According to Maple there is closed form in terms of Hurwitz zeta.

a:='a':su1:=sum(1/(1+a^2*(n+1/2))^(3/2),n=0..infinity):latex(su1);

$$ \left( 2\,\zeta \left( 0,3/2,1/2\,{\frac {2+3\,{a}^{2}}{{a}^{2}}} \right) \sqrt {4+2\,{a}^{2}}+\zeta \left( 0,3/2,1/2\,{\frac {2+3\,{a}^{2}}{{a}^{2}}} \right) \sqrt {4+2\,{a}^{2}}{a}^{2}+4\,{\it csgn}\left( a \right) {a}^{3} \right) {\it csgn} \left( a \right) {a}^{-3}\left( 2+{a}^{2} \right) ^{-1}{\frac {1}{\sqrt {4+2\,{a}^{2}}}} $$

convert(su1,string); "(2*Zeta(0,3/2,1/2*(2+3*a^2)/a^2)*(4+2*a^2)^(1/2)+Zeta(0,3/2,1/2*(2+3*a^2)/a^2)*(4+2*a^2)^(1/2)*a^2+4*csgn(a)*a^3)*csgn(a)/a^3/(2+a^2)/(4+2*a^2)^(1/2) 

$\zeta(n,z,v)$ is the $n$-th derivative of Hurwitz zeta.

Here is an idea. Try to bound your series by simple expressions. For $a>0,k \ge 1$ define

$$ f(a,k)= \sum_{n=0}^\infty \frac{1}{(1+a^2 (n+1/2)^2) ^{k}} $$

Your series is $f(a,\frac32)$. All terms are positive and for all partial sums we have $ f(a,1) > f(a,\frac32) > f(a,2)$. This will hold for the infinite series assuming they converge.

$$ f(a,1) = 1/2\,\pi \,\tanh \left( {\frac {\pi }{a}} \right) {a}^{-1}$$ $$ f(a,2) = -1/4\,\pi \, \left( \pi \, \left( \cot \left( 1/2\,{\frac {\pi \, \left( 2\,i-a \right) }{a}} \right) \right) ^{2}-a\tanh \left( { \frac {\pi }{a}} \right) +\pi \right) {a}^{-2}$$

These bounds appear easier to analyze.

I believe the bounds as $a \to 0^+$ are $ \frac{\pi}{2 a} > f(a,\frac32) > \frac{\pi}{4a}$.