I am trying to evaluate
$$ \int \frac{\tan^{3/2}\left(x\right)} {1 - \sin\left(x\right)}\,dx \label{1}\tag{1} $$
I tried using Weierstrass substitution. > **The Weierstrass substitution**, ( named after K.Weierstrass $\left(~1815~\right)$ ), is a substitution used in order to convert trigonometric functions rational expressions to polynomial rational expressions. Integrals of this type are usually easier to evaluate.
This substitution is constructed by letting: $$t = \tan\left(\frac{x}{2}\right) \iff x = 2\arctan(t) \iff dx = \frac{2}{t^2+1}$$
Using basic trigonometric identities it is easy to prove that: $$\cos x = \dfrac{1 - t^2}{1 + t^2}$$
$$\sin x = \dfrac{2t}{1 + t^2}$$
Using this substitution we end up to this integral:
$$ 2 \int \frac{(2t)^{\frac{3}{2}}(1+t^2)}{(1-t^2)^{\frac{3}{2}}(t^2-2t+1)}\,dt$$
Which is clearly not easier to evaluate than $(1)$.
I also tried other standard trigonometric substitutions such as $u = \cos(x)$, $u = \sin(x)$, $u=\tan(x)$ with no better luck.
At last I can't see any trigonometric identities that could simplify the fraction.
Any ideas on how to evaluate this integral?