# Solve the integral int(1+x+sqrt(1+x^2))(sqrt(x+1)+sqrt(x))dx

Solve the integral
$\int \frac{1+x+\sqrt{1+{x}^{2}}}{\sqrt{x+1}+\sqrt{x}}\phantom{\rule{thinmathspace}{0ex}}dx$
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Plutbantonavv
One can be brought to the form $\int {u}^{a}\left(1-u{\right)}^{b}\mathrm{d}u$ which is discussed in this answer of mine:
$\begin{array}{rcl}J& =& \int \sqrt{x}\sqrt{1+{x}^{2}}\mathrm{d}x\stackrel{u={x}^{2}}{=}\frac{1}{2}\int {u}^{-1/4}\left(1+u{\right)}^{1/2}\mathrm{d}u\\ & =& \frac{2}{3}{u}^{3/4}\cdot {}_{2}{F}_{1}\left(-\frac{1}{2},\frac{3}{4};\frac{7}{4};-u\right)+\text{const.}=\frac{2}{3}{x}^{3/2}\cdot {}_{2}{F}_{1}\left(-\frac{1}{2},\frac{3}{4};\frac{7}{4};-{x}^{2}\right)+\text{const.}\end{array}$
The other integral is an elliptic integral:
$\begin{array}{rcl}I& =& \frac{4}{15}\sqrt{2\alpha }\left(\alpha \mathrm{F}\left(\mathrm{arcsin}\left(\frac{x-\frac{1}{\alpha }}{x+\alpha }\right),-{\alpha }^{2}\right)-\frac{1}{\alpha }\mathrm{E}\left(\mathrm{arcsin}\left(\frac{x-\frac{1}{\alpha }}{x+\alpha }\right),-{\alpha }^{2}\right)\right)\\ & & +\frac{2}{15}\sqrt{1+x}\sqrt{1+{x}^{2}}\left(3x+1+\frac{4}{x+\alpha }\right)+\text{const.}\end{array}$
where $\alpha =\sqrt{2}+1$ and
$\mathrm{E}\left(\varphi ,m\right)={\int }_{0}^{\varphi }\sqrt{1-m{\mathrm{sin}}^{2}\phi }\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\phi ,\phantom{\rule{1em}{0ex}}\mathrm{F}\left(\varphi ,m\right)={\int }_{0}^{\varphi }\frac{\mathrm{d}\phi }{\sqrt{1-m{\mathrm{sin}}^{2}\phi }}$
It can be evaluated using the Jacobi elliptic functions substitution:
$\mathrm{sn}\left(t,-{\alpha }^{2}\right)=\frac{x-\frac{1}{\alpha }}{x+\alpha }$
as described in Byrd and Friedman.
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Kymani Hatfield
For any real number of x,
When $|x|\le 1$
$J=\int {x}^{\frac{1}{2}}\left(1+{x}^{2}{\right)}^{\frac{1}{2}}$
$=\int \sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!{x}^{2n+\frac{1}{2}}}{{4}^{n}\left(n!{\right)}^{2}\left(1-2n\right)}dx$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!{x}^{2n+\frac{3}{2}}}{{4}^{n}\left(n!{\right)}^{2}\left(1-2n\right)\left(2n+\frac{3}{2}\right)}+C$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!{x}^{2n+\frac{3}{2}}}{{4}^{n}\left(n!{\right)}^{2}\left(1-2n\right)\left(2n+\frac{3}{2}\right)}+C$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!{x}^{2n+\frac{3}{2}}}{{2}^{2n-1}\left(n!{\right)}^{2}\left(1-2n\right)\left(4n+3\right)}+C$
$I=\int \left(1+x{\right)}^{\frac{1}{2}}\left(1+{x}^{2}{\right)}^{\frac{1}{2}}$
$=\int \left(x+1{\right)}^{\frac{1}{2}}\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!{x}^{2n}}{{4}^{n}\left(n!{\right)}^{2}\left(1-2n\right)}dx$
$=\int \left(x+1{\right)}^{\frac{1}{2}}\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!\left(x+1-1{\right)}^{2n}}{{4}^{n}\left(n!{\right)}^{2}\left(1-2n\right)}dx$
$=\int \sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{2n}\frac{\left(-1{\right)}^{n-k}\left(\left(2n\right)!{\right)}^{2}\left(x+1{\right)}^{k+\frac{1}{2}}}{{4}^{n}\left(n!{\right)}^{2}k!\left(2n-k\right)!\left(1-2n\right)}dx$
$=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{2n}\frac{\left(-1{\right)}^{n-k}\left(\left(2n\right)!{\right)}^{2}\left(x+1{\right)}^{k+\frac{3}{2}}}{{4}^{n}\left(n!{\right)}^{2}k!\left(2n-k\right)!\left(1-2n\right)\left(k+\frac{3}{2}\right)}+C$
$=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{2n}\frac{\left(-1{\right)}^{n-k}\left(\left(2n\right)!{\right)}^{2}\left(x+1{\right)}^{k+\frac{3}{2}}}{{2}^{2n-1}\left(n!{\right)}^{2}k!\left(2n-k\right)!\left(1-2n\right)\left(2k+3\right)}+C$
When $|x|\ge 1$
$J=\int {x}^{\frac{1}{2}}\left(1+{x}^{2}{\right)}^{\frac{1}{2}}$
$=\int {x}^{\frac{1}{2}}{\left({x}^{2}\left(\frac{1}{{x}^{2}}+1\right)\right)}^{\frac{1}{2}}$
$=\int {x}^{\frac{3}{2}}{\left(1+\frac{1}{{x}^{2}}\right)}^{\frac{1}{2}}$
$=\int {x}^{\frac{3}{2}}\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!{x}^{-2n}}{{4}^{n}\left(n!{\right)}^{2}\left(1-2n\right)}dx$
$=\int \sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!{x}^{\frac{3}{2}-2n}}{{4}^{n}\left(n!{\right)}^{2}\left(1-2n\right)}dx$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left(2n\right)!}{{2}^{2n-1}\left(n!{\right)}^{2}\left(2n-1\right)\left(4n-5\right){x}^{2n-\frac{5}{2}}}+C$