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I am looking for a technique to solve an indefinite integral of $$ \int \frac{dx}{\sqrt{ax^3+bx^2+cx+f}} $$

I honestly have no idea where to start with this and I cannot find anything like this in an integral table.

Actually I was thinking maybe taylor expand?

Thank you

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I believe you are out of luck, see en.wikipedia.org/wiki/… –  Will Jagy Oct 9 '12 at 23:04
    
I should add that x is a substitution for 1/r with r being the radial component of the equations of motion for 2D orbits –  Ricardo Oct 9 '12 at 23:21
    
Are you sure you want to solve an indefinite integral of that integral? I suspect that what you actually want to solve is the indefinite integral itself. –  joriki Oct 10 '12 at 0:05
    
yes you are correct joriki. I want to just solve that indefinite integral. –  Ricardo Oct 10 '12 at 8:09
    
@Ricardo: There's an edit link underneath the question. If comments lead to a clarification of the question, the question should be updated so that people don't have to read through the comments to understand it. –  joriki Oct 10 '12 at 10:36

2 Answers 2

Assuming that the polynomial has no repeated roots, this is an elliptic integral and can not be expressed in terms of elementary functions.

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yeah, I'm not sure if that's the route I want to go. The question asked (in my classical mechanics homework) is.. The addition to the potential energy $$V =\frac{k}{r}$$ of a small correction δV(r) makes the bounded orbits deviate from closed; after each turn, the perihelion shifts by a small angle δϕ. Find for $$δV=\frac{-\gamma}{r^3}$$ –  Ricardo Oct 10 '12 at 8:11

You can certainly do a Taylor expansion, which in this case is simply a binomial expansion. $$(ax^3+bx^2+cx+f)^{-1/2}=f^{-1/2}(1+(Cx+Bx^2+Ax^3))^{-1/2}=f^{-1/2}(1-(1/2)(Cx+Bx^2+Ax^3)+(3/8)(Cx+Bx^2+Ax^3)^2+\cdots)$$ where I've written $A,B,C$ for $a/f,b/f,c/f$, respectively. Now you can multiply out those powers of $Cx+Bx^2+Ax^3$ and integrate term-by-term.

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that is what I ended up doing. although I'm not sure if that's exactly what the homework problem was asking. I took the taylor expansion around the point x=0 which $$x=\frac{1}{r}$$ so around the point $$r=\infty$$ –  Ricardo Oct 10 '12 at 7:56

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