$\int \frac{\sec x}{\sqrt{\sin (2 x+\alpha)+\sin \alpha}} d x$ is equal to

$\sqrt{2 \sec \alpha(\tan x+\tan \alpha)}+C$

Let

$I=\int \frac{\sec x}{\sqrt{\sin (2 x+\alpha)+\sin \alpha}} d x$

$\Rightarrow I =\int \frac{\sec x}{\sqrt{2 \sin (x+\alpha) \cos x}} d x$

$\Rightarrow I =\frac{1}{\sqrt{2}} \int \frac{\sec ^2 x}{\sqrt{\tan x \cos \alpha+\sin \alpha}} d x$

$\Rightarrow I=\frac{1}{\sqrt{2} \cos \alpha} \int \frac{1}{\sqrt{\tan x \cos \alpha+\sin \alpha}} d(\tan x \cos \alpha+\sin \alpha)$

$\Rightarrow I=\frac{1}{\sqrt{2} \cos \alpha} \times 2 \sqrt{\tan x \cos \alpha+\sin \alpha}+C$

$\Rightarrow I=\sqrt{2 \sec ^2 \alpha(\tan x \cos \alpha+\sin \alpha)}+C$

$\Rightarrow I=\sqrt{2 \sec \alpha(\tan x+\tan \alpha)}+C$.