Arclength Parameterization of the Trefoil Knot I would like to find an arclength parameterization of the trefoil knot
The parameterizations I can find are:

$(sin(t) + 2sin(2t),$ $cos(t) - 2cos(2t),$ $sin(3t))$

and

$((2+cos(3t))cos(2t),$ $(2+cos(3t))sin(2t),$ $sin(3t))$

for $t \in [0,2\pi)$
writing $t = t(\theta)$, the magnitude of the derivatives wrt. $\theta$ of these two parameterizations are:
$| \frac{d}{d\theta}  (sin(t) + 2sin(2t),$ $cos(t) - 2cos(2t),$ $sin(3t)) |$ 
= $| (cos(t) + 4cos(2t),$ $-sin(t) + 4cos(2t),$ $3cos(3t))\frac{dt}{d\theta} |$ 
= $\sqrt{cos^2(t) + sin^2(t) + 16[cos^2(2t) + sin^2(2t)] + 8[cos(t)cos(2t) - sin(t)sin(2t)] + 9cos^2(3t)}\frac{dt}{d\theta}$
= $\sqrt{17 + 8cos(3t) + 9cos^2(3t)}\frac{dt}{d\theta}$
and similarly
$|\frac{d}{d\theta}((2+cos(3t))cos(2t),$ $(2+cos(3t))sin(2t),$ $sin(3t))|$
= $\sqrt{25 + 16cos(3t) + 4cos^2(3t)}\frac{dt}{d\theta}$
I need a solution to, for example 
$\int \sqrt{17 + 8cos(3t) + 9cos^2(3t)}dt$ so that I can get $t$ in terms of $\theta$
Wolfram alpha doesn't like either of these integrals, and I can see no way to solve them.
There are two types of answers to this question: one would solve one of these integrals, another would tell me how to change the parameterization, reasonably, so that the integral, and the obtained formula for t, is solvable.
I guess the third is to tell me that this question isn't solvable like this.
My thoughts on the latter: Below, I make the integral solvable by changing the parameterization, but I can't solve for $t(\theta)$
The freedom in the parameterizations can be expressed as:

$(sin(t) + Asin(2t),$ $cos(t) - Acos(2t),$ $Bsin(3t))$

where $A > 1$ and $B > 0$
which gives integrand:
$\sqrt{1 + 4A^2 + 4Acos(3t) + 9B^2cos^2(3t)}$
To eliminate the $\sqrt{}$ we want some factorization:

$(1 + 4A^2 + 4Acos(3t) + 9B^2cos^2(3t)) = (3Bcos(3t) + \lambda)^2$

where $\lambda^2 = 1 + 4A^2$ and $6B\lambda = 4A$
so $\lambda = \frac{2A}{3B} $
so $\frac{4A^2}{9B^2} = 1 + 4A^2$
ie. $4A^2\frac{9B^2 - 1}{9B^2} = -1$ or $4A^2 = \frac{9B^2}{1 - 9B^2} > 4$ 
so $9B^2 > 4/5$ and I will choose $9B^2 = 9/10$ giving:

$B = \frac{1}{\sqrt{10}}$
$A = 3/2$
$\lambda = \sqrt{10}$

ie. the trefoil $(sin(t) + \frac{3}{2}sin(2t),$ $cos(t) - \frac{3}{2}cos(2t),$ $\frac{1}{\sqrt{10}}sin(3t))$
has arclength parameterization given by $t(\theta)$ where:
$\int (\frac{3}{\sqrt{10}} cos(3t) + \sqrt{10}) dt = \theta$
$\frac{sin(3t)}{\sqrt{10}} + t\sqrt{10} = \theta$
But I want $t$ in terms of $\theta$

Since both integrands here are square roots of quadratic polynomials in $\cos3t$, they are reducible to elliptic integrals through the method I detailed here. There will be an extra algebraic/trigonometric term here, but it simplifies quite a bit because of preceding substitutions. I did all of the work in Mathematica and my notebook can be found here.
The end results for $0\le y\le\frac\pi3$ (all other values of $y$ follow by symmetry) are
$$\int_0^y\sqrt{9\cos^23t+8\cos3t+17}\,dt=L+\sqrt g\left(n\left(\frac{F(\varphi,m)-E(\varphi,m)}m\right)+\frac{16}{137}(1-m)\Pi(n,\varphi,m)\right)\\
\cos\varphi=\frac{Au-1}{A-u},u=\cos3y,L=\frac13\left(\frac{\sqrt{(1-u^2)(9u^2+8u+17)}}{u-A}+\frac43\tan^{-1}\sqrt{\frac{1-u^2}{u^2+8u/9+17/9}}\right)\\
A=\frac{-13-3\sqrt{17}}4,g=\frac{46359+11297\sqrt{17}}{192},m=\frac12-\frac2{3\sqrt{17}},n=\frac12-\frac{13}{6\sqrt{17}}$$
$$\int_0^y\sqrt{4\cos^23t+16\cos3t+25}\,dt=L+\sqrt g\left(n\left(\frac{F(\varphi,m)-E(\varphi,m)}m\right)+\frac{16}9(1-m)\Pi(n,\varphi,m)\right)\\
\cos\varphi=\frac{Au-1}{A-u},u=\cos3y,L=\frac13\left(\frac{\sqrt{(1-u^2)(4u^2+16u+25)}}{u-A}+4\tan^{-1}\sqrt{\frac{1-u^2}{u^2+4u+25/4}}\right)\\
A=\frac{-29-3\sqrt{65}}{16},g=\frac{-195+99\sqrt{65}}{512},m=\frac12-\frac7{2\sqrt{65}},n=\frac12-\frac{29}{6\sqrt{65}}$$
where all elliptic integral arguments follow Mathematica/mpmath conventions.
