Rectifying the ABC Lissajous curve with hypergeometric functions

Question

After successfully rectifying Fermat's spiral and the lituus, I set myself a new challenge: find the arc length of the Lissajous curve that forms the basis for the logo of the Australian Broadcasting Corporation.

The specific curve I want to rectify is $(\sin t,\cos3t)$ for $0\le t\le2\pi$. The arc length integral is $$\int_0^{2\pi}\sqrt{\cos^2t+(3\sin3t)^2}\,dt=13.065417\dots$$ After some substitutions I got my integral to $$2\int_{-1}^1\sqrt{\frac{18t^3-14t-5}{t^2-1}}\,dt\tag1$$ But this is ultimately a fifth-degree polynomial under the square root, which means that elliptic integrals will not work. However, seeing this fellow answer using the hypergeometric $_2F_1$ makes me believe that there is a solution to my problem using hypergeometric functions (not necessarily $_2F_1$).

Does the integral $(1)$ have a solution in terms of (generalised) hypergeometric functions? More generally, can integrals $\int R(t,\sqrt{P(t)})\,dt$, where $R$ is any rational function and $P$ any polynomial, be solved using hypergeometrics?

Of course, if all else fails, there's numerical integration.

Answer 1

This is not an answer.

Using $t=\sin ^{-1}(x)$, I end with the ugly $$I=4\int_0^1 \sqrt{\frac{144 x^6-216 x^4+80 x^2+1}{1-x^2} }\,dx$$ Now $x=\sqrt y$ $$I=2\int_0^1 \sqrt{\frac{144 y^3-216 y^2+80 y+1}{y\,(1-y)} }\,dy$$ I really do not see what I could do with this.

Using the inverse symbolic calculator (have a look here), the closest result seems to be

$$10 \times \, _1F_1\left(\frac{3}{50};\frac{11}{36};1\right)$$ but the difference is $5.04\times 10^{-8}$

Answer 2

This is also not an answer

I could argue that the length of $z=\sin t+i\cos 3t$ is the same as that of $z=\cos t+i\cos 3t$ (as well as show that that is true numerically). Then, I can express $z$ strictly in terms $x=\cos t$ as follows:

$$ z=x+i(4x^3-3x) $$

Now I can proceed to the arc length.

$$ s=\int_0^{2\pi} |\dot z| dt\\ \dot z=\dot x +i(12x^2\dot x-3\dot x)\\ \begin{align} |\dot z|&=\sqrt{\dot x^2+(12x^2-3)^2\dot x^2}\\ &=\dot x\sqrt{144x^4-72x^2+10}\\ \end{align} $$ $$ s=2\int_{-1}^1 \sqrt{144x^4-72x^2+10}\ dx $$

So, this is integrable according to Mathematica (Integrate[2Sqrt[144x^4-72x^2+10],{x}] or {x,-1,1}). The analytic solution is in terms of the arcsinh and elliptic integrals of the first and second kinds. The numerical solution is the same as the original. Not what you were looking for I suppose.