Evaluating : $\int \frac{\sec x-\tan x}{\sqrt{\sin^2x-\sin x}} \mathrm{d}x$ As a part of a bigger question, I was asked to evaluate the integral : 

$$\int \frac{\sec x-\tan x}{\sqrt{\sin^2x-\sin x}} \mathrm{d}x$$

Here's what I tried:
(Please bear with me, it gets quite lengthy)
$$\int \frac{\sec x-\tan x}{\sqrt{\sin^2x-\sin x}} \mathrm{d}x$$
$$=\int \frac{1-\sin x}{\cos x \sqrt{\sin^2x-\sin x}}\mathrm{d}x$$
$$=\int \frac{(1-\sin x) \cos x }{\sqrt{\sin^2 x-\sin x}(1-\sin^2 x)}\mathrm{d}x$$
$$=\int \frac{\cos x}{(\sqrt{\sin^2x -\sin x}(1+\sin x)}\mathrm{d}x$$
Substituting $\sin x= t$, we're left with a comparatively good-looking integral:
$$\int \frac {\mathrm{d}t}{(1+t)\sqrt{t^2-t}}$$
Well, this integral looks simple and maybe is, but I'm having real trouble evaluating it :
$$\frac12\int \frac{t+1-(t-1)}{(1+t)\sqrt{t^2-t}}\mathrm{d}t$$
$$=\frac12\left[\int \frac{\mathrm{d}t}{\sqrt{t^2-t}}-\int \frac{t-1}{\sqrt{t^2-t}}\mathrm{d}t\right]$$
Now this is getting longer than I expected it to. Can anyone help me find a shorter and quicker solution to this problem?
Thanks in advance.

Letting $\sin x=\sec^2 \theta$, we have
$$
\begin{aligned}I&=\int \frac{2 \sec ^2 \theta \tan \theta d \theta}{\left(1+\sec ^2 \theta\right) \sec \theta \tan \theta}\\&= 2 \int \frac{d(\sin \theta)}{2-\sin ^2 \theta}\\&= \frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2}+\sin \theta}{\sqrt{2}-\sin \theta}\right|+C\\&= \frac 1{\sqrt{2}} \ln \left|\frac{\sqrt{2 \sin x}+\sqrt{\sin x+1}}{\sqrt{2 \sin x}-\sqrt{\sin x+1}}\right| +C \end{aligned}
$$
