Query on $\int \frac{dx}{\cos x \cos (2x+a)}$ . 
$$\int \frac{dx}{\cos x \cos (2x+a)}$$

MY APPROACH:
I tried using partialization of fraction as follows :
$$ -\frac{1}{\cos a} \int \frac{dx}{\cos x} + \frac{1}{\cos \frac{\pi-2a}{4}} \int \frac{dx}{\cos(2x+a)}$$
$$=-\frac{1}{\cos a} \cdot\ln|\sec x+\tan x| +\frac{1}{\cos \frac{\pi-2a}{4}} \cdot \frac{1}{2} \cdot \ln|\sec (2x+a)+\tan(2x+a)| + c $$
I am dubious since I've never faced partialization of trigonometric fractions.  Is it plausible?
Any other methods?

Hint: Use that $$\cos(x)\cos(2x+a)=\frac{1}{2}\left(\cos(x+a)+\cos(3x+a)\right)$$
Then use that
$$\cos(x)=\frac{1-t^2}{1+t^2}$$ and $$dx=\frac{2tdt}{1+t^2}$$ the so-called Weierstrass substitution.
I will post you the solution for your work!
$$\sec (a) \left(-\sqrt{2} \sqrt{\sin (a)-1} \tan
   ^{-1}\left(\frac{\sqrt{\sin (a)-1} \sec
   \left(\frac{x}{2}\right) \left(\sin
   \left(\frac{a+x}{2}\right)+\cos
   \left(\frac{a+x}{2}\right)\right)}{\sqrt{2}
   \left(\cos \left(\frac{a}{2}\right)-\sin
   \left(\frac{a}{2}\right)\right)}\right)-\sqrt{2}
   \sqrt{\sin (a)+1} \tanh ^{-1}\left(\frac{\sqrt{\sin
   (a)+1} \sec \left(\frac{x}{2}\right) \left(\cos
   \left(\frac{a+x}{2}\right)-\sin
   \left(\frac{a+x}{2}\right)\right)}{\sqrt{2}
   \left(\sin \left(\frac{a}{2}\right)+\cos
   \left(\frac{a}{2}\right)\right)}\right)+\log
   \left(1-\tan \left(\frac{x}{2}\right)\right)-\log
   \left(\tan \left(\frac{x}{2}\right)+1\right)\right)$$
