$$ \int_{x_1} ^{x_2}\sqrt{1+f^{'}(x)^2}dx$$
I would separately determine limits $x_1, x_2 $ as well as $x_3$(vertex) of the parabola
$y= a x^2+b x+c$
getting length before inserting limits:
$ \frac{1}{4a}\left[(2ax+b)+\sqrt{1+(2ax+b)^2}+\log\left((2ax+b)+\sqrt{1+(2ax+b)^2}\right)\right] $
Is it so far correct? Much appreciated.
By translating the parabola in such a way that the $x$-coordinate of the vertex is zero, it is trivial that we may assume $y=ax^2$, hence we just need to check that: $$ \int_{0}^{u}\sqrt{1+4a^2 x^2}\,dx = \frac{u}{2}\sqrt{1+4a^2 u^2}+\frac{1}{4a}\operatorname{arcsinh}(2au).$$ It is interesting to notice that the last integral is related with the area of the circle (it is just its hyperbolic counterpart): $$ \int_{0}^{u}\sqrt{1-x^2}\,dx = \frac{u}{2}\sqrt{1-u^2}+\frac{1}{2}\arcsin u.$$ First piece on the RHS given by the area of a triangle, second piece given by the area of a circular sector.