}}{\tanh ^{-1}\left(\sqrt{\frac{1+c}{1-c}}\tan (x)\right)}$$ and the problem becomes easy (assuming that you enjoy complex numbers).
