If the Gauss hypergeometric function
$F(1, 3/2, 5/2; z^2) = 3 [\tanh^{1} (z) –z]/z^3$
what is the corresponding result for $F(1,15/8,23/8;z^8) $?



There is a special form for hypergeometric functions of the form $$ F(1,\beta,1+\beta,z) = \beta z^{\beta}B(z,\beta,0) $$ when $\beta = p/q$ is a rational number (where $B$ is the incomplete beta function). Write $$ \frac{z^\beta}{\beta}F(1,\beta,1+\beta,z) = \sum_{k\geq0} \frac{z^{\beta+k}}{\beta+k} = q\sum_{k\geq0}\frac{(z^{1/q})^{p+k q}}{p+k q} = q\sum_{k\geq0}\frac{(z^{1/q})^k}{k}[k\bmod q=p]. $$ The point is that you can use $$ \log(1x) = \sum_{k\geq1} \frac{x^k}{k} $$ and $$ \sum_{0\leq l < q} \zeta^{a l} = q[a\bmod q=0], \qquad \zeta=e^{2\pi i/q}, \quad a\in\mathbb{Z} $$ to write $$ \sum_{0\leq l < q} \zeta^{l p}\log(1\zeta^l x) = \sum_{k\geq1}\sum_{0\leq l < q} \frac{\zeta^{lp+lk}x^k}{k} = \sum_{k\geq1} \frac{x^k}{k}q[k\bmod q=p]. $$ So the closed form is $$ \frac{z^\beta}{\beta}F(1,\beta,1+\beta,z) = \sum_{0\leq l < q}\zeta^{lp}\log(1\zeta^l z^{1/q}), \qquad \beta=\frac pq, \quad p,q\in\mathbb{Z}. $$ I would be interested to know how this form should be generalized to irrational $\beta$. 

