We have,
$$
\begin{aligned}
& =\int \frac{\cos ^3 x+\cos ^5 x}{\sin ^2 x+\sin ^4 x} d x \\
& =\int \frac{\cos ^2 x \cdot \cos x\left(1+\cos ^2 x\right)}{\sin ^2 x\left(1+\sin ^2 x\right)} d x \\
& =\int \frac{\left(1-\sin ^2 x\right)\left(2-\sin ^2 x\right) \cos x}{\sin ^2 x\left(1+\sin ^2 x\right)} d x
\end{aligned}
$$
Substitute $\sin x=t$
$$
\begin{aligned}
& \Rightarrow \quad \cos x d x=d t \\
& =\int \frac{\left(1-t^2\right)\left(2-t^2\right)}{t^2\left(1+t^2\right)} d t=\int \frac{2-3 t^2+t^4}{t^2\left(1+t^2\right)} d t \\
& =\int \frac{t^2\left(1+t^2\right)-4\left(t^2+1\right)+6}{t^2\left(1+t^2\right)} d t \\
& =\int\left(1-\frac{4}{t^2}+\frac{6}{t^2\left(t+t^2\right)}\right) d t \\
& =\left\lceil 1-\frac{4}{t^2}+6\left(\frac{1}{t^2}-\frac{1}{1+t^2}\right)\right] d x
\end{aligned}
$$
$\begin{aligned} & =\int\left(1+\frac{2}{t^2}-\frac{6}{1+t^2}\right) d t \\ & =t-2 t^{-1}-6 \tan ^{-1} t+c \\ & =\sin x-2(\sin x)^{-1}-6 \tan ^{-1}(\sin x)+c\end{aligned}$