The tetragonal distortion of the Schwarz P minimal surface has a Weierstrass–Enneper parametrisation of $g(w)=w,f(w)=(w^8+\lambda w^4+1)^{-1/2}$ and a Bonnet angle of $\pi/2$ where $\lambda<-2$. This surface can be decomposed into "catenoids" between square rings:
I have worked out that the aspect ratio (height over base side) of this catenoid is $\sqrt{-ps}\frac ZX$ where $$p=\lambda+\sqrt{\lambda^2-4},s=\sqrt{2-\lambda}$$ $$X=K\left(\frac12+\frac1s\right)$$ $$Z=F\left(\sin^{-1}\sqrt{\frac2{2-p}},1-\frac{p^2}4\right)$$ (Here the last argument of elliptic integrals is the parameter $m$.)
I am unable to simplify $Z$ to a complete elliptic integral – I think it can be done, since the formula for the aspect ratio of the unit cell of the conjugate T (tetragonal D) surface only involves complete integrals. Can simplification on $Z$ be done?
In particular, can the Landen and Gauss transformations be applied to $Z$ to reduce it?
In Carlson symmetric form: $$Z=F\left(\sin^{-1}\sqrt{\frac2{2-p}},1-\frac{p^2}4\right)=\sqrt{\frac2{2-p}}R_F\left(\frac p{p-2},-\frac p2,1\right)$$ Now apply a descending Landen transformation (DLMF 19.22.18, .23) to get a zero as one argument, then scale to make another argument one: $$=\sqrt{\frac2{2-p}}R_F\left(0,\frac{4p}{p-2},\frac{2-p}2\right)=\frac2{2-p}R_F\left(0,\frac{-8p}{(2-p)^2},1\right)$$ $R_F$ may now be converted back into a complete Legendre elliptic integral. Its parameter is quite complicated in $p$ alone, but substituting its definition $p=\lambda+\sqrt{\lambda^2-4}$ gives a surprisingly simple result: $$\boxed{Z=\frac2{2-p}K\left(1+\frac{8p}{(2-p)^2}\right)=\frac2{2-p}K\left(\frac{\lambda+2}{\lambda-2}\right)}$$
$K(p^2/4)$ where $p$ has the same meaning appears in the T surface's aspect ratio formula. By the same approach as detailed above, but using an ascending Gauss transformation, I was able to simplify it to $\frac2{2-p}K\left(\frac4{2-\lambda}\right)$.