With Mathematica Version 8.0, I get good result, with no imaginary part.
integrand = q^2/(q^2 + ks^2) (1 - (1/τ^2 - c^2 q^2) τ^2)/(1 + 2 (ω^2 - (1/τ^2 - c^2 q^2)) τ^2 + (ω^2 + (1/τ^2 - c^2 q^2))^2 τ^4); fs = FullSimplify[ integrand, {ks > 0, ω > 0, τ > 0, c > 0}] (* (c^2 q^4)/((ks^2 + q^2) (4 ω^2 + τ^2 (-c^2 q^2 + ω^2)^2)) *) int = Integrate[fs, {q, 0, Infinity}, Assumptions -> {ks > 0, ω > 0, τ > 0, c > 0}] (* (π (4 c^3 ks^3 τ^(3/2) Sqrt[-τ + ( 2 I)/ω] + ω (4 + τ^2 ω^2) (2 + I τ ω + I Sqrt[-2 I - τ ω] Sqrt[2 I - τ ω]) + c^2 ks^2 τ (τ ω (4 + I τ ω + I Sqrt[-2 I - τ ω] Sqrt[ 2 I - τ ω]) - 2 (2 I + Sqrt[-2 I - τ ω] Sqrt[ 2 I - τ ω]))))/(8 c τ^(3/2) Sqrt[-τ + ( 2 I)/ω] (4 ω^2 + τ^2 (c^2 ks^2 + \ ω^2)^2)) *) ceRe = FullSimplify[ ComplexExpand[Re[int], TargetFunctions -> {Re, Im}], {ks > 0, ω > 0, τ > 0, c > 0}] (* (π (4 c^3 ks^3 τ^2 (τ^2 + 4/ω^2)^(1/4) + 2 Sqrt[τ/( 2 + (2 τ ω)/Sqrt[ 4 + τ^2 ω^2])] (ω (4 + τ^2 ω^2) \ (τ ω + Sqrt[4 + τ^2 ω^2]) + c^2 ks^2 τ (-4 + τ ω (τ ω + Sqrt[ 4 + τ^2 ω^2])))))/(8 c τ^2 (τ^2 + 4/ω^2)^( 1/4) (4 ω^2 + τ^2 (c^2 ks^2 + ω^2)^2)) *) Imaginary part is zero.
