I'm currently struggling to grasp how to express the Schwarz-Christoffel transformation for a rectangle in the commonly used form of an elliptic integral of the first kind.$$w(z)=u+vi=\int_{0}^{b1} \frac{A\;dz}{\sqrt{(z^2-a^2)(z^2-b^2)}}= \frac{A}{b} \int_{0}^{1/k} \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}=K(k)+iK(k')$$
with $k=a/b$ and $k'=\sqrt{1-k^2}$
I tried a substitution $t=z/a$ which lead to: $$u+vi=A\int_{0}^{b/a}\frac{a\;dt}{\sqrt{(t^2a^2-1)(t^2a^2-\frac{b^2}{a^2})}}$$
Letting $a=1 \; \,,a<b$ I got: $$A\int_{0}^{1/k}\frac{dt}{\sqrt{(t^2-1)(t^2-\frac{1}{k^2})}}$$
this doesn't seem that far from the desired result but is still missing the factor $A/b$ and I don't see a way to further transform the denominator of the function. Thanks in advance for any kind of suggestion.