$\int \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x=$

$\log _e\left|\tan ^{-1}(\sec x+\cos x)\right|+C$

Let

$I =\int \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x$

$\Rightarrow I =\int \frac{\frac{\sin ^3 x}{\cos ^2 x}}{\left(\cos ^2 x+3+\sec ^2 x\right) \tan ^{-1}(\sec x+\cos x)} d x$

$\Rightarrow I=\int \frac{1}{1+(\sec x+\cos x)^2} \times \frac{\sin x\left(1-\cos ^2 x\right)}{\cos ^2 x} \times \frac{1}{\tan ^{-1}(\sec x+\cos x)} d x$

$\Rightarrow I=\int \frac{1}{\tan ^{-1}(\sec x+\cos x)} \times \frac{1}{1+(\sec x+\cos x)^2} \times(\tan x \sec x-\sin x) d x $

$\Rightarrow I=\int \frac{1}{\tan ^{-1}(\sec x+\cos x)} d\left\{\tan ^{-1}(\sec x+\cos x)\right\}$

$\Rightarrow I=\log \left|\tan ^{-1}(\sec x+\cos x)\right|+C$