Invariants for a single form, several forms.

mV1+nV2 pV3 pV4

mV1+nV2

The ring of SL2-invariants of mV1+nV2 has the following numbers of basic invariants with given degree:

degree count
2 (m choose 2) + (n+1 choose 2)
3 n(m+1 choose 2) + (n choose 3)
4 (m+1 choose 2)(n choose 2)
total (n choose 3) + (m+1 choose 2)(n+1 choose 2) + (m choose 2) + (n+1 choose 2)

This is more or less classical. See also

A. E. Brouwer & M. Popoviciu, SL2-modules of small homological dimension, preprint, Oct. 2010.

pV3

The ring of SL2-concomitants of pV3 has the following numbers of basic covariants with specified multidegree (degree in coeff, degree in x,y):

type deg count p=1 p=2 p=3 p=4 p=5
type 1 (1,3) p 1 2 3 4 5
type 2 (2,4) (p choose 2) 0 1 3 6 10
type 3 (2,2) (p+1 choose 2) 1 3 6 10 15
type 4 (2,0) (p choose 2) 0 1 3 6 10
type 5 (3,3) (p+2 choose 3) 1 4 10 20 35
type 6 (3,1) 2(p+1 choose 3) 0 2 8 20 40
type 7 (4,0) (p+3 choose 4) 1 5 15 35 70
type 8 (4,2) 3(p+2 choose 4) 0 3 15 45 105
type 9 (5,1) 4(p+3 choose 5) 0 4 24 84 224
type 10 (6,0) ? 0 1 10 50 175
total #inv     1 7 28 91 255
total #cov     4 26 97 280 689

The formulas for types 1-8 were given by Peano in

G. Peano, Sui sistemi di forme binarie di egual grado e sistema completo di quante si vogliano cubiche, Atti di Torino 17 (1882) 580-586.

The formulas for types 1-10 were given by Young in

A. Young, The irreducible concomitants of any number of binary quartics, Proc. London Math. Soc. 30 (1899) 290-307.
but his formula for type 10 (namely, (p+4 choose 6)) is wrong. It looks like the correct formula might be

(p+4 choose 6) + 3(p+3 choose 6) + (p+2 choose 6) = (p+2 choose 4)(p+1 choose 2)/3.

pV4

The ring of SL2-concomitants of pV4 has the following numbers of basic covariants with specified multidegree (degree in coeff, degree in x,y):

deg count p=1 p=2 p=3 p=4
(2,0) (p+1 choose 2) 1 3 6 10
(3,0) (p+2 choose 3) 1 4 10 20
(4,0) (p+1 choose 4)+(p+2 choose 4) 0 1 6 20
(5,0) (p choose 5)+2(p+1 choose 5)+3(p+2 choose 5) 0 0 3 20
(6,0) 10(p+2 choose 6) 0 0 0 10
(2,2) (p choose 2) 0 1 3 6
(3,2) (p choose 3)+2(p+1 choose 3) 0 2 9 24
(4,2) 3(p+1 choose 4)+3(p+2 choose 4) 0 3 18 60
(5,2) 3(p+2 choose 5)+2(p+3 choose 5) 0 2 15 60
(1,4) p 1 2 3 4
(2,4) (p+1 choose 2) 1 3 6 10
(3,4) 2(p+1 choose 3) 0 2 8 20
(4,4) 3(p+1 choose 4) 0 0 3 15
(2,6) (p choose 2) 0 1 3 6
(3,6) (p+2 choose 3) 1 4 10 20
total #inv   2 8 25 80
total #cov   5 28 103 305

This is from

A. Young, The irreducible concomitants of any number of binary quartics, Proc. London Math. Soc. 30 (1899) 290-307.