Are these two elliptic integral evaluations identical? I'm reading a paper on the Schwarz D minimal surface, and I'm wondering whether the authors have made a mistake. They evaluate the integral
$$
\int_0^z \frac{2t\;\mathrm{d}t}{\sqrt{t^8-14t^4+1}},
$$
with $z$ inside a domain $\Omega$ which contains no zeros of the octic polynomial, and say it equals
$$
\frac{1}{4}F\left(\arcsin\left(\frac{4z^2}{z^4+1}\right),97-56\sqrt{3}\right),
$$
with $F$ the incomplete elliptic integral of the first kind. However, Mathematica gives me
$$
\frac{\sqrt{-z^4-4 \sqrt{3}+7} \sqrt{\left(4 \sqrt{3}-7\right) z^4+1}}{\sqrt{z^8-14 z^4+1}}
F\left(\arcsin\left(\frac{z^2}{\sqrt{7-4 \sqrt{3}}}\right),97-56 \sqrt{3}\right).
$$
The strange thing is that, after evaluating them numerically at a couple of points inside $\Omega$, the two expressions seem to follow one another quite closely (within about $10^{-3}$ for both the real and the imaginary part) without actually being equal. Is this due to Mathematica-related inaccuracies or are they really unequal. If the latter is the case, which of the two expressions (if any) is correct?
I'm not very knowledgeable about elliptic integrals, so any help would be warmly appreciated. Thanks for reading!
 A: I think the Mathematica answer is better.  The original integrand
$$
\frac{2t}{\sqrt{t^8-14t^2+1}},
$$
is imaginary for $t>0.267$, but the integral in the paper
$$
\frac{1}{4}F\left(\arcsin\left(\frac{4z^2}{z^4+1}\right),97-56\sqrt{3}\right),
$$
remains real until the arcsin reaches $1$, which is when $z=0.4696$.  But the Mathematica result
$$
\frac{\sqrt{-z^4-4 \sqrt{3}+7} \sqrt{\left(4 \sqrt{3}-7\right) z^4+1}}{\sqrt{z^8-14 z^4+1}}
F\left(\arcsin\left(\frac{z^2}{\sqrt{7-4 \sqrt{3}}}\right),97-56 \sqrt{3}\right)
$$
is real until $z=0.267$ again, because of its denominator.
added
the value $0.4696$ for the singularity of the solution of the paper is
a solution of
$$
\frac{4z^2}{z^4+1} = \sin 1
$$
A: (Too long for a comment.) I think it is all because of a typo. The integrand supposedly is 
$$\frac{2t}{\sqrt{t^8-14t^2+1}}\tag1$$
but I think it should be
$$\frac{2t}{\sqrt{t^8-14t^\color{blue}4+1}}\tag2$$
Note that the latter form is an octahedral invariant and is also expressible as,
$$(t^2-i)^8+(2t)^8+(t^2+i)^8=2(t^8-14t^4+1)^2$$
with the imaginary unit $i$.
A: The paper referred to in the question is most likely Gandy et al. (1999), Exact computation of the triply periodic D ('Diamond') minimal surface, Chemical Physics Letters 314, 543–551. It is one of three related papers in which exact coordinates are derived for the Schwarz D/G/P surface family.
The integral (with my correction) appears in the Weierstrass–Enneper parametrisation of the surfaces. After scaling and taking the real part, it gives the $z$-coordinate of a point on a patch which then gets repeated to form the entire surface.
But I can confirm that
$$\frac14F\left(\sin^{-1}\frac{4z^2}{z^4+1},97-56\sqrt3\right)$$
is dead wrong. The authors say they used Byrd and Friedman's Handbook of Elliptic Integrals for Engineers and Physicists – a reference that really helped me with finding the Willmore energy of an ellipsoid – but they must have applied the tables incorrectly, for a more careful calculation yields the correct answer as
$$\int_0^z\frac{2t}{\sqrt{t^8-14t^4+1}}\,dt=\frac1{2+\sqrt3}F\left(\sin^{-1}\frac{z^2}{2-\sqrt3},97-56\sqrt3\right)$$
As it turns out, there are many, many typos in the three papers, including having $Q_6$ and $Q_7$ among $O_1$ and $O_2$ in a table in the gyroid paper and getting the W–E parametrisation equations themselves wrong in the Schwarz D paper.
