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I need to calculate the following ntegral: $$\int \frac{t }{\sqrt{2 t^3 - 3 t^2 + 6 C}} dt$$ where $C$ is a constant to be determined later, so I cannot look for roots of the polynomial in the denominator. I've found that integrals involving $R(t,\sqrt{P(t)})$, where $R$ is a rational function of its arguments and $P$ is a polynomial of degree 3 or 4 with no repeated roots, can be reduced to elliptic integrals. I've also found that it is sometimes done with Moebius transformation, however I can't find any general "walkthrough" and my attempts to express the above integral in terms of elliptic integrals have failed. I'd be grateful for any help.

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  • $\begingroup$ I think that this could be difficult. Even writing $2t^3-3t^2+6C=2(t-r_1)(t-r_2)(t-r_3)$ and later using the product of the roots makes the problem quite hard (at least to me). Moreover, I suspect that you would get more than one elliptic integral as a result. $\endgroup$ – Claude Leibovici Feb 11 at 11:01
  • $\begingroup$ It does look that pretty, at least what WolframAlpha produces. $\endgroup$ – mrtaurho Feb 11 at 11:08
  • $\begingroup$ @mrtaurho. Change $C$ to $c$ in WA. I do not know what it is doing with $C$ : very strange ! $\endgroup$ – Claude Leibovici Feb 11 at 11:22
  • $\begingroup$ @ClaudeLeibovici It does seem to change that much. But yes indeed, the output looks strange anyway. $\endgroup$ – mrtaurho Feb 11 at 11:26
  • $\begingroup$ @mrtaurho. Yes, this is what I did and obtained. I wonder how was interpreted $C$. Any idea ? $\endgroup$ – Claude Leibovici Feb 11 at 11:35
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I think this reduction is in

Hancock, F. H., Elliptic integrals., New York: Wiley, 104 S (1917). ZBL46.0620.06.

But my memory may be wrong, it was many years ago that I read this.

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  • $\begingroup$ Thank you - it's good to know that this book is available online. I'll check it, yet it may take some time :) $\endgroup$ – P. Szeptynski Feb 12 at 15:56
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It seems that a method for expressing a general elliptic integral in terms of elliptic integrals of the $1^{st}$, $2^{nd}$ and $3^{rd}$ kind can be found in H. Hancock - "Lectures on the theory of elliptic functions" , p. 180:

For a general case of integral of type: $$\int\frac{t \; \text{d}t}{\sqrt{a t^3 + 3 b t^2 + 3 c t + d}}$$ we may introduce a substitution: $$t=m \cdot z + n$$ with $$ m=\sqrt[3]{\frac{4}{a}} \;, \; \; \; n=-\frac{b}{a}$$ what results in: $$\int\frac{t \; \text{d}t}{\sqrt{a t^3 + 3 b t^2 + 3 c t + d}}= A \int\frac{\text{d}z}{\sqrt{4 z^3 - g_2 z - g_3}}+ B \int\frac{z\; \text{d}z}{\sqrt{4 z^3 - g_2 z - g_3}}$$ where $A,B,g_2, g_3$ are constants. The first integral on the right-hand side of the above formula is an elliptic integral of the first kind in Weierstrass normal form and may be expressed in terms of Weierstrass $\wp$ function, while the other one is the elliptic integral of the second kind in Weierstrass normal form which may be expressed in terms of Weierstrass $\zeta$-function.

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