A simpler way to evaluate $\int {dx\over \sin^2x\cos^4x}$ 
Evaluate:
  $$
\int {dx\over \sin^2x\cos^4x}
$$

I've started by using identities:
$$
{1\over \sin^2x} = 1+\cot^2x\\
{1\over \cos^2x} = 1+\tan^2x
$$
So the integral becomes:
$$
\begin{align}
I &= \int (1+\cot^2x)(1+\tan^2x)^2dx \\
&=\int (1+\cot^2x)(1+2\tan^2x + \tan^4x)dx \\
&=\int (1 + 2\tan^2x + \tan^4x+\cot^2x + 2 + 2\tan^2x)dx \\
&=\int (3 + 4\tan^2x+\tan^4x+\cot^2x)dx
\end{align}
$$
I know the integral of $\tan^2x$ and $\cot^2x$, while for $\tan^nx$ one could use a reduction formula. But I find this approach too complicated and seek for a simpler one. Another approach I've tried out is expanding $1$ in the nominator to:
$$
\int {\sin^2x + \cos^2x\over \sin^2x\cos^4x}dx
$$
But that was also clumsy.
Is there another approach simpler than the one suggested above? Could this approach be generalized for integrals of the form:
$$
\int {dx\over \sin^{2k}x\cos^{2p}x}
$$
where $k, p \in\Bbb N$?
Thank you!

$I = \displaystyle\int \dfrac{1}{\sin^2{x}\cos^4{x}} dx = \displaystyle\int \sec^2{x}\dfrac{\left(\tan^2{x}+1\right)^2}{\tan^2{x}} dx$.  Now substitute $y = \tan{x}$. 
$I=  \displaystyle\int \dfrac{(y^2+1)^2}{y^2} dy = \frac{y^3}{3}+2y-\frac{1}{y} +C$
