# How to reduce an integral with square root of cubic function into an elliptic integral

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.

• 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. – Claude Leibovici Feb 11 at 11:01
• It does look that pretty, at least what WolframAlpha produces. – mrtaurho Feb 11 at 11:08
• @mrtaurho. Change $C$ to $c$ in WA. I do not know what it is doing with $C$ : very strange ! – Claude Leibovici Feb 11 at 11:22
• @ClaudeLeibovici It does seem to change that much. But yes indeed, the output looks strange anyway. – mrtaurho Feb 11 at 11:26
• @mrtaurho. Yes, this is what I did and obtained. I wonder how was interpreted $C$. Any idea ? – Claude Leibovici Feb 11 at 11:35

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.