A quadratic bezier is expressed as follows, where A, B and C are our control points. We'll assume these are 2d vectors. t represents how far along the curve we are. $\textbf{g(}t\textbf{)} = A (1 - t)^2 + 2 B (1 - t) t + C t^2, t \epsilon [0, 1]$

The arc length of any curve can be expressed as follows $\int \sqrt{\delta x^2 + \delta y^2}$ We can think of $g'(t)$ as follows $\dfrac{\delta x}{\delta t} = g'_{x}(t), \dfrac{\delta y}{\delta t} = g'_{y}(t)$ Therefore $\int \sqrt{\delta x^2 + \delta y^2} = \int \sqrt{\dfrac{\delta x^2}{\delta t^2} + \dfrac{\delta y^2}{\delta t^2}} \delta t = \int \sqrt{(g_{x}'(t))^2 + (g_{y}'(t))^2} \delta t = \int ||g'(t)|| \delta t$

We need to find $||g'(t)||^2$ $g'(t) = 2[(A-2B+C)t + (B - A)]$ Therefore $||g'(t)||^2 = 4(||A - 2B + C||^2 t^2 + (A - 2B + C)\cdot(B - A)t + ||B - A||^2)$

This leavs us with a root of a quadratic polynomial in the following form. $a = 4||A - 2B + C||^2, b = 4(A - 2B + C)\cdot(B - A), c = 4||B - A||^2 \\ \int ||g'(t)|| \delta t = \int \sqrt{a t^2 + b t + c} \delta t$ Dividing by "a" and converting to a depressed qradratic we get $m = \dfrac{b}{a}, n = \dfrac{c}{a}, \sqrt{a} \int \sqrt{t^2 + m t + n} \delta t = \sqrt{a} \int \sqrt{(\dfrac{m}{2} + t)^2 + n - \dfrac{m^2}{4}} \delta t$ We do some u sub giving us a quadratic in the following form $k = n - \dfrac{m^2}{4}, u=\dfrac{m}{2}+t, \delta u = \delta t, \sqrt a \int \sqrt{u^2 + k} \delta u$ For this to work we need to ensure that k is positibe or we're going to have a bad day. We'll assume A, B and C are scaler in this example for simplicity even though they're not. $p = (A - 2B + C), q = (B - C) \\ a = p^2, b = p q, c = q^2 \\ k = \dfrac{c}{a} - \dfrac{b^2}{4a^2} = \dfrac{q^2}{p^2} - \dfrac{p^2q^2}{4p^4} = \dfrac{q^2}{p^2} - \dfrac{q^2}{4p^2} \\ \dfrac{q^2}{p^2} > \dfrac{q^2}{4p^2}$ This proves k is positive. Going back to our integral, we then divide our integral by $\sqrt k$. $\sqrt {a k} \int \sqrt{(\dfrac{u}{\sqrt k} )^2 + 1} \delta u$ More u sub $v = \dfrac{u}{\sqrt k}, \delta v = \dfrac{\delta u}{\sqrt k}, k \sqrt a \int \sqrt{v^2 + 1} \delta v$ Now we can subsitute $v = tan(\Theta)$ $\delta v = sec^2(\Theta) \delta \Theta, k \sqrt a \int \sqrt{tan^2(\Theta) + 1} sec^2(\Theta) \delta \Theta \\ = k \sqrt a \int \sqrt{sec^2(\Theta)} sec^2(\Theta) \delta \Theta \\ = k \sqrt a \int sec^3(\Theta) \delta \Theta$ Now do integration by parts $\int sec^3(\Theta) \delta \Theta = \int sec^2(\Theta) sec(\Theta) \delta \Theta \\= sec(\Theta) tan(\Theta) - \int tan^2(\Theta) sec(\Theta) \delta \Theta \\= sec(\Theta) tan(\Theta) - \int sec^3(\Theta) - sec(\Theta) \delta \Theta$ Now add $\int sec^3(\Theta) \delta \Theta$ to both sides $2 \int sec^3(\Theta) \delta \Theta = sec(\Theta) tan(\Theta) + \int sec(\Theta) \delta \Theta$ Integrating we get $\int sec^3(\Theta) \delta \Theta = \frac{1}{2} sec(\Theta) tan(\Theta) + \frac{1}{2}ln|sec(\Theta) + tan(\Theta)|$ Now substituting in $sec^2 = tan^2 + 1$ we get $\int sec^3(\Theta)\delta\Theta = \frac{1}{2} tan(\Theta) \sqrt{1 + tan^2(\Theta)} + \frac{1}{2} ln|\sqrt{1 + tan^2(\Theta)} + tan(\Theta)|$ Now substituting in $v = tan(x)$ and adding our $k\sqrt a$ back we get the following. $k\sqrt a \int \sqrt{v^2 + 1}\delta v = \frac{k\sqrt a}{2} [v \sqrt{1+v^2} + ln(\sqrt{1+v^2} + v)]$ Finally we need to calculate v and k in terms of A, B, C and t. I might also multipley my integral by two then divide a b and c by four. $P = (A - 2B + C), Q = (B - C)$ $a = P \cdot P, b = P \cdot Q, c = Q \cdot Q, k = \dfrac{3 c}{4 a}, z = \dfrac{b}{2a\sqrt k}, v = z + \dfrac{t}{\sqrt k}$ $arclength = k\sqrt a (v \sqrt{1+v^2} + ln(\sqrt{1+v^2} + v) - z \sqrt{1+z^2} + ln(\sqrt{1+z^2} + z) )$ And we're done