I have a curve in a three dimensional space and I want to get the arc length function. So I defined the curve and its velocity and asked for the integral of the norm of its speed:

alpha[t_] := {Cosh[t], Sinh[t], t}
alphap[t_] := D[alpha[t], t]
Integrate[Norm[alphap[t]], t]

I got back

\[Integral]Sqrt[1 + Abs[Cosh[t]]^2 + Abs[Sinh[t]]^2] \[DifferentialD]t

I tried it with Rubi:

Int[Norm[alphap[t]], t]

and got back:

 Int[Sqrt[1 + Abs[Cosh[t]]^2 + Abs[Sinh[t]]^2], t]

How can I get the system to give me the integral, or is it just that a closed form does not exist?


By default, MMA assumes expressions to be complex:


Sqrt[Abs[x]^2 + Abs[y]^2 + Abs[z]^2]

If you look at Norm[alphap[t]], there's an Abs raised to the second power - hence, for real values, the Abs is redundant. Simply getting rid of it still requires an assumption that $t\in \mathbb{R}$, so let's go directly there:

Integrate[Norm[alphap[t]], t, Assumptions -> t ∈ Reals]

Sqrt[2] Sinh[t]

Also one could Simplify the integrand with the assumption:

Integrate[Simplify[Norm[alphap[t]], t ∈ Reals], t]

Sqrt[2] Sinh[t]

but it's to convoluted compared to Assumptions, and redundant.

| improve this answer | |
  • $\begingroup$ How to I get the "member of" symbol you used? (the backwards E) $\endgroup$ – Gene Naden Jul 18 '18 at 18:23
  • $\begingroup$ Element $\endgroup$ – corey979 Jul 18 '18 at 18:25
  • $\begingroup$ Also: <Esc> elem<Esc> $\endgroup$ – rhermans Jul 20 '18 at 7:55

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