I am evaluating the following term in a series:
$$I_k = \int\!x^{-3(2k+1)}(1+\lambda x^4)^{-1/2}\,\mathrm dx$$
When I plug this into WolframAlpha, I get the following result:
$$I_k = -\frac{x^{-2(1+3k)}}{2(1+3k)}{}_2F_1\left(\frac 12,-\frac 12 (1+3k);\frac 12 (1-3k);-\lambda x^4\right)$$
Here's the issue: for every odd term $(k = 1, 3, 5, ...)$ both the second and third coefficient of the hypergeometric term is a negative integer, leading one to believe $I_1 = I_3 = ... = \widetilde{\infty}$. Even after searching for hypergeometric transformations on NIST's site and in other places, this is a tough thing to get around.
However, when these values for $k$ are put into the original integral and re-submitted to WolframAlpha, I get a polynomial with some extra terms, leading me to believe a transformation does in fact exist.
For example, take $k = 1$:
$$I_1 = \int\!x^{-9}(1+\lambda x^4)^{-1/2}\,\mathrm dx$$
which becomes
$$I_1 = \frac{1}{16}\left[3\lambda^2\left(\ln x^2 - \ln\left[1+\sqrt{1+\lambda x^4}\right]\right) + \left(3\lambda x^4 - 2\right)x^{-8}\sqrt{1+\lambda x^4}\right]$$
All of the odd terms have these two logarithmic terms followed by a polynomial with some square root stuff (nothing wildly different from term to term).
Here is my question: Does anybody have any suggestions for how I would find a general expression for $I_k$? Even if this means one general expression for even terms and another for odd terms, that would work.