Arc lengths for Bezier curves are only closed form for linear and quadratic ones. For cubics, it is not guaranteed to have a closed solution.
The reason is arc length is defined by a radical integral, for which has a closed for only 2nd degree polynomials.
Just for reference: The length of a quadratic Bezier for the points (a,p) (b,q) and (c,r) is
(a^2·(q^2 - 2·q·r + r^2) + 2·a·(r - q)·(b·(p - r) + c·(q - p)) + (b·(p
- r) + c·(q - p))^2)·LN((√(a^2 - 2·a·b + b^2 + p^2 - 2·p·q + q^2)·√(a^2 + 2·a·(c - 2·b) + 4·b^2 - 4·b·c + c^2 + (p - 2·q + r)^2) +
a^2 + a·(c - 3·b) + 2·b^2 - b·c + (p - q)·(p - 2·q + r))/(√(a^2 +
2·a·(c - 2·b) + 4·b^2 - 4·b·c + c^2 + (p - 2·q + r)^2)·√(b^2 - 2·b·c +
c^2 + q^2 - 2·q·r + r^2) + a·(b - c) - 2·b^2 + 3·b·c - c^2 + (p - 2·q
+ r)·(q - r)))/(a^2 + 2·a·(c - 2·b) + 4·b^2 - 4·b·c + c^2 + (p - 2·q + r)^2)^(3/2) + (√(a^2 - 2·a·b + b^2 + p^2 - 2·p·q + q^2)·(a^2 + a·(c -
3·b) + 2·b^2 - b·c + (p - q)·(p - 2·q + r)) - √(b^2 - 2·b·c + c^2 +
q^2 - 2·q·r + r^2)·(a·(b - c) - 2·b^2 + 3·b·c - c^2 + (p - 2·q + r)·(q
- r)))/(a^2 + 2·a·(c - 2·b) + 4·b^2 - 4·b·c + c^2 + (p - 2·q + r)^2)
Where LN is the natural logarithm, and ^ denotes power and √ the square root.
Hence, it should be easier and cheaper approximate the arc by some other rule, like a polygon or an integration scheme like Simpson's rule, because square roots the LN are expensive operations.