So I'm trying to do an integral $$\int \frac{dx}{(\sin x +\cos x)^3}$$
This of course can be written as :
$$\int \frac{dx}{(\sin x+\cos x)(1+2\sin x\cos x)}$$
If now I use the tangent half-angle substitution, we get:
$$\int \frac{2(t^4+2t^2+1)\, dt}{(-t^2+2t+1)(t^4-4t^3+2t^2+4t+1)}=\int \frac{2(t^4+2t^2+1) \,dt}{(-t^2+2t+1)(t^2-2t-1)^2}$$
Which can be done by using partial fraction, but that seems to be a lengthy process. I also tried deviding each term in denominator with $cosx$ so that i would get $\frac{1}{\cos^2 x}$ (derivative of $\tan x$) which got me in an even worse position than the partial fractions solution. I have a feeling there might be a shortcut for solving this, which is based on some trigonometric identity, but I wasn't able to find it.
Hint: Note that $\sin(x) + \cos(x) = \sqrt{2} \sin(x + \pi/4)$. Make an appropriate substitution to end up with a multiple of $$ \int \csc^3(u)\,du $$