Integrand size = 18, antiderivative size = 122 \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=-\frac {F^{c (a+b x)} \coth (d+e x) \text {csch}(d+e x)}{2 e}-\frac {b c F^{c (a+b x)} \text {csch}(d+e x) \log (F)}{2 e^2}+\frac {e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 (d+e x)}\right ) (e-b c \log (F))}{e^2} \]
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Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5599, 5601} \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=\frac {e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right ),e^{2 (d+e x)}\right )}{e^2}-\frac {b c \log (F) \text {csch}(d+e x) F^{c (a+b x)}}{2 e^2}-\frac {\coth (d+e x) \text {csch}(d+e x) F^{c (a+b x)}}{2 e} \]
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Rule 5599
Rule 5601
Rubi steps \begin{align*} \text {integral}& = -\frac {F^{c (a+b x)} \coth (d+e x) \text {csch}(d+e x)}{2 e}-\frac {b c F^{c (a+b x)} \text {csch}(d+e x) \log (F)}{2 e^2}-\frac {1}{2} \left (1-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text {csch}(d+e x) \, dx \\ & = -\frac {F^{c (a+b x)} \coth (d+e x) \text {csch}(d+e x)}{2 e}-\frac {b c F^{c (a+b x)} \text {csch}(d+e x) \log (F)}{2 e^2}+\frac {e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 (d+e x)}\right ) (e-b c \log (F))}{e^2} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(281\) vs. \(2(122)=244\).
Time = 14.38 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.30 \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (-e \text {csch}^2\left (\frac {1}{2} (d+e x)\right )-4 b c \text {csch}(d) \log (F)+\text {csch}(d) \left (-\frac {4 e^2}{b c \log (F)}+4 b c \log (F)\right )+\frac {4 \left (1-\left (1+e^d\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{e},1+\frac {b c \log (F)}{e},-e^{d+e x}\right )\right ) \left (e^2-b^2 c^2 \log ^2(F)\right )}{b c \left (1+e^d\right ) \log (F)}+\frac {4 \left (1+\left (-1+e^d\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{e},1+\frac {b c \log (F)}{e},e^{d+e x}\right )\right ) \left (e^2-b^2 c^2 \log ^2(F)\right )}{b c \left (-1+e^d\right ) \log (F)}-e \text {sech}^2\left (\frac {1}{2} (d+e x)\right )+2 b c \text {csch}\left (\frac {d}{2}\right ) \text {csch}\left (\frac {1}{2} (d+e x)\right ) \log (F) \sinh \left (\frac {e x}{2}\right )+2 b c \log (F) \text {sech}\left (\frac {d}{2}\right ) \text {sech}\left (\frac {1}{2} (d+e x)\right ) \sinh \left (\frac {e x}{2}\right )\right )}{8 e^2} \]
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\[\int F^{c \left (b x +a \right )} \operatorname {csch}\left (e x +d \right )^{3}d x\]
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\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{3} \,d x } \]
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\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {csch}^{3}{\left (d + e x \right )}\, dx \]
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\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{3} \,d x } \]
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\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{3} \,d x } \]
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Timed out. \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {sinh}\left (d+e\,x\right )}^3} \,d x \]
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