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.
 A: $\bf{My\; Solution::}$ Let $\displaystyle I = \int \sqrt{\sec x+\tan x}\cdot \sec^2 xdx = \int\frac{\sqrt{\sec^2 x-\tan^2 x}}{\sqrt{\sec x-\tan x}}\cdot \sec^2 xdx$
Now Integral $\displaystyle I = \int\frac{\sec^2 x}{\left(\sec x-\tan x \right)^{\frac{1}{2}}}dx$
Now Let $\left(\sec x-\tan x \right) = t^2\;,$ Then $\sec x\cdot \left(\tan x-\sec x\right)dx = 2tdt\Rightarrow \sec x\cdot t dx =-2tdt$
So $\sec xdx = -2dt$ So Integral $\displaystyle I = -2\int \frac{\sec x }{t}dt$
Now Using $\displaystyle \left(\sec x- \tan x\right) = t$ and $\displaystyle \left(\sec x+\tan x \right) = \frac{1}{t}$
Using $\left(\sec^2 x-\tan^2 x\right) = 1$
So $\displaystyle 2\sec x = t+\frac{1}{t} = \frac{t^2+1}{t}$
So Integral $\displaystyle I = -\int\frac{t^2+1}{t^2}dt = -t+\frac{1}{t}+\mathbb{C}$
So $\displaystyle I = \int \sqrt{\sec x+\tan x}\cdot \sec^2 xdx = -\left(\sec x-\tan x\right)+\frac{1}{\left(\sec x -\tan x\right)}+\mathbb{C}$
$\displaystyle  = -\left(\sec x-\tan x\right)+\left(\sec x+\tan x\right)+\mathbb{C} = 2\tan x+\mathbb{C}$
A: Hmmm...  It's possible double angle identities will work here.
$$\int\sqrt{\sec x+\tan x}\sec^2xdx=\int\sec^{5/2}x\sqrt{1+\sin x}dx=$$
$$\int\sec^{5/2}x\sqrt{\sin^2\frac x2+\cos^2\frac x2+2\sin\frac x2\cos\frac x2}dx=$$
$$\int(\cos^2\frac x2-\sin^2 \frac x2)^{5/2}(\sin\frac x2+\cos\frac x2)dx=$$
$$\int(\cos\frac x2-\sin\frac x2)^{5/2}(\cos\frac x2+\sin\frac x2)^{7/2}dx$$
Not sure where to go from here.  Maybe someone else can find a way to make this work.
A: 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 $$I=\frac{2}{3} \Big(2 \sin (x)-1\Big) \sec (x) \sqrt{\frac{\sin \left(\frac{x}{2}\right)+\cos
   \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)-\sin
   \left(\frac{x}{2}\right)}}$$ 
