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Math Help - Primitive of (ax^2 + bx + c)^(n + 1/2)

  1. #1
    Super Member Matt Westwood's Avatar
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    Primitive of (ax^2 + bx + c)^(n + 1/2)

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

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

    \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:

    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 u = a x^2 + b x + c etc. but nothing useful happens, and trying to split it into a form with a 2 a x + b is also unfruitful.

    The usual: hints for a substitution or a splitting?
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  2. #2
    Super Member Matt Westwood's Avatar
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    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 \int (\sqrt{z^2 + q})^n

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

    c) Use the standard result \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.
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  3. #3
    Super Member Matt Westwood's Avatar
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    Re: Primitive of (ax^2 + bx + c)^(n + 1/2)

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

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

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

    c) Use the standard result \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.
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