TITLE: Integral $\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}$
QUESTION [20 upvotes]: $$\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}$$
I tried substituting $x=z^2$ ... also $x=\tan^2 \theta$ ... but couldn't solve it either ways... if someone can help then it would be good.

REPLY [5 votes]: Make the change of variables
$$x=\frac12\left(t^2+t^{-2}-2\right),$$
under which
$$\sqrt{x}=\frac{1}{\sqrt2}\left(\frac1t-t\right),\qquad \sqrt{x+2}=\frac{1}{\sqrt2}\left(\frac1t+t\right),\qquad \sqrt{x+1}=\frac{1}{\sqrt2}
\frac{\sqrt{t^4+1}}{t}.$$
The integral thus becomes
$$\sqrt2\int\frac{t^2-t^{-2}}{2+\sqrt{t^4+1}}dt.$$
The integrand being a rational function of $t$ and $\sqrt{\text{poly}_4(t)}$, the result can be expressed in terms of elliptic integrals.