Expressing hyperelliptic integrals through series Suppose I want to evaluate the following integral, but non-numerically:
$$\int_1^2\frac1{\sqrt{x^5-x+1}}\,dx$$
This is, of course, a hyperelliptic integral, which cannot in general be expressed in terms of elementary (or even elliptic) functions. Furthermore, it is quite well-known that the denominator polynomial here is not even solvable in radicals, but just assume we know the roots. (It is far easier to numerically find polynomial roots than to numerically integrate.)
Byrd and Friedman have this to say about the matter:

For the evaluation of hyperelliptic integrals, one must usually resort to direct numerical integration or to the use of complicated series expansions.

This second part has me intrigued. How can I express the integral above using series, hypergeometric or otherwise? Is there a general procedure to derive such a series representation?
Note that if there were just two terms in the polynomial, the integrand would be expressible as a binomial series; the integral itself would then be expressible using the $_2F_1$ function. But there are three terms here, so $_2F_1$ cannot be immediately applied.
Taylor and Padé expansions are not quite desirable either, since they require much work at extended precision and will not converge over the entire domain if truncated.
 A: $$\begin{align}\int_1^2\dfrac{1}{\sqrt{x^5-x+1}}~dx&=\int_1^\frac{1}{2}\dfrac{1}{\sqrt{\dfrac{1}{t^5}-\dfrac{1}{t}+1}}~d\left(\dfrac{1}{t}\right)
\\&=\int_\frac{1}{2}^1\dfrac{1}{t^2\sqrt{\dfrac{t^5-t^4+1}{t^5}}}~dt
\\&=\int_\frac{1}{2}^1\dfrac{\sqrt t}{\sqrt{1-t^4(1-t)}}~dt
\\&=\int_\frac{1}{2}^1\sqrt t\sum\limits_{n=0}^\infty\dfrac{(2n)!t^{4n}(1-t)^n}{4^n(n!)^2}~dt
\\&=\int_\frac{1}{2}^1\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(2n)!C_k^nt^{4n+\frac{1}{2}}(-t)^k}{4^n(n!)^2}~dt
\\&=\int_\frac{1}{2}^1\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^k(2n)!t^{4n+k+\frac{1}{2}}}{4^nn!k!(n-k)!}~dt
\\&=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^k(2n)!t^{4n+k+\frac{3}{2}}}{4^nn!k!(n-k)!\left(4n+k+\dfrac{3}{2}\right)}\right]_\frac{1}{2}^1
\\&=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^k(2n)!}{2^{2n-1}n!k!(n-k)!(8n+2k+3)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^k(2n)!}{2^{6n+k+\frac{1}{2}}n!k!(n-k)!(8n+2k+3)}
\\&=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^k(2n)!}{2^{2n-1}n!k!(n-k)!(8n+2k+3)}-\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^k(2n)!}{2^{6n+k+\frac{1}{2}}n!k!(n-k)!(8n+2k+3)}
\\&=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^k(2n+2k)!}{2^{2n+2k-1}(n+k)!n!k!(8n+10k+3)}-\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^k(2n+2k)!}{2^{6n+7k+\frac{1}{2}}(n+k)!n!k!(8n+10k+3)}\end{align}$$
Which relates to Srivastava-Daoust Function
