How to evaluate $\int\frac{x^ndx}{\sqrt{ax^2+bx+c}}$ for natural $n$

Question

How do we evaluate the following integral? $$\int\frac{x^ndx}{\sqrt{ax^2+bx+c}}$$ I am strictly looking for a solution for natural $n$

I wanted to try a trig substitution $$\int\frac{x^ndx}{\sqrt{ax^2+bx+c}}=\int\frac{x^ndx}{\sqrt{a\left(\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b^2}{4a^2}\right)}}$$$$=\int\frac{(u-\frac{b}{2a})^ndx}{\sqrt{a\left(u^2+\frac{c}{a}-\frac{b^2}{4a^2}\right)}}$$ but I didn't know how to deal with the $(\tan\theta+k)^n$

I can accept an answer with special functions.

Answer 1

$$\int_0^xt^r(t-a)^{-\frac12}(t-b)^{-\frac12}dt\mathop=^{t=ux}\int_0^1 (ux)^r (ux-a)^{-\frac12}(ux-b)^{-\frac12}x du=\frac{x^r}{\sqrt{ab}}\int_0^1 u^{(r+1)-1} \left(-\left(\frac xbu-1\right)\right)^{-\frac12}\left(-\left(\frac xau-1\right)\right)^{-\frac12} x du=-\frac{x^{r+1}}{\sqrt{ab}}\int_0^1 u^{(r+1)-1} \left(1-\frac xbu\right) ^{-\frac12}\left(1-\frac xau\right)^{-\frac12} du$$

Using this Appel F$_1$ integral representation:

$$\text F_1(a;b_1,b_2;c;x,y)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1\frac{t^{a-1}(1-t)^{c-a-1}}{(1-xt)^{b_1}(1-yt)^{b_2}}dt$$

Therefore:

$$\boxed{\int_0^x\frac{t^r}{\sqrt{(t-a)(t-b)}}dt=\frac{x^{r+1}}{\sqrt{ab}(r+1)}\text F_1\left(r+1;\frac12,\frac12;r+2;\frac xa,\frac xb\right)}$$

Shown here in the “alternate forms“ section. For $r\in\frac{\Bbb Z}2$, the Appel function reduces to a rational function including Elliptic F and Elliptic E. Also, for $r\in\Bbb Z$, inverse trigonometric functions appear with a rational function.

Answer 2

Here's how an electrician tries to solve this problem

First, we introduce the following notation

$$X(x)=X=ax^2+bx+c$$

Then we can present the integral in the form

$$I_n=\int\frac{x^ndx}{\sqrt{X}}$$

Now, let's differentiate $x^{n-1}\sqrt{X}$ with respect to $x$

$$\frac{\mathrm{d} }{\mathrm{d} x}(x^{n-1}\sqrt{X})=an\frac{x^n}{\sqrt{X}}+b(n-\frac{1}{2})\frac{x^{n-1}}{\sqrt{X}}+c(n-1)\frac{x^{n-2}}{\sqrt{X}}$$

Here we took into account that $X=ax^2+bx+c$

Integrating the resulting equation with respect to x we can write

$$x^{n-1}\sqrt{X}=anI_n+b(n-\frac{1}{2})I_{n-1}+c(n-1)I_{n-2}$$

Set here $n=1$ $$\sqrt{X}=aI_1+\frac{b}{2}I_0$$

We're almost done.

The only thing left is to calculate

$$I_0=\int\frac{dx}{\sqrt{ax^2+bx+c}}$$

After that we can calculate the integrals $I_1,I_2...$ successively using the recurrence equation.

Further development

Let's introduce a generating function

$$I(s)=I=\sum_{n=0}^{\infty}I_ns^n$$

Now, multiply both sides of the recurrence relation above by $s^n$ and sum over $n\geqslant 1$.

(I'll skip the long but simple algebraic transformations)

We get the following differential equation $$(cs^2+bs+a)\frac{\mathrm{d} I}{\mathrm{d} s}+(cs+\frac{b}{2})I=\frac{\sqrt{X}}{1-xs}$$

Let's remember that $X=ax^2+bx+c$

All the family of the given integral $I_n$ is hidden in this differential equation.

But how to drag them all out, that's the problem!