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I've been dealing with these for a while, and tried different things with no success as of yet:

$$ \int \frac {dx}{(x²+c²)\sqrt{(x-a)(x-b)}} $$

$a$, $b$ and $c$ are real positive numbers.

Trying solving it in Mathematica, if $c=0$, the answer is pretty reasonable, but if $c\neq0$, everything goes bananas, giving imaginary numbers.

Any help, will be appreciated.

magtovi
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  • Looks most similar to some kind of elliptic integral, which Mathematica isn't great with. Including "Assumptions->a>0&&b>0&&c>0" may simplify things a bit, but I wouldn't count on it. Also, are you certain you need the indefinite integral? Elliptic integrals are usually easier if you can specify at least one endpoint. – Semiclassical Sep 23 '14 at 22:37
  • $$ x = u + \left( \frac{a+b}{2} \right) $$ – Will Jagy Sep 23 '14 at 22:49
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    @Semiclassical: Elliptic integrals are not needed. It can be solved in terms of elementary functions, trigonometric and/or logarithmic. – Lucian Sep 23 '14 at 22:50
  • @lucian: Yes, I was just realizing that. In my defense, it is pseudo-elliptic. – Semiclassical Sep 23 '14 at 22:52
  • What is the domain? Looking at the integrand, it seems when $x$ is between $a$ and $b$, the signs of $x-a$ and $x-b$ will be opposite, so you have the square root of a negative number. – Jeppe Stig Nielsen Sep 23 '14 at 23:03
  • See in the vicinity of p. 287 of the freely available 1921 book A Treatise on the Integral Calculus by Joseph Edwards. This book by Edwards is a good place to look for the kind of integral you asked about. The table of contents is sufficiently detailed that I was able to find p. 287 just by looking there. – Dave L. Renfro Sep 24 '14 at 20:47

2 Answers2

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Using Maple I am obtaining the solution in terms of elementary functions

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Juan Ospina
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Take $a > b > 0.$ Let $$\gamma = \left( \frac{a+b}{2} \right)$$ $$\delta = \left( \frac{a-b}{2} \right)$$ Let $$ x = u + \gamma, $$ so the square root part is $$ \sqrt{u^2 - \delta^2}. $$ Next, take $$ u = \delta \cosh t $$

EDIT: The analog of the Weierstrass substitution, when using hyperbolic trig functions, is to take a rational function of $\cosh t$ and $\sinh t$ and turn that into a rational function of another variable. After that one may hope to make progress with partial fractions.

$$ t = \log \left( \frac{1 + w}{1 - w} \right), $$ $$ \cosh t = \frac{1 + w^2}{1 - w^2}, $$ $$ \sinh t = \frac{2w}{1 - w^2}, $$ $$ \tanh t = \frac{2w}{1 + w^2}. $$ $$ d t = \frac{2dw}{1 - w^2}, $$ The original is for ordinary trig functions, http://en.wikipedia.org/wiki/Tangent_half-angle_substitution

Second to the right, and straight on till morning.

That, Peter had told Wendy, was the way to the Neverland; but even birds, carrying maps and consulting them at windy corners, could not have sighted it with these instructions. Peter, you see, just said anything that came into his head.

Typed everything needed to use hyperbolic trig functions in integrals in my permanent Latex file; variable names not consistent with the paragraphs above.

$$ x = \log \left( \frac{1 + w}{1 - w} \right) = 2 \; \mbox{arctanh} \; w, $$ $$ \cosh x = \frac{1 + w^2}{1 - w^2}, $$ $$ \sinh x = \frac{2w}{1 - w^2}, $$ $$ \tanh x = \frac{2w}{1 + w^2}. $$ $$ d x = \frac{2dw}{1 - w^2}. $$ $$ =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= $$ $$ \cosh^2 x - \sinh^2 x = 1 $$ $$ 1 - \tanh^2 x = \mbox{sech}^2 \; x $$ $$ \cosh x + \sinh x = e^x $$ $$ \cosh x - \sinh x = e^{-x} $$ $$ \sinh (x+y) = \sinh x \cosh y + \cosh x \sinh y $$ $$ \cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y $$ $$ \tanh (x+y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y} $$ $$ \sinh \frac{x}{2} = \pm \sqrt { \frac{\cosh x - 1}{2} } $$ $$ \cosh \frac{x}{2} = \sqrt { \frac{\cosh x + 1}{2} } $$ $$ \tanh \frac{x}{2} = \pm \sqrt { \frac{\cosh x - 1}{\cosh x + 1} } = \frac{\cosh x - 1}{\sinh x} = \frac{\sinh x}{\cosh x + 1} $$ $$ \mbox{arcsinh} \; x = \log \left( x + \sqrt{x^2 + 1} \right) $$ $$ \mbox{arccosh} \; x = \log \left( x + \sqrt{x^2 - 1} \right), \; \; \; x \geq 1 $$ $$ \mbox{arctanh} \; x = \frac{1}{2} \; \log \left( \frac{1 + x}{1 - x} \right), \; \; \; |x | < 1 $$

Will Jagy
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