Finding value of $\displaystyle \int\frac{\sec x}{\sqrt{3+\tan x}}dx$
Try: Let $\displaystyle I = \int \frac{\sec x}{\sqrt{3+\tan x}}dx,$ Put $3+\tan x = t^2$
and $\sec^2 xdx =2tdt$
So $\displaystyle I = 2\int \frac{1}{\sqrt{t^4-6t^2+10}}dt$
could some help me how to solve it, Thanks
Hint:
$$I=\int\frac{1}{\cos x \sqrt {3+ tan x}} dx= 2\cos x\frac{1}{2 \cos^2x \sqrt {3+ tan x}} dx$$
$$\frac{1}{2 \cos^2 x \sqrt {3+ tan x}} dx= du$$
$$u=\sqrt {3+ tan x}$$
$$v=2\cos x$$ $$dv=-2\sin x$$
$$I= \cos x \sqrt {3+ tan x} + 2\int \sin x \sqrt {3+ tan x} dx$$
Now take $3+ tan x=t^2 $ and continue.