This answer determines that the arc length of Fermat's spiral $r=k\sqrt\theta$ over $\theta\in[0,t]$ is given by $$\int_0^tk\sqrt{\frac{1}{4\theta}+\theta}\,d\theta=k\sqrt t\ _2F_1\left(-\frac12,\frac14;\frac54;-4t^2\right)$$ On that answer I commented
I challenge you to find the arc length of the lituus [$r=k\sqrt\theta$] using hypergeometric functions, like I did with elliptic integrals in my answer.
The reply came that the answerer already had the answer for that too. I then remembered that MathWorld states that for the general Archimedean spiral $r=k\theta^{1/n}$, if $n>0$ the arc length over $\theta\in[0,t]$ is $$L_n(k,t)=kt^{1/n}\ _2F_1\left(-\frac12,\frac1{2n};1+\frac1{2n};-n^2t^2\right)$$ (If $n<0$ the spiral is unbounded both ways and the length over $\theta\in[a,b]$ where $0<a\le b$ is $L_n(k,a)-L_n(k,b)$).
It's all nice, but how do we derive the hypergeometric expression for $L_n(k,t)$? Please include a full derivation; don't just lean on the results from Mathematica or any other symbolic software.
Let $\theta=\frac x n$ which makes $$\int \sqrt{r^2+(r')^2}\,d\theta=k \left(\frac{1}{n}\right)^{\frac{1}{n}+1}\int x^{\frac{1}{n}-1}\sqrt{x^2+1}\,dx $$ $$x^{\frac{1}{n}-1}\sqrt{x^2+1}=\sum_{m=0}^\infty \binom{\frac{1}{2}}{m} x^{2 m+\frac{1}{n}-1}$$ $$\int x^{\frac{1}{n}-1}\sqrt{x^2+1}\,dx =\sum_{m=0}^\infty \frac{\binom{\frac{1}{2}}{m} }{2 m+\frac{1}{n}}x^{2 m+\frac{1}{n}}=n x^{\frac{1}{n}} \, _2F_1\left(-\frac{1}{2},\frac{1}{2 n};1+\frac{1}{2 n};-x^2\right)$$