<s>[INST] The indefinite integral $ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$ $$ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$$
$$ \int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}$$
$$ \int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\frac{\sin x + \cos x}{\sqrt{\cos 2x}}}$$
$$ \sqrt 2 \int \frac{\sin \left(2x +\alpha+ \frac{\pi}{4}\right) + \sin \left(\alpha -\frac{\pi}{4}\right)}{\cos^3 x\cdot\sqrt{\cos 2x}}$$
How I can do it after this and get rid of square root?
 [/INST] $$I=\int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx,$$
$$I=\cos{\alpha}\int \frac{\sin x}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx+\sin{\alpha}\int \frac{1}{\cos^2 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}dx,$$
$$I=\cos{\alpha} I_1+\sin{\alpha}I_2,$$ where
$$I_1=\int \tan{x}\sqrt{\frac{1+\tan{x}}{1-\tan{x}}} d(\tan{x}),$$
$$I_2=\int \sqrt{\frac{1+\tan{x}}{1-\tan{x}}}d(\tan{x}).$$
Substitute $t=\tan{x}$ in both integrals:
$$I_1=\int t\sqrt{\frac{1+t}{1-t}}dt,$$
$$I_2=\int \sqrt{\frac{1+t}{1-t}}.$$
</s>