What is the the integral of $\sqrt{x^a + b}$?

Question

How do you evaluate $\displaystyle\int\sqrt{x^a + b}\,\,\text{dx}$, where $a \neq 0$ and $a \neq 1$?

For example, how do you evaluate $\displaystyle\int\sqrt{x^2 + 1}\,\text{dx}$? If we let $u=x^2+1$, then $du=2x\,\text{dx}$. We cannot do this because there is no $2x$ in the original function. Of course you cannot let $u=\sqrt{x^2 + 1}$, because $du=\displaystyle\frac{x}{\sqrt{x^2+1}}\,\text{dx}$. There is no $\displaystyle\frac{x}{\sqrt{x^2+1}}$ in the original function. So how do we solve this?

How about $\displaystyle\int\sqrt{x^3 + 1}\,\text{dx}$?

Answer 1

Lots of integrals of that kind cannot be solved in terms of elementary simple functions. Indeed, there is a large class of integrals whose solutions are the so called Jacobi Elliptic Integrals Functions (of the first kind, second kind and third kind).

Generally, you have an integral of the form

$$\int R\left(x, \sqrt{P(x)}\right)\ \text{d}x$$

where $P(x)$ is a polynomial of degree $3$ or $4$ and $R$ denotes a general rational function. For example, you very last integral will be solvable interns of Jacobi Elliptic Functions:

$$\int\sqrt{x^3 + 1}\ \text{d} x$$

is nothing but the integral of the square root of a polynomial of degree three.

It's solution is

$$\int\sqrt{x^3 + 1}\ \text{d}x = \frac{2}{5\sqrt{1 + x^3}}\left(x + x^4 + (-1)^{1/6} 3^{3/4}\sqrt{(-1)^{1/6}\left((-1)^{2/3} + x\right)}\sqrt{+1 (-1)^{1/3}x + (-1)^{2/3}x^2}\ \mathcal{F}(x)\right)$$

Where

$$\mathcal{F}(x) = F\left(\arcsin\left(\frac{-(-1)^{5/6}(1+x)}{3^{1/4}}\right),\ (-1)^{1/3}\right)$$

Is the so called Jacobi Elliptic Function of the first kind.

More on JEF

https://en.wikipedia.org/wiki/Elliptic_integral

General Case

There is a really funny solution for the general case for your integral, namely

$$\int\sqrt{x^a + b}\ \text{d}x$$

Despite it is not solvable in terms of simple functions, a close form solution does exist. Indeed, this integral has a general solution in terms of the so called HyperGeometric Special Functions:

$$\int\sqrt{x^a + b}\ \text{d}x = \frac{x\left(2(b + x^a) ab\sqrt{\frac{b + x^a}{b}}\ _2F_1\left[\frac{1}{2},\ \frac{1}{a},\ 1 + \frac{1}{a},\ -\frac{x^a}{b}\right]\right)}{(2+a)\sqrt{b + x^a}}$$

Where the HypGeo Function

$$_2F_1\left[\frac{1}{2},\ \frac{1}{a},\ 1 + \frac{1}{a},\ -\frac{x^a}{b}\right]$$

can be defined here

https://en.wikipedia.org/wiki/Hypergeometric_function

and here

http://mathworld.wolfram.com/HypergeometricFunction.html

Some values for the HyperGeometric Function here

http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/

And a good article on computing values for $\ _2F_1$ here

http://arxiv.org/pdf/1212.0251v6.pdf

Answer 2

Kim Peek's "funny" hypergeometric solution is really the series solution near $x=0$. We have, for $|x^a/b| < 1$,

$$ \sqrt{x^a + b} = \sqrt{b} \sqrt{1 + x^a/b} = \sqrt{b} \sum_{k=0}^\infty {1/2 \choose k} (x^a/b)^k$$ so integrating term-by-term

$$ \int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{x^{ak+1}}{(ak+1) \; b^{k-1/2}}$$

where $${1/2 \choose k} = \dfrac{\Gamma(3/2)}{\Gamma(k+1)\; \Gamma(3/2-k)} = \dfrac{(2k)!}{(-4)^k (k!)^2 (1-2k)}$$

If you take the definition of the hypergeometric function as a power series, you get exactly this series.

Alternatively, for $|x^a/b| > 1$ you get a different series involving negative powers of $x$:

$$\sqrt{x^a + b} = x^{a/2} \sqrt{1 + b/x^a} = \sum_{k=0}^\infty {1/2 \choose k} b^k x^{(1/2 - k)a}$$ so that

$$ \int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{b^k x^{(1/2-k)a+1}}{(1/2-k)a + 1}$$

which also has a hypergeometric representation.