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?

Let $$I = \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}dx$$
$$I = \cos \alpha\int \frac{\sin x+\cos x\cdot \tan \alpha}{\cos x}\cdot \sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$$
So $$I = \cos \alpha \int (\tan \alpha+\tan x)\sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$$
Now Put $\tan x= t\;,$ Then $\sec^2 xdx = dt$
So $$I = \cos \alpha \int (t+\tan \alpha)\sqrt{\frac{1+t}{1-t}}dt = \cos \alpha \int \frac{(t+\tan \alpha)(t+1)}{\sqrt{1-t^2}}dt$$
So $$I = \cos \alpha \int\frac{t^2+(\tan \alpha +1)t+\tan \alpha}{\sqrt{1-t^2}}dt$$
So $$I = \cos \alpha \int\frac{t^2}{\sqrt{1-t^2}}dt+\cos \alpha \int\frac{(\tan \alpha +1)}{\sqrt{1-t^2}}dt+\cos \alpha\int\frac{\tan \alpha}{\sqrt{1-t^2}}dt$$
In first  put $t=\sin \phi\;,$ Then $dt = \cos \phi d\phi$ and in second $(1-t^2)=u^2\;,$ Then $tdt = -udu$
So $$I = \frac{\cos \alpha}{2} \int (1-\cos 2 \phi)d\phi-\cos \alpha (1+\tan \alpha)\int du+\sin \alpha \cdot \sin^{-1}(t)$$
So $$I = \frac{\cos \alpha}{2}\left[\sin^{-1}(t)-t\sqrt{1-t^2}\right]-\cos \alpha(1+\tan \alpha)\sqrt{1-t^2}+\sin\alpha\cdot  \sin^{-1}(t)+\mathcal{C}$$
So  $$I = \frac{\cos \alpha}{2}\left[\sin^{-1}(\tan x)-t\sqrt{1-\tan^2x}\right]-\cos \alpha(1+\tan \alpha)\sqrt{1-\tan^2x}+\sin \alpha \cdot \sin^{-1}(\tan x)+\mathcal{C}$$
