Find $\int\frac{dx}{\cos^3x-\sin^3x}$ $\int\frac{dx}{\cos^3x-\sin^3x}$

Let $I=\int\frac{dx}{\cos^3x-\sin^3x}=\int\frac{dx}{(\cos x-\sin x)(\cos^2 x+\sin^2 x+\sin x\cos x)}$
But it does not seem to be solved further by this method,so i tried another method.
$I=\int\frac{dx}{\cos^3x-\sin^3x}=\int\frac{\csc^3 xdx}{\cot^3x-1}=\int\frac{\csc^2 x \csc xdx}{\cot^3x-1}=\int\frac{\csc^2 x \sqrt{1+\cot^2x}dx}{\cot^3x-1}$
Put $\cot x=t\implies -\csc^2 x dx=dt$
$I=\int\frac{-\sqrt{1+t^2}dt}{t^3-1}=\int\frac{-\sqrt{1+t^2}dt}{(t-1)(t^2+t+1)}$
But i am stuck here.

$$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x\cos x)=\dfrac{(\cos x-\sin x)\{3-(\cos x-\sin x)^2\}}2$$
Writing $\cos x-\sin x=t$ and using Partial Fraction,
$$\dfrac1{t(3-t^2)}=\dfrac1{3t}+\dfrac t{3(3-t^2)}$$
$$\implies\dfrac3{\cos^3x-\sin^3x}=\dfrac1{\cos x-\sin x}+\dfrac{\cos x-\sin x}{3-(\cos x-\sin x)^2}$$
The first integral can be managed easily
For the second $$\text{as }\int(\cos x-\sin x)dx=\sin x+\cos x$$ and as $$(\cos x-\sin x)^2+(\sin x+\cos x)^2=2$$
write $$3-(\cos x-\sin x)^2=1+(\sin x+\cos x)^2$$ and replace $\sin x+\cos x$ with $u$
