The following may be more clear:
The BCS theory can be applied to 3He, as well as to electrons. However, the Cooper pairs for 3He are much more complex creatures than the those in a conventional superconductor. Due to the hard-core repulsion of the helium nuclei, the two 3He atoms in the Cooper pair feel a greater need to keep away from each other than the electrons do; so while the electrons bind together in a tight, symmetric, s-wave package, the 3He atoms bind together in a loose, anti-symmetric, p-wave package. "P-wave" just means that the pair has angular orbital momentum L=1. An L=1 system has three different quantum-mechanical states, denoted by ml = 1, 0, or -1.
Since all combinations of fermions must be anti-symmetric, the spin angular momentum in this case must be symmetric. There are three symmetric ways to combine the spins of two spin-1/2 fermions, Y = |++>, |-->, or |+-> + |-+>. (Read this as both spin up, both spin down, or the symmetric combination of the two.)
So a 3He Cooper pair has three different possible orbital states, and three different possible spin states. This gives a total of nine different combinations, each of which is weighted in the order parameter Y by a complex number, giving 18 degrees of freedom. This allows the 3He superfluid to behave in much more complex ways than the conventional superconductor, with its two degrees of freedom.
Note the "Due to the hard-core repulsion of the helium nuclei, the two 3He atoms in the Cooper pair feel a greater need to keep away from each other than the electrons do;", an equally handwaved statement to the Wiki's : " These complications arise because helium atoms repel each other much more strongly than electrons".
The last statement I would say is the mundane : "two electrons tied in tandem with other two repel more than one with one", and He3 has two electrons in orbitals. Both try to explain why in the He3 Cooper pair we find the He3 in P waves and not in S waves.