The $\int \frac{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}{{\left({\sin }^5\mathrm{x}+{\cos }^3{\mathrm{xsin}}^2\mathrm{x}+{\sin }^3{\mathrm{xcos}}^2\mathrm{x}+{\cos }^5\mathrm{x}\right)}^2}\mathrm{dx}=$

$\begin{array}{l}\int \frac{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}{{\left({\sin }^3\mathrm{x}+{\cos }^3\mathrm{x}\right)}^2}\mathrm{dx}\\ =\int \frac{{\sin }^2{\mathrm{xcos}}^2\mathrm{x}}{{\cos }^6\mathrm{x}{\left({\tan }^3\mathrm{x}+1\right)}^2}\mathrm{dx}\\ =\int \frac{{\sec }^2{\mathrm{xtan}}^2\mathrm{x}}{{\left(1+{\tan }^3\mathrm{x}\right)}^2}\mathrm{dx}\end{array}$
Put 1+ tan³ x = t
$\begin{array}{l}3{\tan }^2{\mathrm{xsec}}^2\mathrm{xdx}=\mathrm{dt}\\ =\frac{1}{3}\int \frac{1}{{\mathrm{t}}^2}\mathrm{dt}\\ =\frac{1}{3}\left(\frac{-1}{\mathrm{t}}\right)+\mathrm{c}\\ =-\frac{1}{3\left(1+{\tan }^3\mathrm{x}\right)}+\mathrm{c}\end{array}$