I am reading about elliptic integrals, and there is an article showing how to transform general form $\int R(x, \sqrt{P(x)})\,\mathrm{d}x$ with $P(x)$ being a polynomial of degree 3 or 4. It boils down to the three type of elliptic integrals (ie, non-elementary):

$$ \int \frac{1}{\sqrt{P(x)}}\,\mathrm{d}x\tag{$I_0$} $$

$$ \int \frac{x^2}{\sqrt{P(x)}}\,\mathrm{d}x\tag{$I_2$} $$ and $$ \int \frac{1}{(x-b)\sqrt{P(x)}}\,\mathrm{d}x\tag{$H_1$} $$

It is explicitly mentioned that the $I_1$ integral, $\int \frac{x}{\sqrt{P(x)}}\,\mathrm{d}x$, is elementary, but didn't show how to integrate it. I searched the web but found nothing similar. Could someone give a hint how to reduce it to elementary integral?


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    $\begingroup$ Which article ? $\endgroup$ – Lord Shark the Unknown Feb 2 '18 at 6:23
  • $\begingroup$ Are you sure that this is not in the case of repeated roots for $P(x)=0$ ? $\endgroup$ – Claude Leibovici Feb 2 '18 at 6:45
  • $\begingroup$ I think we're all a little skeptical about that integral being elementary (without some additional assumptions). $\endgroup$ – Gerry Myerson Feb 2 '18 at 8:09
  • $\begingroup$ If the polynomial only contains even powers the substitution $x^2=t$ will reduce the degree by two and make the integral elementary $\endgroup$ – Yuriy S Feb 2 '18 at 8:09
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    $\begingroup$ On page 186, the author reduces to the case where the quartic in the denominator is a biquadratic (that is, only even powers of $x$ appear), and in the case (as @Yuriy says) you get something elementary. $\endgroup$ – Gerry Myerson Feb 3 '18 at 12:02

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