<s>[INST] Relation between full elliptic integrals of the first and third kind I am working on a calculation involving the Ronkin function of a hyperplane in 3-space.
I get a horrible matrix with full elliptic integrals as entries. A priori I know that the matrix is symmetrical and that give me a relation between full elliptic integrals of the first and third kind.
I can not find transformations in the literature that explain the relation and I think I need one in order to simplify my matrix.
The relation
With the notation
$\operatorname{K}(k) = \int_0^{\frac{\pi}{2}}\frac{d\varphi}{\sqrt{1-k^2\sin^2\varphi}},$
$\qquad$ $\Pi(\alpha^2,k)=\int_0^{\frac{\pi}{2}}\frac{d\varphi}{(1-\alpha^2\sin^2\varphi)\sqrt{1-k^2\sin^2\varphi}}$
$k^2 = \frac{(1+a+b-c)(1+a-b+c)(1-a+b+c)(-1+a+b+c)}{16abc},\quad a,b,c > 0$
the following is true:
$2\frac{(1+a+b-c)(1-a-b+c)(a-b)}{(a-c)(b-c)}\operatorname{K}(k)+$
$(1-a-b+c)(1+a-b-c)\Pi\left( \frac{(1+a-b+c)(-1+a+b+c)}{4ac},k\right) +$
$\frac{(a+c)(1+b)(1-a-b+c)(-1-a+b+c)}{(a-c)(-1+b)}\Pi\left( \frac{(1+a-b+c)(-1+a+b+c)(a-c)^2}{4ac(1-b)^2},k\right)+$
$(1-a-b+c)(-1+a+b+c)\Pi\left( \frac{(1-a+b+c)(-1+a+b+c)}{4ac},k\right)+$
$\frac{(1+a)(b+c)(-1-a+b+c)(-1+a+b-c)}{(1-a)(c-b)}\Pi\left( \frac{(1-a+b+c)(-1+a+b+c)(b-c)^2}{4ac(1-a)^2},k\right)$
$==0$.
Is there some addition formula or transformation between elliptic integrals of the first and third kind that will explain this? 
 [/INST] Are you sure that $a$, $b$, and $c$ have no relations among them?
I ask this because I attempted to (painstakingly!) copy your proposed equation over to Mathematica (taking care of the fact that the functions in it take the modulus $m=k^2$ as argument), and I keep getting complex results instead of 0.
On the other hand, your problem might actually be easier to solve if you use Carlson's symmetric elliptic integrals. Maybe you can try re-expressing the Legendre forms as Carlson forms?
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