Following the change-in-variable approach...
sol = Integrate[u/(p^2 (u^2 + c)^2) (2 Log[p] + Log[(u + 1)^2 + d]), {u, 0, Infinity},
Assumptions -> {c > 0, d > 0}] /. {c -> b^2/p^2, d -> m^2/p^2};
sol = FunctionExpand[sol, Assumptions -> {m > 0, p > 0, b > 0}] //. {Sqrt[(b^2 m^2)/p^4] -> b m/p^2,
Sqrt[b^2/p^2] -> b/p, ((b^2 m^2)/p^4)^(3/2) -> b^3 m^3/p^6,
Sqrt[(b^10 m^2)/p^12] -> b^5 m/p^6, (b^2/p^2)^(3/2) -> b^3/p^3, (b^2/p^2)^(5/2) -> b^5/p^5};
sol = FullSimplify[sol];
sol = Collect[sol, {\[Pi], Log[b], Log[m], Log[p], Log[1 + m^2/p^2], Log[1 + p^2/m^2]}]
I get the following solution (which certainly could be compacted a bit more):
(p \[Pi])/(2 b ((b + m)^2 + p^2)) +
(2 m p ArcTan[p/m])/(((b - m)^2 + p^2) ((b + m)^2 + p^2)) + ((b^2 - m^2 + p^2) Log[b])/(((b - m)^2 + p^2) ((b + m)^2 + p^2)) +
((2 b^4 - 4 b^2 m^2 + 2 m^4 + 4 b^2 p^2 + 4 m^2 p^2 + 2 p^4) Log[m])/(2 b^2 ((b - m)^2 + p^2) ((b + m)^2 + p^2)) +
((-b^4 + b^2 m^2 - b^2 p^2) Log[1 + m^2/p^2])/(2 b^2 ((b - m)^2 + p^2) ((b + m)^2 + p^2)) +
((-2 b^4 + 2 b^2 m^2 - 2 b^2 p^2) Log[p])/(2 b^2 ((b - m)^2 + p^2) ((b + m)^2 + p^2)) +
(((b^2 - m^2)^2 + 2 (b^2 + m^2) p^2 + p^4) Log[1 + p^2/m^2])/(2 b^2 ((b - m)^2 + p^2) ((b + m)^2 + p^2))
As a way to do some checking:
parms = {b -> 2.3, p -> 4.7, m -> 18.2};
sol /. parms
NIntegrate[a/(a^2 + b^2)^2 Log[(a + p)^2 + m^2] /. parms, {a, 0, Infinity}]
which results in 0.567437 for both.