Find the integral $\int\limits \frac{1}{(\sin x+\cos x)^3}\mathrm dx$ Using tangent half-angle substitution, the integral becomes $$\int\limits \frac{t^4+2t^2+1}{(-t^2+2t+1)^3}\mathrm dt$$ 
This integral can be solved using partial fractions,
$$\frac{t^4+2t^2+1}{(-t+\sqrt 2+1)^3(t+\sqrt 2-1)^3}=\frac{A}{-t+\sqrt 2+1}+\frac{B}{(-t+\sqrt 2+1)^2}+\frac{C}{(-t+\sqrt 2+1)^3}+\frac{D}{t+\sqrt 2-1}+\frac{E}{(t+\sqrt 2-1)^2}+\frac{F}{(t+\sqrt 2-1)^3}$$
What is the quicker way to solve this integral?

HINT:
$$\sin x+\cos x=\sqrt2\cos\left(x-\dfrac\pi4\right)$$
Now use Finding $\int \sec^3x\,dx$
OR
Indefinite integral of secant cubed
