Lower bound for integral $\int_0^\infty (1+2ax)^{-3/2}(1+2bx)^{-n/2} dx$ when $a+nb=1$

Question

One can show that the integral is lower bounded by $1/(2a+1)$ by simply using the $1+x\le \exp(x)$ inequality: \begin{align} \int_0^\infty (1+2ax)^{-3/2}(1+2bx)^{-n/2} dx &\ge \int_0^\infty \exp(-3ax)\exp(-bnx)dx\\ &=\int_0^\infty\exp(-(2a+1)x)dx \\ &= \frac{1}{2a+1} \end{align} However, I believe this is not a very tight bound. For instance assuming that $0 < a < b < 1$, by numerical analysis the integral is above $1+(1/n-a)\frac{(n-1)^2}{n(n+1)}$. Is there a way to prove this tighter bound?