<s>[INST] Evaluation of $ \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$ Evaluation of $\displaystyle \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$
$\bf{My \; Try::}$ Let $\displaystyle I = \int \frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx$
So $\displaystyle I = \int\frac{\sin (x+\alpha)}{\cos x}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx\cdot \sec^2 x$
$\displaystyle I  = \int \frac{\sin x\cdot \cos \alpha+\cos x\cdot \sin \alpha}{\cos x}\cdot \sqrt{\frac{\cos +\sin x}{\cos x-\sin x}}\cdot \sec^2 x dx$
$\displaystyle I = \int\left(\cos \alpha\cdot \tan x+\sin \alpha\right)\cdot \sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$
Now Let $\tan x= t\;,$ Then $\sec^2 xdx = dt$
$\displaystyle I = \int (\cos \alpha\cdot t + \sin \alpha)\sqrt{\frac{1+t}{1-t}}dt$
Now How can I solve after that 
Help me 
Thanks
 [/INST] Assuming what you have done up till the last part is correct, these hints I think will help you go further:
(1) Multiplying and dividing by $\sqrt{1+t}$
(2) Knowing the derivative of  $\sinh^{-1} (\sqrt{t-1}) $ and $\sqrt{t+1}\sqrt{t-1}$
(3) Using partial integration
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