Is there an equivalent elliptic integral (with solution) in the Byrd-Friedmann's elliptical integral Handbook? This is the integral:
$$ \int \frac{1}{\sqrt{(x-p)(x-q)(x-̅q)}}dx $$
where we have three roots p>0 and q=a+ib and q̅=a-ib (complex conjugated of q) ; a,b>o and $i=\sqrt{-1}$. The roots come from a cubic polynomial and I am looking for a solution incl. steps. For instance, in my previous question and thanks to the members of math.stack, I solved a similar integral containing a quartic polynomial at the denominator (therefore four roots): I used the solution proposed by the integral 260.00 of the mentioned book. The integral has been classified as an incomplete elliptical integral of the first order. What about this case?
Any proposal would be very appreciated. Thank you so much for your support. Mark.
The Byrd-Friedman formula 239.00 covers the case $$ \int_a^y \frac{dt}{\sqrt{(t-a)[(t-b_1)^2+a_1^2]}} = g F(\varphi,k);\quad y>a $$ as an elliptic integral of the first kind. In your case $(x-q)(x-\bar q)=(x-\Re q)^2+\Im^2 q$. The auxiliary variable are (note that $a$ here is your $p$) $$ A^2\equiv (b_1-a)^2+a_1^2. $$ $$ g\equiv 1/\surd A. $$ $$ k^2=(A+b_1-a)/(2A) $$ $$ \cos\varphi = \frac{A+a-y}{A-a+y}. $$ and a complementary formula 241.00 $$ \int_y^\infty \frac{dt}{\sqrt{(t-a)[(t-b_1)^2+a_1^2]}} = g F(\varphi,k);\quad \infty> y\ge a $$ where $$ A^2\equiv (b_1-a)^2+a_1^2. $$ $$ g\equiv 1/\surd A. $$ $$ k^2=(A+b_1-a)/(2A) $$ $$ \cos\varphi = \frac{y-a-A}{y-a+A}. $$ See also https://catalog.hathitrust.org/Record/000615533 .