I would like to compute the length of a parametric curve
$$ x(\xi) = -\frac{1}{2} \; x_1 \;\xi (1 - \xi) + x_2 \;(1 - \xi^2) + \frac{1}{2} \;x_3 \;\xi (1 + \xi) \\ y(\xi) = -\frac{1}{2} \;y_1 \;\xi (1 - \xi) + y_2 \;(1 - \xi^2) + \frac{1}{2} \;y_3 \;\xi (1 + \xi) $$
that passes between three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ in the limits $\xi = [-1, 1]$, using the formula:
$$ s = \int_{-1}^1 \sqrt{\left(\frac{dx}{d\xi}\right)^2 + \left(\frac{dy}{d\xi}\right)^2} d\xi $$
My approach I've taken is to substitute the differentials
$$ \frac{dx}{d\xi} = \frac{1}{2}(a\xi + b) \\ \frac{dy}{d\xi} = \frac{1}{2}(c\xi + d) $$
into the formula for the curve length which can now be written as
$$ s = \frac{1}{2}\int_{-1}^1 \sqrt{e \xi^2 + f \xi + g} \; d\xi $$
where
$$ \begin{align} a &= 2 x_1 - 4x_2 + 2 x_3 \\ b &= x_3 - x_1 \\ c &= 2 y_1 - 4y_2 + 2 y_3 \\ d &= y_3 - y_1 \\ e &= a^2 + c^2 \\ f &= 2(ab + cd) \\ g &= b^2 + d^2 \\ \end{align} $$
Finally I use the formula for the integral of the square root of a polynomial given here and arrive at
$$ s = \frac{\sqrt e}{4} \left[ (\xi + h) \left( \sqrt{(\xi + h)^2 + k} \right) + \log\left( \frac{\sqrt{(\xi + h)^2 + k} + (\xi + h)}{\sqrt k} \right) k \right] $$
where
$$ \begin{align} h &= \frac{f}{2e} \\ k &= \left(\frac{g}{e} - \frac{f}{4e^2} \right) \\ \end{align} $$
Unfortunately the results of my calculation indicate that the above formula isn't correct.
The most worrying aspect of this solution is that $k \rightarrow \infty$ as $e \rightarrow 0$.