Simple -- get enough capacitors to handle the worst-case peak energy, and use a switching converter to emulate an inductance.

1H at 10A is E = 0.5 * (1H) * (10A)^2 = 50J of stored energy, equivalent to a fist-sized electrolytic capacitor. We can calculate the volume of air gap required to store as much energy in an inductor:
v = 2 E mu_0 / B^2
This is useful because we know B from the materials we will use: transformer steel saturates around 1.2T. Putting that in, we get 87 266 mm^3.
Note that v is inverse with B, so we could simply choose a large B and get what we want; the problem is, we can't get B that high using magnetic materials (the best materials saturate up to 2T, which does help), and to push B that high without materials requires massive magnetization, i.e. lots of ampere-turns -- we need orders of magnitude more copper, or incur orders of magnitude higher resistive losses.
The resistivity of copper is actually a limiting factor here, and it turns out, a relative permeability of 20-60 is typically best suited to it. If we had a better conductor, we could get by with less, or with a perfect conductor perhaps no core at all. Unfortunately silver is only a few percent better than copper, so we cannot find a solution at room temperature; copper does have a generous tempco and could be cooled to LN2 temperatures, if the refrigeration losses aren't terrible (you might want to check the price on LN2 truckloads and dewars though

). Or the perfect conductor does, in fact, exist -- but we still need LN2 temperatures or below to make use of superconductors.
If we suppose an air gap of 20 mm, that's a cross-sectional area of 4 363 mm^2, or a 66 mm square. That's about two MOT sized cores, stacked.
This is probably a good lesson on the X-Y problem: you have problem X (unknown), and a tentative solution Y (designing a compact inductor that stores a metric shitload of energy). If you could only refine the solution to problem Y, you'd be done; it seems simple enough. But as it turns out, Y is, in fact, impossible (if your volume and cost requirements are completely rigid), so you will have to reconsider problem X instead.
Very likely, for example -- my initial suggestion is comically overcomplicated, but with the slightest rearrangement (e.g. using a parallel filter capacitor instead of a series filter inductor), perhaps it becomes a nearly trivial solution. We just don't know until you've given us some information about your underlying application.

Tim