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Can one do the indefinite integral of

$$\frac{1}{(ax^2+bx+c)^n}, a, b, c \in \mathbb{R}$$

quickly, without resorting to those *awful* recursion relations:

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$$\eqalign{\int \dfrac1{(ax^2+bx+c)^n}dx&=\dfrac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}

\\&+\dfrac{(2n-3)2a}{(n-1)(4ac-b^2)}

\int\dfrac1{(ax^2+bx+c)^{n-1}}dx+C.}$$

or without using tricks such as writing $ax^2+bx+c$ as $x^2+1$ or something like that?

Hardy (P. 11-12) seems to claim that we can write the denominator as $[(x-\alpha)(x-\alpha^*)]^n$ and then re-write $\frac{1}{(ax^2+bx+c)^n}$ via partial fractions and integrate a bunch of $\frac{1}{(x-\alpha)^j}$ and $\frac{1}{(x-\alpha^*)^j}$ terms, but wont you end up with stuff that is necessarily complex, similar to this.

I think you’re supposed to end up with something like this:

which gives only real numbers, but I’m not getting that 🙁 Furthermore the answer you get when you use recursion relations involves $\arctan$ which is obviously not here 🙁

So:

What is the integral of

$$\frac{1}{(ax^2+bx+c)^n} = \frac{1}{(x-\alpha)^n(x-\alpha^*)^n}, a, b, c \in \mathbb{R}$$

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I was a bit confused about what you did and did not want to involve, but in my opinion the simplest way to do this is using differentiation.

Note that:

$${\frac { \left( -1 \right) ^{n-1} }{ \left( n-1 \right) !}}{\frac {\partial }{\partial c^{n-1}}}\frac{1}{ a{x}^{2}+bx+c

} = \frac{1}{ \left(a{x}^{2}+bx+c

\right)^n}

$$

and that:

$$\int \! \frac{1}{ a{x}^{2}+bx+c

}{dx}=2\,\arctan \left( {

\frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c

a-{b}^{2}}}}$$

hence:

$$\int \! \frac{1}{ \left(a{x}^{2}+bx+c

\right)^n}{dx}=

{\frac { \left( -1 \right) ^{n-1} }{ \left( n-1 \right) !}}{\frac {\partial }{\partial c^{n-1}}}

\left( 2\,\arctan \left( {\frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}}

\right) {\frac {1}{\sqrt {4\,ca-{b}^{2}}}} \right)

$$

You can then use the generalized Leibniz formula to obtain a polynomial plus an arctan part, although as far as I can tell the arctan part will always be there as it begins the *awful* recursion.

**Update**

In light of the discussion in the comments of this answer, perhaps something like this would be better…

Note that:

$${\frac {1}{ \left( x-\mu \right) ^{1

+n} \left( x-\nu \right) ^{1+n}}}=\frac{1}{(n!)^2}{\frac {\partial ^{2\,n}}{\partial {\nu}^{n}\partial {\mu}^{n}}}

\left( {\frac {1}{ \left( x-\mu \right) \left( x-\nu \right) }}

\right) \tag{1}$$

and:

$${\frac {1}{ \left( x-\mu \right) \left( x-\nu \right) }}={\frac {1}{ \left( –

\nu+\mu \right) \left( x-\mu \right) }}-{\frac {1}{

\left( x-\nu \right) \left( -\nu+\mu \right) }} \tag{2}$$

hence:

$$\begin{aligned}

\int{\frac {1}{ \left( x-\mu \right) ^{1+n} \left( x-\nu \right) ^{1+n}}}{dx}=

&\frac{1}{(n!)^2}{\frac {\partial ^{2\,n}}{\partial {\nu}^{n}\partial {\mu}^{n}}}\frac{1}{\left( -\nu+\mu \right)}\int

\!{

\frac {1}{ \left( x-\mu \right) }}-{\frac {1}{ \left( x-\nu \right) }}{dx}\\

&\frac{1}{(n!)^2}{\frac {\partial ^{2\,n}}{\partial {\mu}^{n}\partial {\nu}^{n}}}

\left({\frac {\ln

\left( x-\mu \right) }{-\nu+\mu}} -{\frac {\ln \left( x-\nu \right) }{-\nu+\mu}}\right) \tag{3}

\end{aligned}

$$

Then note that, for example:

$$\frac{1}{(n!)^2}{\frac {\partial ^{2\,n}}{\partial {\mu}^{n}\partial {\nu}^{n}}}\frac {\ln \left( x-\mu \right) }{-\nu+\mu}=\frac{1}{n!}{\frac {\partial }{\partial \mu^n}} \left( {\frac {\ln \left( x-\mu \right) }{ \left( –

\nu+\mu \right) ^{1+n}}} \right) \tag{4}

$$

and that the second term on the right of $(3)$ is just $\mu \leftrightarrow\nu$. The generalized Leibniz formula states:

$${\frac {d}{d \mu^n}}\left( f\left( \mu \right) g \left( \mu \right) \right)

=\sum _{m=0}^{n}{n\choose m} \left( {\frac {d^{m}}{d{\mu}^{m}}}f \left( \mu \right) \right)\left( {\frac {d^{n-m}}{d{\mu}^{n-m}}}g \left( \mu \right)\right) \tag{5}$$

apply $(5)$ to the right hand side of $(4)$, using the numerator as $f$ and the denominator as $g$. Pull out the zeroth order term from the sum as this contains the $\log$ term. Shift down the index on the remaining sum by $1$, applying the derivative once, to the log term in the sum. For the remaining derivatives you will need the useful formula:

$${\frac {\partial ^{k}}{\partial {\mu}^{k}}} \frac{1}{\left( –

\nu+\mu \right) ^{1+n}} ={\frac { \left( -1

\right) ^{k} \left( n+k \right) !}{n!\, \left( -\nu+\mu \right) ^{1+n

+k}}} \tag{6}$$

After some lengthy algebra, the end result is:

$$\int{\frac {1}{ \left( x-\mu \right) ^{1+n} \left( x-\nu \right) ^{1+n}}}{dx}= \frac{\left( -1 \right) ^{n}}{\left( -\nu+\mu \right) ^{1+2\,n}} \left[{2\,n\choose n}\left(\ln \left( x-\mu\right)-\ln \left( x-\nu \right)\right)-P_n(\mu,\nu,x)\right] $$

where:

$$P_n(\mu,\nu,x)=\sum _{m=1}^{n}\frac{{2\,n-m\choose n}}{m} \left[ \left( {\frac {-\nu+\mu}

{-x+\mu}} \right) ^{m}- \left( {\frac {\nu-\mu}{-x+\nu}} \right) ^{m}

\right] \tag{7}$$

The polynomial is of generalized hypergeometric form.

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