# Thread: Primitive of (ax^2 + bx + c)^(n + 1/2)

1. ## Primitive of (ax^2 + bx + c)^(n + 1/2)

No. 295 in my list of difficult integrals ...

Integral of the general power of $\displaystyle a x^2 + b x + c$: the object here is to establish the following reduction formula:

$\displaystyle \int \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x = \frac {\left({2 a x + b}\right) \left({a x^2 + b x + c}\right)^{n + \frac 1 2} } {4 a \left({n + 1}\right)} + \frac {\left({2 n + 1}\right) \left({4 a c - b^2}\right)} {8 a \left({n + 1}\right)} \int \left({a x^2 + b x + c}\right)^{n - \frac 1 2} \ \mathrm d x$

What I've tried: Integration by Parts, which gets me to:

$\displaystyle x \left({a x^2 + b x + c}\right)^{n + \frac 1 2} - \left({2 n + 1}\right) a \int x^2 \left({a x^2 + b x + c}\right)^{n - \frac 1 2} \ \mathrm d x - \frac {\left({2 n + 1}\right) b} 2 \int x \left({a x^2 + b x + c}\right)^{n - \frac 1 2} \ \mathrm d x$

... I waded further into this morass but I lost my boots.

I have also tried substitutions $\displaystyle u = a x^2 + b x + c$ etc. but nothing useful happens, and trying to split it into a form with a $\displaystyle 2 a x + b$ is also unfruitful.

The usual: hints for a substitution or a splitting?

2. ## Re: Primitive of (ax^2 + bx + c)^(n + 1/2)

I might have an angle on this:

a) Complete the square to get the integral in the form $\displaystyle \int (\sqrt{z^2 + q})^n$

b) Substitute $\displaystyle u = z^2$ to get it in the form $\displaystyle \int \frac {(\sqrt{u + q})^n} {\sqrt u}$

c) Use the standard result $\displaystyle \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x$

but it's late and I need to sleep on it.

3. ## Re: Primitive of (ax^2 + bx + c)^(n + 1/2)

Originally Posted by Matt Westwood
I might have an angle on this:

a) Complete the square to get the integral in the form $\displaystyle \int (\sqrt{z^2 + q})^n$

b) Substitute $\displaystyle u = z^2$ to get it in the form $\displaystyle \int \frac {(\sqrt{u + q})^n} {\sqrt u}$

c) Use the standard result $\displaystyle \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x$

but it's late and I need to sleep on it.
This actually worked (more or less, I've got a factor of 2 that I need to lose), so feel free to consider this one solved.