I am looking for explicit formulas for the four basic invariants $I\_4, I\_8, I\_{12}, I\_{18}$ of a generic binary quintic form, either given in the shape


$$\displaystyle F(x,y) = ax^5 + 5bx^4y + 10cx^3y^2 + 10dx^2y^3 + 5exy^4 + fy^5$$


or


$$\displaystyle F(x,y) = ax^5 + bx^4y + cx^3y^2 + dx^2y^3 + exy^4 + fy^5.$$


I. Dolgachev seems to have given exactly such formulas in his book "Lectures on Invariant Theory", but it appears that his formulas are not correct. For example, his claim that the discriminant of the binary quintic is given by the expression $I\_4^2 - 128I\_8$, with $I\_4, I\_8$ given on page 151, does not hold (checked with computer algebra). Further, his formula for $I\_{18}$ on page 151 does not appear to be right either, since the formula is not invariant under the map $(a,b,c,d,e,f) \mapsto (f,e,d,c,b,a)$. 



Explicit formulas for invariants of binary quintic forms
Here's another way to do it that you might find useful:


Recall that $\mathrm{SL}(2,\mathbb{C})$
acts on the polynomial ring $\mathbb{C}[x,y]$ 
by linear substitution in $x$ and $y$,
making the subspace $V\_d\subset \mathbb{C}[x,y]$, consisting
of polynomials homogeneous of degree $d$ in $x$ and $y$, into
an irreducible $\mathrm{SL}(2,\mathbb{C})$-representation of
dimension $d{+}1$.


Define a bilinear pairing $\langle,\rangle\_p:\mathbb{C}[x,y]\times 
\mathbb{C}[x,y]\to\mathbb{C}[x,y]$ 
for $p\ge 0$ by the formula
$$
\langle u,v\rangle\_p 
= \frac1{p!}\sum\_{k=0}^p (-1)^k{p\choose k}
\frac{\partial^pu}{\partial x^{p-k}\partial y^k}
\frac{\partial^pv}{\partial x^{k}\partial y^{p-k}}\,.
$$
For example, $\langle u,v\rangle\_0 = uv$ 
and $\langle u,v\rangle\_1 = u\_xv\_y-u\_yv\_x$. 


The bilinear pairings $\langle,\rangle\_p$ 
are $\mathrm{SL}(2,\mathbb{C})$-equivariant,
and they restrict to $\langle,\rangle\_p: V\_a\times V\_b\to V\_{a+b-2p}$
to be nonzero as long as $p\le\mathrm{min}(a,b)$.  

These expressions $\langle u,v\rangle\_p$ 
are called `transvectants' in the classical literature.


If 
$$u = u\_{-5}\,x^5 + u\_{-3}\,x^4y + u\_{-1}\,x^3y^2 + u\_{1}\,x^2y^3
+ u\_{3}\,xy^4 + u\_5\,y^5\in V\_5
$$ 
is a quintic, 
then it is not difficult to check that the quantities
$$
I\_4(u) = \langle u^2,u^2\rangle\_{10},\quad 
I\_8(u) = \langle u^4,u^4\rangle\_{20},\quad\text{and}\quad
I\_{12}(u) = \langle u^6,u^6\rangle\_{30}
$$
are independent $\mathrm{SL}(2,\mathbb{C})$-invariant polynomials
of degrees $4$, $8$, and $12$, respectively. The invariant polynomial
$$
I\_{18}(u) = \langle \langle u^5,u^6\rangle\_{10},u^7\rangle\_{35}
$$
is nonzero, and, since $18$ is not a multiple of $4$, 
it is not expressible as a polynomial in $I\_4(u)$, $I\_8(u)$, 
and $I\_{12}(u)$,
though its square, which is of degree 36, can be written as
a polynomial in these three lower-degree invariants.
(It is, of course, a classical result 
that these four invariant polynomials generate 
the ring of $\mathrm{SL}(2,\mathbb{C})$-invariants on $V\_5$ 
and that they are subject only to this single relation.)


These polynomials are not particularly nice when written out
in terms of the coefficients $u\_j$. For example, $I\_4(u)$
is a sum of $12$ monomials, $I\_8(u)$ is a sum of $71$ monomials,
$I\_{12}(u)$ is a sum of $252$ monomials, and $I\_{18}(u)$
is a sum of $848$ monomials.


Finally, in this normalization, the discriminant of $u$ is
a constant multiple of 
$$
I\_8(u) - 92610\,I\_4(u)^2,
$$
which has only $59$ monomials when expanded in the $u\_j$.


