If you have found a parametrization $\vec \alpha(s)$ of a curve $C$ for which $\int \lVert \vec \alpha\,'(s)\rVert \mathrm ds$ cannot be expressed in terms of elementary functions, does it make sense to search for a different parametrization?
This question was inspired by this specific example. I feel like the answer is no, and here is a sketchy proof:
Suppose we have two parametrizations of the curve $C$, say $\vec \alpha (s)$ and $\vec \mu (t)$ valid on $s\in[a,b]$ and $t\in[c,d]$, with continuous derivatives. Then there is some function $f$ such that $\vec \alpha(s)=\vec\mu(f(s))$. Now if $\vec \alpha\,'(s)$ and $\vec \mu \, '(t)$ are never zero (which I believe we can assume without loss of generality) then we have $\vec \alpha \, ' (s)=\vec \mu\,'(f(s))f'(s)$ (the existence of $f'(s)$ is guaranteed in this case).
Furthermore, since $f$ is differentiable it is continuous. Now the fact that $\vec \alpha$ and $\vec \mu$ are differentiable implies continuity of each function $\alpha_i$, where $\vec \alpha(s)=(\alpha_1,\,\alpha_2,\,\dots,\alpha_n)$. Similarly, each $\mu_i$ is continuous, and for all $t$ at least one of $\mu_i(t)\ne 0$. Hence we have that $f'(s)=\frac {\alpha_i'(s)}{\mu_i'(f(s))}$.
There must be some overlap in non-zero functions, however. I'm not sure how to write this formally, but for example: if we have that $\mu_1'(f(s))$ is the only nonzero $\mu_i'(f(s))$ on some interval $(x,y)$, then $f'(s)=\frac{\alpha_1'(s)}{\mu_1'(f(s))}$ on $(x,y)$. Now suppose $\mu_1'(f(s))=0$ outside this interval. Then on $[y,z)$ there must be some $i=k$ such that $\mu_k'(f(s))$ is nonzero. Since $\mu_k'(f(s))$ is continuous, it must be nonzero on an open interval (including $y$), and therefore $\mu_1'(f(s))$ is not the only nonzero function on $(x,y)$.
This overlap allows us to conclude that $f'(s)$ must be continuous (that was the whole point of the above argument) and nonzero. Thus we either have that $f'(s)>0$ on $[a,b]$ or $f'(s)<0$ on $[a,b]$. Either way, we have that $\int \lVert \vec \alpha\,'(s)\rVert \mathrm ds =\pm\int \lVert \vec \mu\,'(f(s))\rVert f'(s)\mathrm ds=\pm\int\lVert\vec\mu\,'(t)\rVert\mathrm dt$ and so if the arc length using the parametrization $\vec \alpha$ cannot be expressed in terms of elementary functions, the new parametrization cannot either.
I'm not positive we can assume that $\vec\alpha\,'(s)$ and $\vec \mu\,'(t)$ are never zero without loss of generality, but it seems to make sense. Actually, looking at the result I believe we could argue that it wouldn't make a difference if they were equal to zero at some points.
It also occurred to me that the argument yielding continuity of $f'(s)$ was not really needed, it was obvious that $f'(s)$ was at least piecewise continuous and we can reach our conclusion with that weaker result as well.