Integral $\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))dx$ 
Evaluate
  $$\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))\mathrm dx$$

I tried the following ideas:
$u=a+b\tan(x)$, $du=b\sec^2(x)dx$
$$\frac{1}{b}\int \cos^4(x)\cos(u)du$$
using $1+\tan^2(x)=\sec^2(x)$
$$b^3\int \frac{\cos(u)du}{(u-a)^4}$$
I am stuck at this point. Any help?

$$I=\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))\mathrm dx\overset{x\to -x}=\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a-b\tan(x))\mathrm dx$$
Summing up the two integrals from above gives us:
$$ 2I= 2\cos a \int_{-\pi/2}^{\pi/2} \cos^2 x \cos(b\tan x) dx\Rightarrow I=2\cos a\int_0^{\pi/2}\cos^2 x\cos(b\tan x) dx$$
$$\overset{\tan x=t}=2\cos a\int_0^\infty \frac{\cos(b t)}{(1+t^2)^2}dt\overset{bt=x}=2b^3 \cos a\int_0^\infty \frac{\cos x}{(b^2+x^2)^2}dx$$
Consider the following integral, found here: $$I(b)=\int_0^\infty \frac{\cos x}{b^2+x^2}dx=\frac{\pi}{2b e^{b}}\Rightarrow I'(b)=-2b\int_0^\infty \frac{\cos x}{(b^2+x^2)^2}dx$$
$$\Rightarrow I= -2b^3\cos a\cdot \frac{1}{2b}I'(b)= \frac{\pi}{2}\cos a\frac{b+1}{ e^b}$$
