derivative of hypergeometric function I am doing an integral in Mathematica and I find the solution contains derivatives of hypergeometric functions. I would like (ideally) a simple analytic form for these. I have tried HypExp mathematica package but there is no derivative capability. 
I would like to know how to simplify the following functions:
$\text{Hypergeometric2F1}^{(0,0,1,0)}\left(1,1,2,-\frac{2}{a-2}\right)$
$\text{Hypergeometric2F1}^{(0,1,0,0)}\left(1,1,2,-\frac{2}{a-2}\right)$
Thanks
 A: A closed form in terms of the non-elementary dilogarithm is available for the first expression:
$$S_1:= \frac{d}{du} {}_2F_1(1,1,u,-x)\big|_{u=2} = \frac{d}{du}\Big(1+ \sum_{k=1}^\infty \frac{k!}{(u)_k} (-x)^k \Big)\big|_{u=2}$$
where the definition of Gauss's hypergeometric has been used in terms of the Pochhammer symbol, and $(1)_k=k!$  Taking the derivative of the reciprocal of $(u)_k=\Gamma(u+k)/\Gamma(u)$ and evaluating it in terms of the digamma function,
$$S_1=\sum_{k=1}^\infty \frac{k!}{(k+1)!} (-x)^k \big(1-\gamma-\psi(k+2) \big) =$$
$$= \frac{1}{x}\sum_{k=2}^\infty\frac{1}{k}\big(\gamma+\psi(k+1) -1\big) (-x)^k =
\frac{1}{x}\sum_{k=1}^\infty\frac{1}{k}\big(H_k - 1\big)(-x)^k$$
where in the last step the harmonic sum symbol has been used.  Both sums are well-known (Mathematica knows them) so the answer is
$$ S_1 = \frac{1}{x} \Big(-\frac{\pi^2}{6}-\log{x}\,\log{(1+x)} + \log^2(1+x) + \text{Li}_2\big(\frac{1}{1+x}\big)+\log{(1+x)} \Big) $$
Similar manipulations yield the second expression,
$$S_2:= \frac{d}{du} {}_2F_1(1,u,2,-x)\big|_{u=1}=\sum_{k=1}^\infty \frac{H_k}{k+1}(-x)^k=$$
$$=-\,\frac{\log^2(1+x)}{2x} $$
Again, once the sum has been simplified to the last expression on the penultimate line, Mathematica knows the answer.
