Hyperelliptic integral: Piecewise solution (handbook) vs General solution (Wolfram)

Question

Recently, I posted a question asking for hints on how to solve the integral $$\int \frac{{\rm d}x}{-\sqrt{x^6+x^2+a}}$$ for which I received an amazing answer. The answer helped me reduce the integral to an elliptic integral which could be solved using the Handbook of Elliptic integrals for scientists and engineers by Byrd and Friedman. The solution though, required information on $a$ as well as the roots of the polynomial in order to choose the right formula.

If you ask Wolfram, to solve this integral it seems to use a general formula that requires no assumptions. Furthermore, while the amplitude of the elliptic function is an arccos in the former, in the later it is an arcsin.

Is there somewhere where I can find the source of this difference or a general formula for such integral that requires no partition of the domain of integration?

Answer 1

Doing the same as @Paul Enta in his answer $$I=-\int \frac{dx}{\sqrt{x^6+x^2+a}}=-\frac 12 \int \frac{dt}{\sqrt{t} \sqrt{t^3+t+a}}$$

Writing $$t^3+t+a=(t-\alpha)(t-\beta)(t-\gamma)$$ simplifies the problem and, back to $x$ $$I=\frac{1}{\sqrt{\gamma \, (\alpha -\beta )}}\,\,F\left(\sin ^{-1}\left(\sqrt{\frac{(\beta -\alpha ) (x^2-\gamma )}{(\beta -\gamma ) (x^2-\alpha )}}\right)|\frac{\alpha (\gamma -\beta )}{\gamma (\alpha -\beta )}\right)$$ which looks a bit nicer than the general formula given by Wolfram Alpha (which does not know the roots).

Now, concening the roots of $t^3+t+a=0$, since $\Delta=-(4+27a^2)$, one is real and it is given by

$$\alpha=-\frac{2}{\sqrt{3}} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3 \sqrt{3} }{2}a\right)\right)$$ from which $\beta$ and $\gamma$ are esily obtained from the sum and product of the roots.