Is there any other way to solve this difficult integral $ \int \frac{1}{ (a^2 \cos ^2 x + b ^ 2 \sin ^2x) ^2} \ dx $ $$
 \int \frac{1}{ (a^2 \cos ^2 x + b ^ 2  \sin ^2x) ^2} \ dx
$$
So this is the question . 
The solution given in book is to divide numerator and denominator by $\cos ^4x$ and then substitute $\tan x = t$ in the resulting integrand.
Other way of doing this was to substitute $b \tan x = a \tan t$. 
So I was thinking is not there any other way to solve this as it seems to a complicated problem as the given methods are very lengthy while solving.
Any simpler\shorter method anyone could think of ?

Note that
$$I(p,q)=\int \frac{dx}{ a^2\cos ^2 x + b^2 \sin ^2x}
=\int \frac{d(\tan x)}{ a^2+b^2\tan ^2 x }
=\frac1{ab}\tan^{-1}\bigg({\frac ba}\tan x \bigg)
$$
and evaluate
\begin{align}
&\int \frac{1}{ (a^2\cos ^2 x + b^2 \sin ^2x)^2}dx
=  -\frac12\left(\frac{I’_a}a+\frac{I’_b}b\right)\\
=&\ \frac1{2ab}\left(\frac1{a^2}+\frac1{b^2}\right)\tan^{-1}\left(\frac ba\tan x \right)
+ \frac12\left(\frac1{a^2}-\frac1{b^2}\right)\frac{\tan x}{b^2\tan^2x+a^2}
\end{align}
