Antiderivative of a trigonometric integral $$\int\frac{\sin^3(x)}{(\cos^4(x)+3\cos^2(x)+1)\cdot\arctan(\sec(x)+\cos(x))}$$
I am unable to manipulate this integral. The actual integral in question is definite with limits from $0$ to $\frac{\pi}{2}$, but I feel it can't be calculated without knowing the antiderivative.
My initial try was to take $\sec(x)+\cos(x) $ as $t$, but It didn't solve.
Please give some suggestions!

Instead try $$\begin{align}&t=\tan^{-1}(\sec x +\cos x)\\& dt=\frac{\sec x \tan x -\sin x}{(\sec x +\cos x)^2+1}dx\\&=\frac{\sin x(\sec^2 x-1)}{\cos^2 x+3+\sec^2x}dx=\frac{\sin^3x}{\cos^4x+3\cos^2x+1}dx\end{align}$$
Then the integral simply becomes $$\int \frac1tdt=\log t=\log\arctan(\sec x+\cos x)$$
