\(I=4 \int \frac{\cos 2 x \cdot(2 \sin 2 x \cos 2 x)}{(1+\cos 2 x)^2\left(1+\cos ^2 2 x\right)} d x\)
Put \(\cos 2 x=t\)
\(\Rightarrow -2 \sin 2 x d x=d t\)
\(\therefore \quad I=-4 \int \frac{t^2}{(1+t)^2\left(1+t^2\right)} d t\)
Split into partial fractions
\(\begin{aligned}
I & =-4 \int\left[\frac{1}{2(1+t)^2}-\frac{1}{2(1+t)}+\frac{t}{2\left(1+t^2\right)}\right] d t \\
& =-2\left[-\frac{1}{1+t}-\log (1+t)+\frac{1}{2} \log \left(1+t^2\right)\right] \\
& =\frac{2}{2 \cos ^2 x}+2 \log (1+\cos 2 x)-\log \left(1+\cos ^2 2 x\right)+c \\
& =\sec ^2 x+2 \log (1+\cos 2 x)-\log \left(1+\cos ^2 2 x\right)+c \\
& =\sec ^2 x+\log \frac{(1+\cos 2 x)^2}{\left(1+\cos ^2 2 x\right)}+c
\end{aligned}\)