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Elliptic integrals are integrals of the form $\int R\left[ t, \sqrt{P(t)} \,\right] \, dt$, where $P(t)$ is a polynomial of the third or fourth degree and $R$ is a rational function.

The above is the definition of an elliptic integral, I understand the definition. But then based on the definition why is the following function an elliptic integral:

$$f(x)=\int^{\frac{L}{15}}_{0} \sqrt{{(x^{2}-A)}^{2}+B}\ dx$$

I have been told the above function is an elliptical integral, however according to the definition it does not satisfy all criteria. The integrand is a polynomial of the fourth degree under a square root. However, the function is not rational. So then how can it be an elliptic integral?

Does it get somehow expanded so that integrand becomes rational? My end goal is to find the solution to that integral analytically, but currently I am stuck.

I could not find anything helpful online, I would appreciate some help. Thanks :)

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  • $\begingroup$ $R(x,y)=y$ and $y(x)=\sqrt{(x^2-A)^2+B}$. $\endgroup$
    – Ian
    Oct 22, 2020 at 18:23
  • $\begingroup$ Why do you think it is not rational ? $\endgroup$
    – user65203
    Oct 22, 2020 at 18:43
  • $\begingroup$ @YvesDaoust I thought it would not be rational since the integrand has no fraction $\endgroup$
    – user10764803
    Oct 22, 2020 at 18:59
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    $\begingroup$ @user10764803 Since $1$ is a polynomial, polynomials are rational functions. $\endgroup$
    – Ian
    Oct 22, 2020 at 20:56
  • $\begingroup$ Rational functions are a superset of polynomials. I have never seen this, but if you wanted to insist that the denominator must be non-trivial, you could say proper rational function. $\endgroup$
    – user65203
    Oct 23, 2020 at 7:20

1 Answer 1

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The function is rational when $\sqrt{P(t)}$ is treated like a variable, so it is elliptic. (There's also the condition that $P(t)$ has no repeated roots, otherwise a factor may be taken out and the integral becomes elementary.)

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  • $\begingroup$ Why is it that when $\sqrt{P(t)}$ is treated as a variable the function is rational? That does not change the fact that there is no fraction in the integrand $\endgroup$
    – user10764803
    Oct 22, 2020 at 19:01
  • $\begingroup$ @user10764803 It can be written as $\frac\dots1$, right? $\endgroup$
    – Parcly Taxel
    Oct 22, 2020 at 19:56

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