How to evaluate $\int \frac{\tan^{3/2}\left(x\right)}{1 - \sin\left(x\right)} \,\mathrm{d}x$? 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?
 A: We can show fairly straightforwardly that this reduces to an elliptic integral, which cannot be an elementary function: put $ x = \arctan(u^2) $. Then $ dx = 2u/(1+u^4) \, du $, $\tan x = u^2$ and $\sin x = u^2/\sqrt{1+u^4}$, and rationalising implies that the integral becomes
$$ \int \bigg( 2u^4 + \frac{2u^6}{\sqrt{1+u^4}} \bigg) \, du , $$
and we just need to worry about the second term. It so happens that this was one of the earliest integrals Liouville considered when he became interested in when an integral is algebraic (See Lützen's Joseph Liouville 1809–1882 pp. 374ff. for the details). An integration by parts reduces us to $ \int \frac{u^2}{\sqrt{1+u^4}} \, du $, which is known to not be elementary (see either Liouville's work, or Ritt's book Integration in finite terms). Thus the "elementary part" is
$$ \frac{2}{5} ( u^5 + u^3 \sqrt{1+u^4}) , $$
while the non-elementary part is the elliptic integral
$$ - \frac{6}{5} \int \frac{u^2}{\sqrt{1+u^4}} \, du = \frac{6}{5}\sqrt{i} ( F(\arcsin(\sqrt{i}u) \mid -1) - F(\arcsin(\sqrt{i}u) \mid -1) . $$
One could write in terms of $x$ again, but there doesn't seem much point.
A: First, a small historical precision : most books call the tangent half-angle substitution the Weierstrass $(1815-1897)$ substitution while in fact the technique appears in Euler's work $-1707-1783).
Concerning the possible substitutions, what @Chappers proposed is probably the best since leading to a real closed form expression in terms of elliptic integrals.
Just to continue with what I early wrote in comments, letting $x=\sin ^{-1}(u)$, we end with
$$I=\int \frac{\tan^{\frac{3}{2}}(x)}{1-\sin(x)} \,dx=\int \frac{u^{3/2}}{(1-u)^{9/4} \,(u+1)^{5/4}}\,du$$
$$I=\frac{2 u^{3/2} (3 u-2)}{5 (1-u)^{5/4}\, (u+1)^{1/4}}+\frac{4}{5} u^{3/2} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{7}{4};u^2\right)$$ which I have not been able to simplify further.
Remarks
Looking again at @Chappers's answer, I have the filling that there are a few minot typo's.
$$x= \tan ^{-1}\left(u^2\right)\implies dx=\frac{2 u}{u^4+1}\,du \qquad \text{and} \quad \sin(x)=\frac{u^2}{\sqrt{u^4+1}}$$ making
$$I =2\int  \left(u^4+\frac{u^6}{\sqrt{u^4+1}}\right)\,du$$ making
$$I=\frac{2}{5} u^3 \left(u^2+\sqrt{u^4+1}\right)+$$ $$\frac{6}{5} (-1)^{3/4} \left(E\left(\left.i \sinh ^{-1}\left((-1)^{1/4}
   u\right)\right|-1\right)-F\left(\left.i \sinh ^{-1}\left((-1)^{1/4}
   u\right)\right|-1\right)\right)$$
