This question was written because I have come across a few arc length functions which cannot be evaluated. This also has happened to me when trying to solve some problems of my own. Too often an arc length integral is non elementary causing a conflict with the integration of the function. The only way to take an arc length integral for any function would just be to use the Riemann Sum definition of the integral and maybe painstakingly put it into an alternate form. For example: $$\mathrm{Arclength(f(x), a\le x\le b)=\int_a^b \sqrt{\bigg(\frac{d}{dx}f(x)\bigg)^2+1}\ dx}$$$$\mathrm{=\lim_{N\to \infty}\frac{b-a}{N}\sum_{n=1}^N \sqrt{\bigg[\frac{d}{dx}f\bigg(x+n\frac{b-a}{N}\bigg)\bigg]^2+1}}$$
Let: $$\mathrm{\bigg[\frac{d}{dx}f(x)\bigg]^2+1=T(f(x))=T(f)} $$ This substitution naturally leaves the question of:
$$\mathrm{\sqrt{T(f)}=\sum_{n=u}^vt_n(x)}$$ In conclusion, what is the value of $\mathrm{u,v, and \ t_n(x)}$? One could always try using binomial series, but that would require |T(f)-1|<1 or simply 0<T(f)<2.
My question is what would would be a way to evaluate any arc length function such that we get rid of the square root perhaps through a series expansion? Easily, a substitution may be done to cancel the square root term, but would probably make the integrand more complicated. I have given a well written question, context and some attempts, so please correct me and give me feedback!
You can make series expansions around $x=c$ such that $(a \leq c \leq b)$.
Let $$g(x)=f'(x)=\sum_{n=0}^\infty \alpha_n (x-c)^n$$ For conveniency, let $y=x-c$ and we shall need to truncate the expansion to some order $$g(x)=\sum_{n=0}^p \alpha_n y^n+O(y^{p+1})$$ this would give $$\sqrt{1+g(x)^2}=\sum_{n=0}^p \beta_n y^n+O(y^{p+1})$$ but the relation between the coefficients become quite complex.
For example, using $p=3$,this would give $$\beta_0=\sqrt{1+\alpha_0^2}\qquad \beta_1=\frac {\alpha_0 \alpha_1}{\beta_0}\qquad \beta_2=\frac {\alpha_1^2+2\alpha_0\alpha_2\beta_0^2}{2\beta_0^3}$$ $$\beta_3=\frac {-\alpha_0 \alpha_1^3+2 \beta_ 0^2 \alpha_2 \alpha_1+2 \alpha_0 \beta_ 0^4 \alpha_3 }{2\beta_0^5}$$
Let us try for $$\int_1^3 \sqrt{1+e^{2x}}\,dx$$ making the expansion around $x=2$. It will gives for the integrand $$\sqrt{1+e^4}+\frac{e^4 (x-2)}{\sqrt{1+e^4}}+\frac{e^4 \left(2+e^4\right) (x-2)^2}{2 \left(1+e^4\right)^{3/2}}+\frac{e^4 \left(4+2 e^4+e^8\right) (x-2)^3}{6 \left(1+e^4\right)^{5/2}}+O\left((x-2)^4\right)$$ and the integration will lead to $$L=\frac{6+14 e^4+7 e^8}{3 \left(1+e^4\right)^{3/2}}\sim 17.3975$$ while the exact result is
$$L=-\sqrt{1+e^2}+\sqrt{1+e^6}+\tanh ^{-1}\left(\sqrt{1+e^2}\right)-\tanh ^{-1}\left(\sqrt{1+e^6}\right)\sim 17.5243$$ This make a relative error of $0.72$%.
Adding one term to the expansion would give $$L=\frac{40+176 e^4+272 e^8+188 e^{12}+47 e^{16}}{20 \left(1+e^4\right)^{7/2}}\sim 17.5214$$ reducing the relative error to $0.017$% ($43.4$ times lower than the previous)
Edit
As asked in comments, I had a look at the problem of the arc length of function $$f(x)=\tan[\log(x)]$$ asked in this question and I chose for the inteval of integration $\left(\frac e2,\frac {3e}2\right)$. As anyone could expect, the expansion of $$g(x)=\frac{\sec ^2[\log (x)]}{x}$$ was done around $x=e$ to order $(2n+1)$. Below are given the results $$\left( \begin{array}{cc} n & \text{result} \\ 0 & 4.37301 \\ 1 & 5.59580 \\ 2 & 6.15811 \\ 3 & 6.37893 \\ 4 & 6.47506 \\ 5 & 6.51552 \\ 6 & 6.53216 \\ 7 & 6.53943 \\ 8 & 6.54242 \\ 9 & 6.54368 \end{array} \right)$$
while the "exact" value is $6.54461$.