TITLE: Brumer quintic polynomials - is there a general formula for the roots?
QUESTION [5 upvotes]: There exist a family of quintic polynomials, called Brumer's polynomials (or Kondo-Brumer), which have the form:
$$x^5+(a-3)x^4+(-a+b+3)x^3+(a^2-a-1-2b)x^2+bx+a,~~~a,b \in \mathbb{Q}$$
According to Wikipedia these polynomials are solvable in radicals.

Is there a general formula for roots of these polynomials? Or at least the closed form for some special cases?

I searched the web, but only found papers discussing the group properties (for example here) or other properties of Brumer's polynomials. Nothing about the roots.
Edit
I'm starting a bounty, and I would like either of these things:

General solution (at least one root), depending on $a,b$ - only if such a solution exists, and is short enough to write here in closed form.
Some methods for obtaining this solution - again, if it will lead to a form of the solution more compact and simpler than the general way to solve an arbitrary solvable quintic.
Solutions to some special cases (for some values of $a,b$ with $b \neq 0$)
A proof that no such simple solution is possible for this family of quintics.

REPLY [2 votes]: (Addendum to my answer.) Since the primary objective of the OP is to find solvable quintics with a "simple" solution, then we can add the "depressed" multi-parameter family,
$y^5+10cy^3+10dy^2+5ey+f = 0\tag{1}$ 
where the coefficients obey the quadratic in $f$,
$(c^3 + d^2 - c e) \big((5 c^2 - e)^2 + 16 c d^2\big) = (c^2 d + d e - c f)^2
\tag{2}$
Solve for $f$. Define this quintic's Lagrange resolvent as,
$$(z^2+u_1z-c^{5})(z^2+u_2z-c^{5}) = 0$$
where the $u_i$ are the two roots of the quadratic,
$$u^2-fu+(4c^5-5c^3e-4d^2e+ce^2+2cdf) = 0$$
then the solution to $(1)$ is,
$y = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\tag{3}$
Note that appropriate choices of the $3$ free parameters $c,d,e$ can yield rational $f$.

Example: A particular case is the Lehmer quintic,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x + 1=0$$
Let $x = (y-n^2)/5$ to transform it to depressed form $(1)$. Its transformed coefficients then obey $(2)$.