How to find $\int \frac {\sec x}{1+ \csc x} dx $? 
How to find $\int \frac {\sec x}{1+ \csc x} dx  $ ?

Well it reduces to $ \int \frac {\sin x}{\cos x (1+\sin x)} dx $ .Any hints next ?
I'm looking for a short and simple method without partial fractions if possibe.

Hint:
$$ \int \frac {\sin x}{\cos x (1+\sin x)} dx =\int\frac{\sin x\cos x}{\cos^2x(1+\sin x)}dx=\int\frac{\sin x\cos x}{(1-\sin^2 x)(1+\sin x)}dx\int\frac{\sin x\cos x}{(1+\sin x)^2(1-\sin x)}dx$$
Take $u=\sin x$ and use partial fractions.
