# Evaluate $\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx$

Question:

Solve the integral $$\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx$$

My solution:

Multiply both the numerator and denominator by $\sqrt{x+1}-\sqrt{x}$. This changes the integral to

\begin{align} \int&\left(1+x+\sqrt{1+x^2}\right)\left(\sqrt{x+1}-\sqrt{x}\right)\,dx\\ &= \int{(1+x)^{3/2}}\,dx-\int{\sqrt{(1+x)(1+x^2)}}\,dx-\int{\sqrt{x}(1+x)}\,dx-\int{\sqrt{x}\sqrt{1+x^2}}\,dx\\ & = \frac{2}{5}(1+x)^{5/2}-I-\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}-J \end{align}

where $I = \int{(1+x)^{1/2}(1+x^2)^{1/2}}\,dx$ and $J = \int{x^{1/2}(1+x^2)^{1/2}}\,dx$.

How can I solve the integrals $I$ and $J$?

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The integrands in $I$ and $J$ are of the form, square root of a cubic. In general, such antiderivatives cannot be expressed in terms of the usual functions of intro calculus, but instead require what are called "elliptic functions". – Gerry Myerson Jan 20 '13 at 12:18
Thanks Gerry Myerson ... would you like to give me an link or document about Elliptical function and how can i solve these type of integral using elliptical function – juantheron Jan 20 '13 at 17:12
I think if you do a websearch on "elliptic integral" or "elliptic function" you will find loads of material is available. You don't so much "solve" the integral using elliptic functions, as define the elliptic functions to be the solutions of this kind of integral. Once you've looked over the literature on the topic, if you have any questions about what you have found, I encourage you to post new questions to m.se. – Gerry Myerson Jan 20 '13 at 22:29

One can be brought to the form $\int u^a (1-u)^b \mathrm{d}u$ which is discussed in this answer of mine: $$\begin{eqnarray} J &=& \int \sqrt{x} \sqrt{1+x^2} \mathrm{d}x \stackrel{u=x^2}{=} \frac{1}{2} \int u^{-1/4} (1+u)^{1/2} \mathrm{d} u \\ &=& \frac{2}{3} u^{3/4} \cdot {}_2F_1\left(-\frac{1}{2}, \frac{3}{4}; \frac{7}{4}; -u\right) +\text{const.} = \frac{2}{3} x^{3/2} \cdot {}_2F_1\left(-\frac{1}{2}, \frac{3}{4}; \frac{7}{4}; -x^2\right) +\text{const.} \end{eqnarray}$$ The other integral is an elliptic integral: $$\begin{eqnarray} I &=& \frac{4}{15} \sqrt{2\alpha} \left(\alpha \operatorname{F}\left( \arcsin \left( \frac{x-\frac{1}{\alpha}}{x+\alpha}\right) ,-\alpha^2\right) -\frac{1}{\alpha} \operatorname{E}\left(\arcsin \left( \frac{x-\frac{1}{\alpha}}{x+\alpha}\right) ,-\alpha^2\right)\right) \\ && + \frac{2}{15} \sqrt{1+x}\sqrt{1+x^2} \left(3x+1 + \frac{4}{x+\alpha}\right) + \text{const.} \end{eqnarray}$$ where $\alpha = \sqrt{2}+1$ and $$\operatorname{E}\left(\phi, m\right) = \int_0^\phi \sqrt{1-m \sin^2\varphi}\, \mathrm{d}\varphi, \quad \operatorname{F}\left(\phi, m\right) = \int_0^\phi \frac{\mathrm{d}\varphi}{\sqrt{1-m \sin^2\varphi}}$$ It can be evaluated using the Jacobi elliptic functions substitution: $$\operatorname{sn}\left(t, -\alpha^2\right) = \frac{x-\frac{1}{\alpha}}{x+\alpha}$$ as described in Byrd and Friedman.

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Thanks Gerry Myerson and also Thanks to Sasha. – juantheron Jan 22 '13 at 3:10
@juantheron : could you please mark this as the answer? – Arjang Sep 1 '13 at 11:43
@Sasha: But why $I$ can express in terms of incomplete elliptic integral of the first kind and incomplete elliptic integral of the second kind as you claimed? – Harry Peter Sep 14 '13 at 4:29

For any real number of $x$ ,

When $|x|\leq1$ ,

$J=\int x^\frac{1}{2}(1+x^2)^\frac{1}{2}$

$=\int x^\frac{1}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n}}{4^n(n!)^2(1-2n)}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n+\frac{1}{2}}}{4^n(n!)^2(1-2n)}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n+\frac{3}{2}}}{4^n(n!)^2(1-2n)\left(2n+\dfrac{3}{2}\right)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n+\frac{3}{2}}}{2^{2n-1}(n!)^2(1-2n)(4n+3)}+C$

$I=\int(1+x)^\frac{1}{2}(1+x^2)^\frac{1}{2}$

$=\int(x+1)^\frac{1}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n}}{4^n(n!)^2(1-2n)}dx$

$=\int(x+1)^\frac{1}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!(x+1-1)^{2n}}{4^n(n!)^2(1-2n)}dx$

$=\int(x+1)^\frac{1}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{-n}(2n)!C_k^{2n}(-1)^{2n-k}(x+1)^k}{4^n(n!)^2(1-2n)}dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n-k}((2n)!)^2(x+1)^{k+\frac{1}{2}}}{4^n(n!)^2k!(2n-k)!(1-2n)}dx$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n-k}((2n)!)^2(x+1)^{k+\frac{3}{2}}}{4^n(n!)^2k!(2n-k)!(1-2n)\left(k+\frac{3}{2}\right)}+C$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n-k}((2n)!)^2(x+1)^{k+\frac{3}{2}}}{2^{2n-1}(n!)^2k!(2n-k)!(1-2n)(2k+3)}+C$

When $|x|\geq1$ ,

$J=\int x^\frac{1}{2}(1+x^2)^\frac{1}{2}$

$=\int x^\frac{1}{2}\left(x^2\left(\dfrac{1}{x^2}+1\right)\right)^\frac{1}{2}$

$=\int x^\frac{3}{2}\left(1+\dfrac{1}{x^2}\right)^\frac{1}{2}$

$=\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-2n}}{4^n(n!)^2(1-2n)}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{\frac{3}{2}-2n}}{4^n(n!)^2(1-2n)}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{\frac{5}{2}-2n}}{4^n(n!)^2(1-2n)\left(\dfrac{5}{2}-2n\right)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{2n-1}(n!)^2(2n-1)(4n-5)x^{2n-\frac{5}{2}}}+C$

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This is not a closed-form expression. – user64494 Sep 21 '13 at 5:24