Yes, if they had then those particular elliptic integrals would have a closed form. Some degenerate classes of elliptic integrals do have closed forms, in precisely the same way that you can determine whether certain subsets of programs halt despite the halting problem in general being insoluble! I think you misunderstand what is meant by the statement "elliptic integrals in general have no closed form".
No, I don't think I misunderstand, and I don't think my reduction is invalid. I take a Bezier curve, and reduce the expression of its arc length to an elliptic integral. If this particular elliptic integral has no closed form, that means that this particular Bezier curve doesn't have closed form arc length parametrization and also that Bezier curves in general don't admit such parametrization. Because that's what "X in general is not Y" means: that there is at least one X which is not Y.
I agree with everything you've written in this reply, but originally you were appealing to the fact that elliptic curves in general don't have closed forms ("they don't, though") to prove your point, which isn't valid because we're talking about a strict subset of them here.
