Brumer quintic polynomials - is there a general formula for the roots? 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.
 A: Given the solvable Kondo-Brumer quintic, 
$$x^5 + (a - 3)x^4 + (-a + b + 3)x^3 + (a^2 - a - 1 - 2b)x^2 + b x + a = 0\tag1$$
One in fact can make explicit formulas for these. For simplicity, assume $a=1$, so

$$x^5 - 2x^4 + (b + 2)x^3 + (-1 - 2b)x^2 + b x + 1=0\tag2$$

Define the four roots $z_i$ of its quartic Lagrange resolvent,
$$z^2 + \tfrac{1}{2}\big((597 + 225 b + 200 b^2) \color{blue}+ 5\sqrt{5} (43 + 33 b + 20 b^2)\big)z +{c_1}^5=0\tag3$$
$$z^2 + \tfrac{1}{2}\big((597 + 225 b + 200 b^2) \color{blue}- 5\sqrt{5} (43 + 33 b + 20 b^2)\big)z +{c_2}^5=0\tag4$$
which was factored into two quadratics for convenience, and,
$$c_1=-\tfrac{1}{2}\big((2+5b)\color{blue}-\sqrt{5}(4-b)\big)\tag5$$
$$c_2=-\tfrac{1}{2}\big((2+5b)\color{blue}+\sqrt{5}(4-b)\big)\tag6$$
Note the constant term of the quartic is a nice fifth power,
$$(c_1 c_2)^5=(5b^2+15b-19)^5$$
We can then give the relatively "simple" solution,
$$x =  \frac{1}{5}\Big(2+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\Big)\tag7$$

Example: Let $b=-1$, then,

$$x^5 - 2x^4 + x^3 + x^2 - x + 1=0$$
a quintic which was also solved by Ramanujan. Its resolvent using $(3),(4)$ is,
$$z^4 + 572z^3 + 70444z^2 + 1600203z - 29^5=0$$
then,
$$x= \frac{1}{5}\Big(2+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\Big) = -0.90879\dots$$
If all the $z_i$ are real, as in the example, then it is a simple matter of taking fifth roots of real numbers which will then yield a real root of the quintic.
A: The roots of any (irreducible) solvable quintic can be found, using methods due to George Paxton Young in 1888. An explicit 3-page formula based on those methods was given by Daniel Lazard in 2004. Source: https://en.wikipedia.org/wiki/Quintic_function
A: (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 $(2)$.
