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?

As $\int(\sin x+\cos x)dx=\sin x-\cos x$ and $(\sin x-\cos x)^2+(\sin x+\cos x)^2=2$
let $\sin x-\cos x=u\implies(\sin x+\cos x)^2=2-u^2$
$$\int\dfrac{dx}{(\sin x+\cos x)^3}=\int\dfrac{(\sin x+\cos x)dx}{(\sin x+\cos x)^4}=\int\dfrac{du}{(2-u^2)^2}$$
$$=\int\dfrac1u\cdot\dfrac u{(2-u^2)^2}du$$
$$=\dfrac1u\int\dfrac u{(2-u^2)^2}du-\int\left(\dfrac{d(1/u)}{du}\int\dfrac u{(2-u^2)^2}du\right)du$$
$$=\dfrac1{2u(2-u^2)}-\dfrac14\int\dfrac2{u^2(2-u^2)}du$$
Now, $$\int\dfrac2{u^2(2-u^2)}du=\int\dfrac{u^2+(2-u^2)}{u^2(2-u^2)}du=\int\dfrac{du}{2-u^2}+\int\dfrac{du}{u^2}$$
