# 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.

• Even WolframAlpha chokes on it – egreg May 9 '15 at 13:00
• Do you have some reason for thinking it can be done? – Gerry Myerson May 9 '15 at 13:02
• $-\int \frac{2 \cot^2 \theta d\theta} {1+cos \theta + \sqrt{\cos 2\theta}}$ , that's all i could get. – Mann May 9 '15 at 13:03
• Just curious ... Where did your find this problem? – Mark Viola May 9 '15 at 15:26
• Does someone know the transformation which brings it to a form which looks more lof the elliptic type? – tired May 10 '15 at 15:45

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.
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}