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.
 A: 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.
A: 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.
