<s>[INST] Evaluate the closed form of $\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)$ Last question of this form
I am very curious to what is the closed form of:
Assume where $a,b,c > 0$ 

$$\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)\tag1$$

$$b\sin^2x+c\cos^2x$$
$$=b(1-\cos^2 x)+c\cos^2 x$$
$$=b-(b-c)\cos^2 x$$
$$\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b-(b-c)\cos^2x}\mathrm dx\tag2$$
$u=a\tan^2x\implies du=2a\tan x\sec^2 x dx$
${u-a\over a}=\sec^2 x$
$${\sqrt{a}\over 2}\int_{0}^{\infty}{\arctan(u)\over \sqrt{u}(bu+ac-2ab)}\mathrm du\tag3$$
$u=y^2\implies du=2ydy$
$$\sqrt{a}\int_{0}^{\infty}{\arctan(y^2)\over by^2+ac-2ab}\mathrm dy\tag4$$
$b=P$ and $ac-2ab=Q$
Take the form of: $$\int_{0}^{\infty}{\arctan(y^2)\over Py^2+Q}\mathrm dy\tag5$$
Apply $arctan(y)$ series
$$\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\int_{0}^{\infty}{y^{4n+2}\over Py^2+Q}\mathrm dy\tag6$$
 [/INST] We will use Fourier transformation and Parseval's identity.
Say that
\begin{equation*}
\hat{g}(\xi) = \int_{-\infty}^{\infty}e^{-i\xi x}g(x)\, dx .
\end{equation*}
Then the Fourier transform of $g(x) = \arctan\dfrac{2}{x^2}$ will be 
\begin{equation*}
\hat{g}(\xi) = 2\pi\dfrac{\sin \xi}{\xi}e^{-|\xi|}\tag{1}
\end{equation*}
(integration by parts followed by residue calculus).
We are now prepared for the given integral. We start with the substituition $y=\tan x$.
\begin{equation*}
f(a,b,c) = \int_{0}^{\infty}\dfrac{\arctan(ay^2)}{by^2+c}\, dy =\left[y =\dfrac{\sqrt{2}}{z\sqrt{a}}\right] = \dfrac{1}{c\sqrt{2a}} \int_{-\infty}^{\infty}\dfrac{1}{z^2+d^2}\arctan\dfrac{2}{z^2}\, dz \tag {2}
\end{equation*}
where $d=\sqrt{\dfrac{2b}{ac}}$. But Parseval's identity combined with (1) give
\begin{gather*}
\int_{-\infty}^{\infty}\dfrac{1}{z^2+d^2}\arctan\dfrac{2}{z^2}\, dz =[\text{ Parseval ]} = \dfrac{1}{2\pi}\int_{-\infty}^{\infty}\dfrac{\pi}{d}e^{-d|\xi|}\, 2\pi\dfrac{\sin \xi}{\xi}e^{-|\xi|}\, d\xi = \\[2ex] \dfrac{2\pi}{d}\dfrac{1}{2\pi}\int_{-\infty}^{\infty}\pi e^{-(1+d)|\xi|}\, \dfrac{\sin \xi}{\xi}\, d\xi = [\text{ Parseval, H = Heaviside ]} = \\[2ex]\dfrac{2\pi}{d}\int_{-\infty}^{\infty}\dfrac{1+d}{x^2+(1+d)^2}\dfrac{1}{2}({\rm H}(\xi+1)-{\rm H}(\xi-1))\, d\xi = \dfrac{\pi}{d}\int_{-1}^{1}\dfrac{1+d}{x^2+(1+d)^2} = \\[2ex]\dfrac{2\pi}{d}\arctan\dfrac{1}{1+d}.
\end{gather*}
Finally this and (2) give
\begin{equation*}
f(a,b,c) = \dfrac{\pi}{\sqrt{bc}}\arctan\left(\dfrac{\sqrt{ac}}{\sqrt{ac}+\sqrt{2b}}\right).
\end{equation*}
</s>