Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$

Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv  =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)^2+2}\sqrt{2} dw$$$$= \frac{1}{2} \int \frac{1}{\sqrt{2}\left(w^2+1\right)}dw$$$$ =  \frac{1}{2\sqrt{2}}\arctan \left(w\right) + C$$$$=  \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C$$
Therefore,
$$\int \frac{1}{\cos 2x+3} dx = \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C $$

That's a decent solution but I am wondering if there are any other simpler ways to solve this (besides Weierstass). Can you come up with one?

HINT:
Using Euler's formula, we have $\cos(2x)=\frac12(e^{i2x}+e^{-i2x})$.
Now make the substitution $z=e^{i2x}$ with $dx=\frac1{i2z}\,dz$
