Hyperelliptic integral involving square root of sextic polynomial

Question

I am interesting in finding the integral of $$ I = \int \frac{{\rm d}x}{-\sqrt{x^6+x^2-c}}$$ Where $c \leq x^6 + x^2$ such that the integral is always real. I was trying to look at a reduction of 579.00 from the Handbook of Elliptic integrals for scientists and engineers by Byrd and Friedman, i.e., $$ \int\frac{{\rm d}x}{a_0x^6 + a_1x^5 + a_2+x^4 + a_3x^3 + a_2x^2 + a_1x + a_0}, $$ But in my case $c \neq -1$ in general. I know Wolfram can solve this equation, but I am interested in learning how to apply the reduction and subsequent integration. Moreover, I am unsure on how to check that the roots of the polynomial form three pairs of points of an involution.

Answer 1

To evaluate \begin{equation} I(u,v)=\int_u^v\frac{dx}{\sqrt{x^6+x^2-c}} \end{equation} it is sufficient to calculate \begin{equation} J(Y)=\int_0^Y\frac{dx}{\sqrt{x^6+x^2-c}} \end{equation} (by parity, we can take $Y>0$). By changing $t=x^2$ \begin{equation} J(Y)=\frac12\int_0^y\frac{dt}{\sqrt{t(t^3+t-c)}} \end{equation} where $Y=\sqrt y$. The polynomial $P(t)=t^4+t^2-c t$ in the denominator has a degree 4, the integral can thus be expressed in terms of an elliptic integral of the first kind.

The polynomial $t^3+t-c$ is strictly increasing, it has thus one real root $t=\alpha$ and two complex ones, $t=\gamma$ and $t=\bar{\gamma}$. Moreover, the sign rule for the product of the roots indicates that $\alpha$ and $c$ have the same sign ($\gamma\bar{\gamma}>0$). In the following we take $c>0$, thus $\alpha>0$. Another discussion must be done in the opposite case. Then \begin{equation} P(t)=t(t-\alpha)(t-\gamma)(t-\bar{\gamma}) \end{equation} Now, we must discuss the value of $y$ with respect to $\alpha$.

If $0<y\le\alpha$ \begin{equation} J(Y)=\frac{1}{2i}\int_0^y\frac{dt}{\sqrt{t(\alpha-t)(t-\gamma)(t-\bar{\gamma})}} \end{equation} Expression 259.00 of the "Handbook of Elliptic integrals for scientists and engineers" by Byrd and Friedman (with $a=\alpha,b=0$) gives the result \begin{equation} J(Y)=\frac{g}{2i}F(\varphi,k) \end{equation} with \begin{align} g&=(AB)^{-1/2};\hspace{0.5cm}A^2=(\alpha-b_1)^2+a_1^2;\hspace{0.5cm}B^2=b_1^2+a_1^2\\ k^2&=\frac{\alpha^2-(A-B)^2}{4AB};\hspace{0.5cm}\varphi=\cos^{-1}\frac{(\alpha-y)B-yA}{(\alpha-y)B+yA}\\ b_1&=\Re\gamma;\hspace{0.5cm}a_1=\Im\gamma \end{align} Now if $0<\alpha<y$ \begin{equation} J(Y)=\left( \int_0^\alpha+\int_\alpha^y \right)\frac{dt}{\sqrt{t(t-\alpha)(t-\gamma)(t-\bar{\gamma})}} \end{equation} The first integral is obtained from the above formula and, from the expression 260.00 \begin{align} \int_\alpha^y \frac{dt}{\sqrt{t(t-\alpha)(t-\gamma)(t-\bar{\gamma})}}=gF(\varphi,k) \end{align} Here \begin{equation} k^2=\frac{(A+B)^2-\alpha^2}{4AB};\hspace{0.5cm}\varphi=\cos^{-1}\frac{(A-B)y+\alpha B}{(A+B)y-\alpha B} \end{equation} the other symbols are the same as above.

Section 3.145 of Gradshteyn and Ryzhik can also be useful.