Consider the following pentacube (front and back view shown.)
I have used Burr Tools to determine that 24 of these will NOT fit in a 5x5x5 box. According to my notes when working on this problem a while ago 22 will fit in a 5x5x5 box. But the question of whether 23 will is unknown to me.
Burr tools is unable to answer this question with at least a day's worth of computing. My guess is that this problem is more difficult than other packings because:
The pentacube fits in a 2x2x2 box and has no rotational symmetry, thus the pentacube can be in 24*4*4*4=1536 different positions within a 5x5x5 cube.
Two copies of the pentacube can form the same shape (which fits in a 2x2x3 box) in two different ways. Thus, the program may be trying possibilities which are equivalent to previous possibilities.
Any idea if 23 of these will fit in a 5x5x5 box?
It's not possible to fit 23 of these pentacubes in a 5x5x5 box. I show that were such a packing to exist, the top and bottom halves of the packing must have some cubes in common, an impossibility. This proof uses some results from Burr Tools, I've included the relevant file here. To simplify the notation let us call this pentacube, $A$.
Since $A$ fits in a 2x2x2 box, no matter what orientation $A$ is in, it occupies part of exactly two layers. Thus, we consider the 5 5x5 layers of the box. Let us number the layers 1 through 5.
Some of these pentacubes will be in layers 1 and 2, or 2 and 3. Call these the Bottom. The rest will be in 3 and 4, or 4 and 5. Call these the Top. Burr tools verifies that a 3x5x5 block cannot contain 13 $A$ (but can contain 12.) Therefore to pack 23 $A$ in a 5x5x5 box, 12 must be in the Bottom and 11 in the Top (or the other way around, but by symmetry it doesn't matter.)
A packing of 12 $A$ in the 3x5x5 Bottom must cover each of the highlighted 4 cubes in layer 3, as Burr tools verifies that 12 $A$ will not fit in a 3x5x5 box with even one of these cells vacant.
Since these 4 cells must be part of the Bottom packing, they must not be part of the Top. However Burr tools shows that 11 $A$ will not fit in the 3x5x5 Top section with those four cells excluded. Thus there is no packing of 23 $A$ in a 5x5x5 box.