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This has probably been asked countless times before, but is there a particularly elegant method of evaluating$$\displaystyle I=\int\frac{dr}{r\sqrt{ar^2+br+c}}$$ ,with $a,b>0,c<0$?

Background: The integral occurs when calculating the angle, $\phi$, of a body in an attractive central field as a function of $r$, the distance from to centre of the field (although it could be a one-to-many function). If the body has mass $m$, angular momentum $M$, and total energy $E_0$, then
$$\displaystyle \phi=\int\frac{Mdr}{r^2\sqrt{2m\left(E_0-U(r)\right)-\frac{M}{r^2}}}$$, with potential energy $$ U(r)=\frac{-\alpha}{r}.$$ Of course the solution will be the polar equation $r(\phi)$ of an ellipse (which immediately suggests the form of a fairly cumbersome substitution), but I'm more interested in the methods than the solution.

More generally: is there a method of evaluating $$\displaystyle I=\int\frac{dr}{r^n\sqrt{ar^2+br+c}}$$ , or, better still, $$\displaystyle I=\int\frac{dr}{r^n\sqrt{p_n(r)}}?$$

Meow
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  • May be I missed something, but what is wrong with $$ \frac{1}{\sqrt{-c}} \arctan\left(\frac{1}{2\sqrt{-c}} \frac{2 c+ b r}{\sqrt{a r^2 + b r + c}}\right)? $$ – gammatester Aug 29 '13 at 13:52
  • @gammatester that's probably fine (the solution I've seen uses $\arccos$ instead), could you perhaps answer using the methods you used to obatin this? – Meow Aug 29 '13 at 13:57
  • Ah, now I see, that I did miss somethong. With evaluate you mean, the derivation of an expression for your integral, not the evaluation of the expression. Anyway, I got the result by letting Maple do the work with a twist. If you simply enter the integrand you get an expression with $\log$ (similar to the result of Wolfram Alpha), you have to add the assumption $c<0$ to get the $\arctan$. – gammatester Aug 29 '13 at 14:10
  • @gammatester is 'evaluate' a nonstandard verb to use in this case, then? – Meow Aug 29 '13 at 14:15
  • I don't know because English is not my first language, in numerical computing it seems to be used in the sense compute the value of an expression. – gammatester Aug 29 '13 at 14:20
  • If we are working in the reals, the answers "look" wildly different depending on whether $a\lt 0$, $=0$, $\gt 0$, and on whether $b^2-4ac$ is positive, $0$, or negative. And $a=0$ breaks up into some (trivial) special cases! – André Nicolas Aug 29 '13 at 14:40
  • In this case, though, $a>0$ and $b^2-4ac>0$. – GEdgar Aug 29 '13 at 14:42

2 Answers2

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For the first question: the general method is: complete the square for the quadratic under the radical, then make the trigonometric substitution suggested by the result. The same method applies to the "more generally" clause.

For the "better still" clause: With a polynomial of degree 3 or 4 under the radical, this will be an elliptic integral, and not (in general) elementry. For degree 5 or higher it is even worse: a so-called "abelian integral" but (in general) not elementary and not an elliptic integral.

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How to evaluate $\int \frac{dx}{(x-a)\sqrt{x^2-b}}$ ?
Say $b>0$, write $b=\beta^2$, the integral is $$ \int \frac{dx}{(x-a)\sqrt{x^2-\beta^2}} . $$ Use substitution $x = \beta \sec \theta$ to get $$ \int\frac{\beta\sec\theta \tan\theta \, d\theta}{(\beta\sec\theta - a)\sqrt{\beta^2 \tan^2\theta}\;} = \int\frac{\sec\theta\,d\theta}{\beta\sec\theta - a} = \int\frac{d\theta}{\beta-a\cos\theta} , $$ an elementary integral.

GEdgar
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  • I'm sorry to be so late in replying, but for some reason I can't see which substitution to use when evaluating $\int \frac{dx}{(x-a)\sqrt{x^2-b}}$: could you help? – Meow Sep 28 '13 at 16:54
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Start with the substitution $r=1/u$.

larry
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