Integrand size = 14, antiderivative size = 50 \[ \int e^{a+b x} \text {csch}(c+d x) \, dx=-\frac {2 e^{a+c+b x+d x} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{2 d},\frac {1}{2} \left (3+\frac {b}{d}\right ),e^{2 (c+d x)}\right )}{b+d} \]
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Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {5601} \[ \int e^{a+b x} \text {csch}(c+d x) \, dx=-\frac {2 e^{a+b x+c+d x} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{2 d},\frac {1}{2} \left (\frac {b}{d}+3\right ),e^{2 (c+d x)}\right )}{b+d} \]
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Rule 5601
Rubi steps \begin{align*} \text {integral}& = -\frac {2 e^{a+c+b x+d x} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{2 d},\frac {1}{2} \left (3+\frac {b}{d}\right ),e^{2 (c+d x)}\right )}{b+d} \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int e^{a+b x} \text {csch}(c+d x) \, dx=-\frac {2 e^{a+c+(b+d) x} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{2 d},\frac {1}{2} \left (3+\frac {b}{d}\right ),e^{2 (c+d x)}\right )}{b+d} \]
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\[\int {\mathrm e}^{b x +a} \operatorname {csch}\left (d x +c \right )d x\]
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\[ \int e^{a+b x} \text {csch}(c+d x) \, dx=\int { \operatorname {csch}\left (d x + c\right ) e^{\left (b x + a\right )} \,d x } \]
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\[ \int e^{a+b x} \text {csch}(c+d x) \, dx=e^{a} \int e^{b x} \operatorname {csch}{\left (c + d x \right )}\, dx \]
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\[ \int e^{a+b x} \text {csch}(c+d x) \, dx=\int { \operatorname {csch}\left (d x + c\right ) e^{\left (b x + a\right )} \,d x } \]
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\[ \int e^{a+b x} \text {csch}(c+d x) \, dx=\int { \operatorname {csch}\left (d x + c\right ) e^{\left (b x + a\right )} \,d x } \]
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Timed out. \[ \int e^{a+b x} \text {csch}(c+d x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x}}{\mathrm {sinh}\left (c+d\,x\right )} \,d x \]
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