Integrate $\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$ Integrate: $$\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$$  
My attempt :
I substituted $\sec{x} + \tan{x} $ as $t^2$
Then, $$ (\sec{x} \cdot \tan{x} + \sec^2x) dx =2tdt$$
$$\sec{x}(  \tan{x} + \sec{x}) dx =2tdt$$  
$$\sec{x}\cdot t^2 dx =2tdt$$
But there is  $\sec{x}$ left outside.

If we use Weierstrass substitution, the problem becomes$$I=\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx=2 \int \frac{ \sqrt{\frac{1+t}{1-t}} \left(1+t^2\right)}{\left(1-t^2\right)^2}\,dt$$ Now, setting $$\sqrt{\frac{1+t}{1-t}}=z$$ the integral becomes $$I=\int (z^2+\frac{1}{z^2})\,dz=\frac{z^3}{3}-\frac{1}{z}$$ Now, back to $x$, we get $$I=\frac{2}{3} \Big(2 \sin (x)-1\Big) \sec (x) \sqrt{\frac{\sin \left(\frac{x}{2}\right)+\cos
   \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)-\sin
   \left(\frac{x}{2}\right)}}$$ 
