Legendre-type relations for generalized complete elliptic integrals by Shingo Takeuchi

Legendre’s relation for the complete elliptic integrals of the first and second kinds is generalized. The proof depends on an application of the generalized trigonometric functions and is alternative to the proof for Elliott’s identity.

The entire paper is available here.

Evaluate the Integral \(\int_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}} \)

How to evalute
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution

The integral is of the form of the polynomial under the radical having 1 real and 2 complex roots and the lower limit of integration equal to the real root. So we have
\begin{equation}
\int\limits_{a}^{x} \left( (t-a)[(t-r)^{2} + s^{2}] \right)^{-1/2}\, dt
= \frac{1}{\sqrt{m}} \mathrm{cn}^{-1}\left(\frac{m-(t-a)}{m+(t-a)},k \right)
\end{equation}
where
\begin{equation}
m^{2} = (r – a)^{2} + s^{2} \quad \mathrm{and} \quad k^{2} = \frac{1}{2} – \frac{a-r}{2m}
\end{equation}
\(\mathrm{cn}^{-1}z\) is the inverse of one of the Jacobi elliptic functions and \(k\) is the modulus.

To factor the polynomial inside of the radical, we note that the roots are the cube roots of 1, thus
\begin{align}
t^{3}-1 &= (t-t_{1})(t-t_{2})(t-t_{3}) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]-i\frac{\sqrt{3}}{2} \right) \left(\Big[t+\frac{1}{2}\Big]+i\frac{\sqrt{3}}{2} \right) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]^{2} + \frac{3}{4}\right)
\end{align}

We now have
\begin{equation}
a = 1 \quad r = -\frac{1}{2} \quad s = \frac{\sqrt{3}}{2} \quad m^{2} = 3 \quad k^{2} = \frac{1}{2} – \frac{3}{4\sqrt{3}}
\end{equation}
and thus
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
= 3^{-1/4} \mathrm{cn}^{-1}\left(\frac{\sqrt{3} + 1 – x}{\sqrt{3} – 1 + x},\sqrt{\frac{1}{2} – \frac{3}{4\sqrt{3}}} \right)
\end{equation}