Transformation of the integrand of an elliptic integral to the general form of an elliptic curve First, some background.
An ellipse is defined by the following equation:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\implies y=b\sqrt{1-\frac{x^2}{a^2}}.$
To get the arclength of the ellipse, we use calculus to obtain:
$4\int_{0}^{a}\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}dx.$
We introduce a change of variables $x=at$ to get:
$4a\int_{0}^{1}\sqrt{1+\frac{a^2b^2t^2}{a^2(a^2-a^2t^2)}}dt=4a\int_{0}^{1}\sqrt{\frac{1-t^2(1-b^2/a^2)}{1-t^2}}dt.$
Let $e^2=1-\frac{b^2}{a^2}$ be the square of the eccentricity of the ellipse, then the integral becomes:
$\int_{0}^{1}\sqrt{\frac{1-e^2t^2}{1-t^2}}dt.$
This is an elliptic integral.
Let $u=u(t)=\sqrt{\frac{1-e^2t^2}{1-t^2}}$ be the integrand, then:
$u^2(1-t^2)=1-e^2t^2.$
I believe that this equation defines an elliptic curve of the form:
$y^2=x^3+ax+b$
Such that:
$4a^3+27b^2\neq0.$
How does one transform the equation in terms of $u$ into an elliptic curve?

$y^2=(1+e^2t^2)/(1+t^2)$
If $2\ne 0$,
$y=z/(1+t^2)$ gives $z^2=(1+e^2t^2)(1+t^2)=e^2(t^2+a)^2+b$
$z=w+ie(t^2+a)$ gives $w^2 +2wie(t^2+a)=b$
$t=u/w$ gives $w^2 +2ieu^2/w+2wiea=b$
or $$w^3+2w^2iea-bw=-2ieu^2$$
If $e\ne 0,3\ne 0$  it can be transformed into $U^2= W^3+AW+B$
