Example: The order 3 Bezier Curve
Let g(t) be the length² of the tangent:
And .
The arc length of the order 3 Bezier curve is: .
The integral will not bow to formal manipulation. But f(t) can be easily evaluated at a series of points. Using these points, we can approximate f(x) by polynomials which are easily integrated.
Here is a function explorer for selecting and arranging the interpolating polynomials.

NOTES:

The Bezier curve may have at most two cusps where the tangent reaches a local minimum.
Here, the cusps can be given by the red control points. Since these points present the most difficulty for polynomial approximation, I have built the test function so that both cusps exist, and can be made arbitrarily sharp (drag either red point toward the x-axis).

The Lagrange polynomial is the ordinary polynomial through n points (order n-1).
The Hermitian polynomial matches f(x) and its first derivative at each control point. For n points, the curve is order 2n-1.

If you think you have a good approximation, remember to drag the red points around. Do you have a plan for convergence?

The second graphics view provides information about the error for the selected subinterval.

GGB will agree that two numbers are equal if the relative difference is roughly of the order 10^(-8).