I find myself needing to compute the integral $$\int_{x = 0}^X\left(1+x^2\right)^{8/3}\mathrm{d}x$$ for real, positive $X$. Gradshteyn and Ryzhik (2007 edition) paragraph 2.202 implies that the answer can't be expressed in terms of elementary functions, but I really don't mind it being expressed in terms of special functions. Does this integral look familiar to anyone as the definition of a special function, please?
(If it helps, I found substitutions that convert the integral to $\int\cosh^{19/3}\left(\theta\right)\mathrm{d}\theta$ or $\int\sec^{22/3}\left(\theta\right)\mathrm{d}\theta$; or repeated integration by parts could reduce it to $\int_{x = 0}^Xx^6\left(1+x^2\right)^{-1/3}\mathrm{d}x$.)
According to Maple, it's $$X \; {}_{2}{{{F_{1}^{}}}}\! \left(-\frac{8}{3},\frac{1}{2};\frac{3}{2};-X^{2}\right)$$
And more generally,
$$ \int (1+x^2)^p \; dx = x \; {}_{2}{{{F_{1}^{}}}}\! \left(-p,\frac{1}{2};\frac{3}{2};-x^{2}\right) + c$$
EDIT:
From the definition, $$x\; {}_2F_1\left(-p,1/2; 3/2; -x^2\right) = \sum_{n=0}^\infty \frac{\Gamma(n-p) (-1)^n x^{2n+1}}{\Gamma(-p) (2n+1) n!} $$ Differentiate term-by-term and you get $$ \sum_{n=0}^\infty \frac{\Gamma(n-p) (-1)^n x^{2n}}{\Gamma(-p) n!} = \sum_{n=0}^\infty \frac{\Gamma\left(p+1\right) x^{2n}}{\Gamma\left(p-n+1\right) n!} = (1+x^2)^p $$ by the binomial series.