Using Rubi, you can investigate a bit further, what steps are required to find an antiderivative

<< Rubi`
int = Steps[Int[r^2 Sin[r^-3 * c], r]]

First, you can show that the solution is indeed one valid antiderivative of your expression:

In[31]:= D[int, r]

Out[31]= r^2 Sin[c/r^3]

Now, you can look at the limits of your integration bounds

In[32]:= Limit[int, r -> 0, Direction -> "FromAbove"]

Out[32]= ConditionalExpression[-(1/6) c (2 Log[c] - Log[c^2]),  c \[Element] Reals]

This looks good, but this here not

In[33]:= Limit[int, r -> Infinity]

Out[33]= c \[Infinity]

Therefore, I suspect your version 10.4 result is not correct and the integral does not converge.

Thank you! Plotting the integrand also suggests a function that increases monotonically as r goes to infinity.

This brings me to my original problem, a 2D integral of the form

Integrate[
 r^2 Sin[a r^-3 (1 - 3 z^2)], {r, 0, \[Infinity]}, {z, -1, 1}, 
 Assumptions -> {a \[Element] Reals}]

Mathematica fails to find an answer. But numerically, it appears that this double integral converges nicely

In[56]:= R = 100; NIntegrate[
 r^2 Sin[(1 - 3 z^2) r^-3], {r, 0, R}, {z, -1, 1}]

Out[56]= 0.106513

with a value

In[59]:= 4/27 (3 - Sqrt[12] ArcCoth[Sqrt[3]]) // N

Out[59]= 0.106513

Is there a way to convince Mathematica to solve the 2D integral? I suppose the integral over z needs to be done first, since the r integral (for any z except [PlusMinus]1/Sqrt[3]) diverges.

Thanks! Yes, the r integral diverges, but the 1-3z^2 term seems to rescue convergence in the 2D integral. This term alternates signs twice between z = -1 and +1, so it's plausible that this is possible.

This is a funny situation to be in: to know the answer (helped by that faulty answer from Mathematica 10.4), but to have no clue how to get there, and no tool that might show the way.