$\int \frac{\left({\sin }^{2}x{\cos }^{2}x\right)dx}{{\left({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x\right)}^2}$

$\begin{array}{l}I=\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{\left({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x\right)}^2}\\ =\int \frac{{\sin }^{2}x\cdot {\cos }^{2}xdx}{{\left[\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^3+{\cos }^{3}x\right)\right]}^2}\\ =\int \frac{{\sin }^{2}x\cdot {\cos }^{2}xdx}{{\left({\sin }^{3}x+{\cos }^{3}x\right)}^2}=\int \frac{{\tan }^{2}x\cdot {\sec }^{2}x}{{\left(1+{\tan }^{3}x\right)}^2}dx\\ \text{ Put }1+{\tan }^{3}xdx=t\Rightarrow 3{\tan }^{2}x{\sec }^{2}xdx=dt\\ I=\frac{1}{3}\int \frac{3{\tan }^{2}x{\sec }^{2}xdx}{{\left(1+{\tan }^{3}x\right)}^2}=\frac{1}{3}\int \frac{dt}{t^2}=-\frac{1}{3t}+c\\ \Rightarrow I=-\frac{1}{3\left(1+{\tan }^{3}x\right)}+c\end{array}$