Arc length of $x^n$ found using Hypergeometric function and series. Alternate representations and solution verification needed.

Take a look at the over 111000 values for the hypergeometric function. This shows that many results can be derived from using this genius function. Now for the arc length formula derivation. An arc length function for the arc length of $x^n$ would be messy, so we want something like this using the main arc length formula, not for others like this or this one, but rather for y=f(x). Series expansion for the hypergeometric function. There may be a small typo in here:

$$\mathrm {S(n,x)=\text{Arc length with respect to x}(\text{monomial power function})=}$$ $$\mathrm{\int\sqrt{1+x^n}dx=\int\sqrt{1+\left(y’(x)\right)^2}\implies y=\pm\int x^{n/2}dx=c\pm \frac{2x^{\frac n2+1}}{n+2}\implies \text{S(n,x)=[Arc length with respect to x]}\left(c\pm \frac{2x^{\frac n2+1}}{n+2}\right)=}$$ $$\mathrm{x \ _2F_1\left(-\frac12, \frac1n,\frac1n+1,-x^n\right)=\left\{\sum_{k=0}^\infty \frac{(-1)^k\left(-\frac12\right)_k\left(\frac1n\right)_k x^{kn+1}}{\left(\frac1n+1\right)_kk!},|-x^n|\le 1\ \text{or} \ \frac{Γ\left(\frac1n +\frac12\right)Γ\left(\frac1n+1\right)x^{\frac n2}}{Γ\left(\frac1n\right)Γ\left(\frac1n+\frac32\right)}\sum_{k=0}^\infty \frac{(-1)^k\left(-\frac12\right)_k\left(-\frac1n-\frac12\right)_k}{\left(\frac12-\frac1n\right)_kk! x^{kn-1}} +C\sum a_n (0)_k,|-x^n|>1,n\not\in -\frac{2}{2\Bbb Z+1} \right\}=}$$ $$\mathrm{\left\{-\frac1{2\sqrt\pi}\sum_{k=0}^\infty \frac{(-1)^kΓ\left(k-\frac12\right) x^{kn+1}}{(kn+1)k!},|-x^n|\le1\ \text{or}\ \frac{x^{\frac n2+1}}{\sqrt\pi}\sum_{k=0}^\infty \frac{(-1)^k Γ\left(k-\frac12\right)}{(2kn-n-2)k!x^{kn}},\quad |-x^n|>1,n\not\in -\frac{2}{2z+1},z\in\Bbb Z \right\}}$$

This large expression would be useful for seeing the sum version of the series. This brings up the question of if this is the correct summation expansion using the main definition of the hypergeometric function. If there was a mistake somewhere, I would like to know where and have it corrected in an answer. Assuming this form is correct using the linked “main definition”, then what is another representation of this S(n,x) function?

Your hypergeometric representation for $S$ is indeed correct. Another way to obtain this is to write $$ \begin{align} S &=\int(1+x^n)^{1/2}\,\mathrm dx\\ &\overset{u=x^n}{=}\frac{1}{n}\int u^{1/n-1}(1+u)^{1/2}\,\mathrm du\\ &=\frac{1}{n}\int_0^u z^{1/n-1}(1+z)^{1/2}\,\mathrm dz\\ &\overset{z=ut}{=}u^{1/n}\frac{1}{n}\int_0^1 \frac{t^{1/n-1}(1-t)^{(1+1/n)-1/n-1}}{(1-(-u)t)^{-1/2}}\,\mathrm dt, \end{align} $$ and then use DLMF 15.6.1 to identify the remaining integral as that of the hypergeometric function $F(-1/2,1/n;1+1/n;-u)$. Then, reintroducing $u=x^n$ gives $$ S=x\,{_2F_1}\biggl({-\frac{1}{2},\frac{1}{n}\atop 1+\frac{1}{n}};-x^n\biggr). $$ As for a general procedure for reducing this to simpler functions you are in good company. Of course you have the list of reduction formulae available through Wolfram functions, which will give you results in terms of more elementary functions for specific values of $n$. As for other potentially interesting forms you can use differential formula in the linked question above to show $y(z)=F(-1/2,1/n;1+1/n;z)$ satisfies $$ zy^\prime+\tfrac{1}{n}y=\tfrac{1}{n}\sqrt{1-z},\quad y(0)=1. $$ If $n=1$ this gives $$ zy^\prime+y=\sqrt{1-z}\implies \partial_z(zy)=\sqrt{1-z}, $$ which immediately lends to finding a simpler expression for $S$.

Another form for $S$ that may be of interest to you is that terms of the incomplete beta function, which can be found directly with this relation.