Arc length of general polynomial I am coding a computer program and I am in need of calculating the arc length of a curve described by a polynomial $P(x)=ax^n+bx^{n-1}+\dots+cx+d$ within $[x_0, x_1]$, where $n\ge2$.
Fast methods for $P'(x)$ and $\int P(x) \, dx$ calculations are available and can be used at will.
I have tried expanding $\displaystyle\int_{x_0}^{x_1}\sqrt{1+(P'(x))^2} \, dx$ without success. However, it can be used any other formula or method. The big issue here is getting a result - or approximation of it - fast.
 A: In general, the best thing to do is use numerical integration methods.  But if you want a series expansion, and $P'(x) > 1$ everywhere, you could try
$$ \sqrt{1+y^2} = y \sqrt{1/y^2 + 1} = y + \dfrac{1}{2y} - \dfrac{1}{8y^3} + \dfrac{1}{16 y^5} - \dfrac{5}{128 y^7} + \dfrac{7}{256 y^9} + \ldots $$
EDIT:
If speed is the issue, then numerical integration is definitely the way to go.
Simpson's Rule will provide reasonably good results and is very fast and robust.
A: Numerical integrations seems to me the only practical solution. Taylor series built for $P(x)$ at $x=x_0$ or $x=\frac{x_0+x_1}{2}$ lead to nightmares; the same for Pade approximants.
I should personally use sophisticated Runge-Kutta integrations or other similar methods. 
Added later
Suppose that $x_1$ is close to $x_0$ (you can make this subdividing your real interval into smaller pieces). Now, expand $P(x)$ as a second order Taylor series at $x=\frac{x_0+x_1}{2}$. So, basically $$P(x) \simeq a + b x + c x^2$$ and now compute the arc length $$\displaystyle\int\sqrt{1+(P'(x))^2} \, dx=\int\sqrt{1+(b+2 c x)^2} \, dx$$ $$\displaystyle\int\sqrt{1+(P'(x))^2} \, dx=\frac{(b+2 c x)\sqrt{(b+2 c x)^2+1} +\sinh ^{-1}(b+2 c x)}{4 c}$$ Let me try with very bad conditions:$$P(x)=1+x+x^2+x^3+x^4+x^5+x^6$$ and let us compute the arc length between $4$ and $5$. The approximation I suggest leads to $13557$ while the solution is $14070$.
