Tedious Trig Integral: $\int\limits_{0}^{\pi/2}{\frac{\theta (1+\sin^2\theta)\cos\theta}{(1+3\sin^2\theta)(3+\sin^2\theta)}d\theta}$ 
$$\int\limits_{0}^{\pi/2}{\frac{\theta \left( 1+\sin^{2}\theta \right)\cos \theta }{\left( 1+3\sin^{2}\theta \right)\left( 3+\sin^{2}\theta \right)}d\theta }$$

Attempt: I tried making the substitution $u = \sin(\theta)$ so that the integral could become easier to split. 
I then ended up with http://www.wolframalpha.com/input/?i=integrate+arcsin(x)(1%2Bx%5E2)%2F((1%2B3x%5E2)(3%2Bx%5E2))+from+0+to+1 and tried to use a from of partial fractions (using a system of equations) to arrive at the answer. 
I  somehow got $A = -\pi/16$ and $B = 3\pi/16$, from the system $3A + B = 0$ and $A + B = \pi/8$. But my answer was off by a bit, which is why I'm curious what to do next. Any help would be appreciated!
 A: It is not complete answer, but hopefully will help to approach this task.
First of all, let us generalize integral:
$$
I(b)=\int\limits_{0}^{\pi/2}{\frac{\arccos\left(b\cos(\theta)\right) \left( 1+\sin^{2}\theta \right)\cos \theta }{\left( 1+3\sin^{2}\theta \right)\left( 3+\sin^{2}\theta \right)}d\theta },\quad b\in[0,1]
$$
For $b=1$ it appears to be initial one. 
Next, derivate with respect to $b$. (Change order of integration/derivation).
$$
\frac{\partial I }{\partial b}=-\int\limits_{0}^{\pi/2}{\frac{ \left( 1+\sin^{2}\theta \right)\cos^{2} \theta }{\sqrt{1-b^2\cos^{2}(\theta)}\left( 1+3\sin^{2}\theta \right)\left( 3+\sin^{2}\theta \right)}d\theta }
$$
Notice that argument is an even function, thus
$$
\int\limits_{0}^{\pi/2}=\int\limits_{-\pi/2}^{0}\to\frac{\partial I }{\partial b}=-\frac{1}{2}\int\limits_{-\pi/2}^{\pi/2}
$$ 
Make shift $\theta\to\theta+\pi/2$. (Change $\sin^2(\theta)\to\cos^2(\theta),$ $\cos^2(\theta)\to\sin^2(\theta)$)
$$
\frac{\partial I }{\partial b}=-\frac{1}{2}\int\limits_{0}^{\pi}{\frac{ \left( 1+\cos^{2}\theta \right)\sin^{2} \theta }{\sqrt{1-b^2\sin^{2}(\theta)}\left( 1+3\cos^{2}\theta \right)\left( 3+\cos^{2}\theta \right)}d\theta }
$$
Once again, notice that shift $\theta\to\theta+\pi$ don't change the integral, thus we can write
$$
\frac{\partial I }{\partial b}=-\frac{1}{4}\int\limits_{0}^{2\pi}{\frac{ \left( 1+\cos^{2}\theta \right)\sin^{2} \theta }{\sqrt{1-b^2\sin^{2}(\theta)}\left( 1+3\cos^{2}\theta \right)\left( 3+\cos^{2}\theta \right)}d\theta }
$$
At this point we can use theory of residuals. Corresponding contour integral after some simplifications
$$
\frac{\partial I }{\partial b}=\frac{i}{6b}\int\limits_{|z|=1}\frac{(z^4+6z^2+1)(z^2-1)^2}{\sqrt{-z^4-1+z^2\left(2+\frac{4}{b^2}\right)}(z^4+\frac{10}{3}z^2+1)(z^4+14z^2+1)}dz
$$
Next step should be calculating of residuals, but I stubmle here, having $0$ everywhere (probably, miscalculations). 
Anyway, idea of further solution is to integrate obtained expression over $b$, calculate constant of integration from $I(0)=\int\limits_{0}^{\pi/2}{\frac{\left( 1+\sin^{2}\theta \right)\cos \theta }{\left( 1+3\sin^{2}\theta \right)\left( 3+\sin^{2}\theta \right)}d\theta }$, which is super easy to compute, and then just take value of $I(1)$.
