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 ?

For the fun of it :-)
\begin{align}
4(\cos^4x+2\sin^4x)+2\sin^2 2x&=4(1-\sin^2x)^2+8\sin^4x+2(2\sin x \cos x)^2\\
&=4-8\sin^2x+4\sin^4x+8\sin^4x+8\sin^2 x (1-\sin^2x)\\
&=4+4\sin^4x
\end{align}
Hence 
\begin{align}
I&=\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{4(\cos^4x+2\sin^4x)+2\sin^2 2x}\,dx\\
&=\frac14\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{1+\sin^4x}\,dx\\
&=\frac12\int_{0}^{\pi/2}\frac{\sin^2x}{1+\sin^4x}\,dx\\
&=\frac12\int_{0}^{\pi/2}\sin^2x\sum_{j=0}^{\infty}(-1)^j\sin^{4j}x\,dx\\
&=\frac12\sum_{j=0}^{\infty}(-1)^j\int_{0}^{\pi/2}\sin^{2\times\frac{4j+3}{2}-1}x\cos^{2\times\frac12-1}x\,dx\\
&=\frac14\sum_{j=0}^{\infty}(-1)^j \text{Beta}\left(\frac{4j+3}{2},\frac12\right)\\
&=\frac14\sum_{j=0}^{\infty}(-1)^j \frac{\Gamma(\frac12)\Gamma(2j+\frac32)}{\Gamma(2j+2)}\\
\end{align}
I am not sure how to pin the last down. I will be happy if someone would share thoughts on this :-)
