I think that does not have finite closed-form expression in terms of very large class of special functions.

$$\sum _{i=1}^{\infty } \frac{\tanh \left(\frac{i \pi }{2}\right)}{i^5}=\\\zeta (5)-2 \sum _{k=1}^{\infty } \text{Li}_5\left(e^{-k \pi }\right)+4 \sum _{k=1}^{\infty } \text{Li}_5\left(e^{-2 k \pi }\right)=\\\zeta (5)-2* \text{A255701}+4* \text{A255702}$$

Where A255701 and A255702

  {NSum[Tanh[(i \[Pi])/2]/i^5, {i, 1, Infinity}, 
    WorkingPrecision -> 31], 
   Zeta[5] - 
    N[Sum[2 PolyLog[5, E^(-k \[Pi])] - 4 PolyLog[5, E^(-2 k \[Pi])], {k,
        1, 1000}], 21]}

 (*{0.9539629221432281032692, 0.9539629221432281032692}*)

Regards M.I.