Fred Lunnon writes:

I suppose you have already consulted Konrad Knopp "Counterexamples in Analysis", reprinted by Dover? WFL

I haven't been able to track down a copy. I did find the book of the same name by Gelbaum and Olmsted, but it didn't offer anything useful. Rich Schroeppel writes:

Do you mean "have a closed form integral"?

Dan Asimov has it right when he replies:

No. The issue is existence, not what form it takes.

(Thanks Dan!) I should remind everybody that "has an antiderivative" is not the same thing as "is integrable". For instance, the function F that's defined as x^2 sin 1/x^2 when x is non-zero and 0 when x is zero has a derivative f that's defined everywhere, so (by definition) f has an antiderivative, but f is neither Riemann integrable nor Lebesgue integrable on [-1,1]. Jim