Factor the denominator: $\sin^5x+\cos^5x+\sin^3x\cos^2x+\cos^3x\sin^2x$

$=(\sin^3x+\cos^3x)(\sin^2x+\cos^2x)=\sin^3x+\cos^3x$

So integral $=\displaystyle\int\dfrac{\sin^2x\cos^2x}{(\sin^3x+\cos^3x)^2}dx$. Divide by $\cos^6x$:

$=\displaystyle\int\dfrac{\tan^2x\sec^2x}{(\tan^3x+1)^2}dx$. Let $t=\tan^3x+1$, $dt=3\tan^2x\sec^2x\,dx$:

$=\dfrac{1}{3}\displaystyle\int\dfrac{dt}{t^2}=\dfrac{-1}{3t}+C=\dfrac{-1}{3(1+\tan^3x)}+C$