Integral $\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}$ $$\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}$$
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.
 A: 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.
A: Possible approach:
Let $0<a$.
$$\begin{align}
\mathcal{I}{\left(a\right)}
&=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}\\
&=\int_{1}^{a+1}\frac{\mathrm{d}y}{\sqrt{y-1}+\sqrt{y}+\sqrt{y+1}};~~~\small{\left[x+1=y\right]}\\
&=\int_{a+1-\sqrt{a(a+2)}}^{1}\frac{\frac{1-z^2}{2z^2}\,\mathrm{d}z}{\frac{1-z}{\sqrt{2z}}+\frac{\sqrt{1+z^2}}{\sqrt{2z}}+\frac{1+z}{\sqrt{2z}}};~~~\small{\left[y-\sqrt{y^2-1}=z\right]}\\
&=\frac{1}{\sqrt{2}}\int_{\frac{\left(\sqrt{a+2}-\sqrt{a}\right)^2}{2}}^{1}\frac{1-z^2}{2+\sqrt{1+z^2}}\cdot\frac{\mathrm{d}z}{z^{3/2}}\\
&=\sqrt{2}\int_{\frac{\sqrt{a+2}-\sqrt{a}}{\sqrt{2}}}^{1}\frac{1-w^4}{2+\sqrt{1+w^4}}\cdot\frac{\mathrm{d}w}{w^2};~~~\small{\left[\sqrt{z}=w\right]}\\
&=\sqrt{2}\int_{\alpha}^{1}\frac{\left(1-w^4\right)\left(2-\sqrt{1+w^4}\right)}{w^2\left(3-w^4\right)}\,\mathrm{d}w;~~~\small{\left[\alpha:=\frac{\sqrt{a+2}-\sqrt{a}}{\sqrt{2}}\right]}\\
&=2\sqrt{2}\int_{\alpha}^{1}\frac{\left(1-w^4\right)}{w^2\left(3-w^4\right)}\,\mathrm{d}w\\
&~~~~~-\sqrt{2}\int_{\alpha}^{1}\frac{\left(1-w^4\right)\sqrt{1+w^4}}{w^2\left(3-w^4\right)}\,\mathrm{d}w\\
&=\frac{2\sqrt{2}}{3}\int_{\alpha}^{1}\left[\frac{1}{w^2}-\frac{2w^2}{3-w^4}\right]\,\mathrm{d}w\\
&~~~~~-\frac{\sqrt{2}}{3}\int_{\alpha}^{1}\left[\frac{1}{w^2}-\frac{2w^2}{3-w^4}\right]\sqrt{1+w^4}\,\mathrm{d}w\\
&=\frac{2\sqrt{2}}{3}\int_{\alpha}^{1}\frac{\mathrm{d}w}{w^2}-\frac{4\sqrt{2}}{3}\int_{\alpha}^{1}\frac{w^2}{3-w^4}\,\mathrm{d}w\\
&~~~~~-\frac{\sqrt{2}}{3}\int_{\alpha}^{1}\left[\frac{1}{w^2}-\frac{2w^2}{3-w^4}\right]\frac{1+w^4}{\sqrt{1+w^4}}\,\mathrm{d}w\\
&=\frac{2\sqrt{2}}{3}\left(\frac{1-\alpha}{\alpha}\right)+\frac{2\sqrt{2}}{3}\int_{\alpha}^{1}\frac{\mathrm{d}w}{\sqrt{3}+w^2}-\frac{2\sqrt{2}}{3}\int_{\alpha}^{1}\frac{\mathrm{d}w}{\sqrt{3}-w^2}\\
&~~~~~-\frac{\sqrt{2}}{3}\int_{\alpha}^{1}\left[3w^2+\frac{1}{w^2}-\frac{8w^2}{3-w^4}\right]\frac{\mathrm{d}w}{\sqrt{1+w^4}}\\
&=\frac{2\sqrt{2}}{3}\left(\frac{1-\alpha}{\alpha}\right)+\frac{2\sqrt{2}}{3}\left[\frac{\arctan{\left(\frac{1}{\sqrt[4]{3}}\right)}-\arctan{\left(\frac{\alpha}{\sqrt[4]{3}}\right)}}{\sqrt[4]{3}}\right]\\
&~~~~~-\frac{2\sqrt{2}}{3}\left[\frac{\operatorname{arctanh}{\left(\frac{1}{\sqrt[4]{3}}\right)}-\operatorname{arctanh}{\left(\frac{\alpha}{\sqrt[4]{3}}\right)}}{\sqrt[4]{3}}\right]\\
&~~~~~-\sqrt{2}\int_{\alpha}^{1}\frac{w^2\,\mathrm{d}w}{\sqrt{1+w^4}}-\frac{\sqrt{2}}{3}\int_{\alpha}^{1}\frac{\mathrm{d}w}{w^2\sqrt{1+w^4}}\\
&~~~~~+\frac{4\sqrt{2}}{3}\int_{\alpha}^{1}\left[\frac{1}{\sqrt{3}-w^2}-\frac{1}{\sqrt{3}+w^2}\right]\frac{\mathrm{d}w}{\sqrt{1+w^4}}.\\
\end{align}$$
I don't have the gas to work out the remaining elliptic integrals right now, but given that they are very nearly in standard form I think there's very good reason to expect a relatively "nice" final form (or at the very least a more digestible expression than the primitive produced by Maple). TBC...
