convert/trigh converts the exponentials in an expression as well as the special functions when possible into hyperbolic functions, that is, into any of $\sinh ,\cosh ,\tanh ,\mathrm{sech},\mathrm{csch},\coth$.

$\exp \left(x\right)$

${\ⅇ}^x$

$\mathrm{convert}\left(\,\mathrm{trigh}\right)$

$\cosh \left(x\right)+\sinh \left(x\right)$

${\mathrm{\pi }}^{\frac{1}{2}}\mathrm{MeijerG}\left(\left[\left[\right]\,\left[\right]\right]\,\left[\left[\frac{1}{2}\right]\,\left[0\right]\right]\,\frac{1}{4}{x}^4\right)\mathrm{MeijerG}\left(\left[\left[\right]\,\left[\right]\right]\,\left[\left[0\right]\,\left[\right]\right]\,-x\right)$

$\sqrt{\mathrm{\pi }}\mathrm{MeijerG}\left(\left[\left[\right]\,\left[\right]\right]\,\left[\left[\frac{1}{2}\right]\,\left[0\right]\right]\,\frac{x^4}{4}\right)\mathrm{MeijerG}\left(\left[\left[\right]\,\left[\right]\right]\,\left[\left[0\right]\,\left[\right]\right]\,-x\right)$

$\mathrm{convert}\left(\,\mathrm{trigh}\right)$

$-\frac{\mathrm{I}\sqrt{x^4}\sinh \left(\mathrm{I}{x}^2\right)\cosh \left(x\right)}{x^2}-\frac{\mathrm{I}\sqrt{x^4}\sinh \left(\mathrm{I}{x}^2\right)\sinh \left(x\right)}{x^2}$