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* <math>\operatorname{tanhc}(z)=2\,{\frac {{{\rm KummerM}\left(1,\,2,\,2\,z\right)}}{(2\,iz+\pi) {\rm KummerM}(1,\,2,\,i\pi -2\,z) e^{2\,z-1/2\,i\pi} }}</math>
*<math>\operatorname{tanhc}(z)=2 \frac {\operatorname{HeunB}(2,0,0,0,\sqrt{2} \sqrt{z})}{( 2iz+\pi) \operatorname{HeunB}( 2,0,0,0,\sqrt{2} \sqrt{1/2\,i\pi -z}) e^{2\,z-1/2\,i\pi}} </math>
* <math>\operatorname{tanhc}(z)= \frac{i{\rm \ WhittakerM}(0,\,1/2,\,2\,z)}{{\rm WhittakerM}(0,\,1/2,\,i\pi -2\,z)} z</math>
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9}}{z}^{3}+{\frac {2}{75}}{z}^{5}-{\frac {17}{2205}}{z}^{7}+{\frac {62
}{25515}}{z}^{9}-{\frac {1382}{1715175}}{z}^{11}+O \left( {z}^{13} \right) )</math>
==Pade approximation==
<math> {\it Tanhc} \left( z \right) = \left( 1+{\frac {7}{51}}\,{z}^{2}+{
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