Difficulty and established result in the classification of Galois groups of quintic polynomials. This article provides a classification of Galois groups of cubic and quartic polynomials and criteria to determine it, then I try to find some paper about the same result in quintic equations. However, I find nothing similar. 
So I am wondering if there is some fundamental difficulty abandoning people to approach to 5 degree, or it is just because it is less interested and no one wants to do it?
 A: It is only more complex. The number of possible Galois groups is larger, and the degree of the resolvants needed to discriminate between candidate Galois groups also get larger.
The basic process to compute the Galois group of a polynomial is always the same :
You pick one expression $f(x_1,\ldots,x_5)$ where $x_i$ are the roots of $P$, and look at the subgroup $H_f$ of $S_5$ stabilizing it. You build the resolvant $\prod (X - \sigma(f(x_i)))$ for $\sigma \in S_5/H_f$. Don't be scared, the degree will only go up to $120$ in the worst case. Each coefficient is completely symmetric so can be expressed in terms of the coefficients of the initial polynomial.
Does the resolvant have a rational root ? If yes, great, you have learned that the Galois group of $P$ is included in some conjugate of $H_f$.
Does the resolvant split over $\Bbb Q$ ? If yes, great, you have learned that the Galois group of $P$ is included in the intersection of all the conjugates of $H_f$.
More precisely, the way the resolvant factors over $\Bbb Q$ tells you of the size of the $G$-orbits of $S_5/H_f$. You can still compute the Galois group without having to decide the factorisation over $\Bbb Q$, and only using the rational root test. You simply don't get as much information for each $H_f$, but that's all.

If you have two expressions corresponding to two groups $H_1 \subset H_2$, and if you get a rational root when doing a resolvant for $H_2$, you are lucky because when building your resolvant for $H_1$, you may restrict the product to range over $H_2/H_1$ and not over $S_5/H_1$. When doing that the coefficients obtained are invariant by $H_2$, so you will need to replace them with a rational root of their own resolvant (in a coherent way if the resolvants have several rational roots. In theory you only need one, and all the others are rational fuctions of the coefficients of $P$ and that one rational value. However the dependencies may be a bit long to compute) . This allows you to get a smaller polynomial

Now, all you have to do is carefully plan and choose which $H_f$ you want to use. To do that you will need a nice picture of all possible conjugacy classes of transitive subgroups $G$ of $S_5$, as well as what each possible choice for $H_f$ can tell you about each case once you have factored the resolvant.
For each transitive conjugacy class of $G$ and each $H$ you can compute the size of the $G$-orbits of $S_5/H$. Once you have this down, you can elaborate a strategy of choosing a sequence of $H$ that discriminates among all possible $G$ as fast as possible.
For example, first you will want to use the biggest possible $H$ so that $S_5/H$ is small. So start with $A_5$. If you pick $\prod_{i<j} (x_i-x_j)$, then its resolvant is $X^2 - D$ where $D$ is the discriminant of $P$. So the first question you have to ask is "is this discriminant a square ?"
A: The irreducible solvable (by radicals) quintic equations in $\mathbb{Z}[X]$ have been characterized by D.S. Dummit (1991) in Solving Solvable Quintics, his theorem being an irreducible quintic is solvable if and only if an explicit sextic integer polynomial has a rational root.
More recently Boswell and Glasser extended this approach to the irreducible sextic case, where solvability is equivalent to a certain tenth degree equation having a rational root.
Because it acts transitively on the roots, the Galois group of an irreducible quintic must be a transitive subgroup of $S_5$, so the only possibilities for an unsolvable quintic turn out to be Galois group $S_5$ or $A_5$.  For a solvable irreducible quintic the possibilities are the Frobenius group $F_{20}$ of order 20, the dihedral group $D_5$ of order 10, or the cyclic group $Z_5$ of order 5.
Following the discussion at MathOverflow, it seems that improved techniques for computing Galois groups are finding their way in to computer algebra software (CAS) packages.
