I need help to evaluate $$\int \frac{1}{\sqrt{1+\tan(x)}}dx$$
I am trying this question by using the formula $\tan(x)=\frac{2\tan(\frac{x}{2})}{1-\tan^{2}(\frac{x}{2})}$. Now I have put this above formula of $\tan(x)$ in the above expression. So, the integral will become $$\int \frac{\sqrt{1-\tan^{2}(\frac{x}{2})}}{\sqrt{1-\tan^{2}(\frac{x}{2})+2\tan(\frac{x}{2})}}dx$$
Now I am thinking of writing $\tan(\frac{x}{2})=\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}$. So, after this the above integral will become $$\int \frac{\sqrt{\cos(x)}}{\sqrt{\cos(x)+\sin(x)}}dx$$
Now after this step I wrote the integral as $$\frac{1}{2}\int \frac{\sqrt{\cos(x)}+\sqrt{\sin(x)}}{\sqrt{\cos(x)+\sin(x)}}dx\;\;+\;\;\frac{1}{2}\int \frac{\sqrt{\cos(x)}-\sqrt{\sin(x)}}{\sqrt{\cos(x)+\sin(x)}}dx$$
But I am facing problem to solve further. Please help me out.
Seems to be a little hard one.
$$\left(\left.\text{dx}^2\right/(1+\text{Tan}[x])\right) \\ \text{:}\quad 1+\text{Tan}[x]:\to y, \ x\to \text{ArcTan}[y-1] , \ \text{dx}\to \frac{\text{dy}}{1+(y-1)^2}) \\ =\frac{\text{dy}^2}{\left(1+(-1+y)^2\right)^2 y}$$
$$\int \frac{1}{\sqrt{y}\left(1+(1-y)^2\right)} \, dy=\frac{\sqrt{2} \text{ArcTan}\left[\frac{1}{\sqrt{-\frac{1+i}{y}}}\right]+(1+i) \text{ArcTan}\left[\frac{1}{\sqrt{-\frac{1-i}{y}}}\right]}{\sqrt{2+2 i}}$$
Mathematica directly
$$-i \left(\frac{\text{ArcTanh}\left[\frac{\sqrt{1+\text{Tan}[x]}}{\sqrt{1-i}}\right]}{\sqrt{1-i}}-\frac{\text{ArcTanh}\left[\frac{\sqrt{1+\text{Tan}[x]}}{\sqrt{1+i}}\right]}{\sqrt{1+i}}\right)$$
This can be simplified to
$$\frac{\text{ArcTan}\left[\frac{\sqrt{1+\text{Tan}[x]}}{\sqrt{i-1}}\right]}{\sqrt{1-i}}+\frac{\text{ArcTan}\left[\frac{\sqrt{1+\text{Tan}[x]}}{\sqrt{-1-i}}\right]}{\sqrt{1+i}}$$