Optimal. Leaf size=126 \[ -\frac {f \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {f \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {2 i f \log \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{a d^2}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d} \]
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Rubi [A] time = 0.15, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5575, 4182, 2279, 2391, 3318, 4184, 3475} \[ -\frac {f \text {PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac {f \text {PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac {2 i f \log \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{a d^2}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d} \]
Antiderivative was successfully verified.
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Rule 2279
Rule 2391
Rule 3318
Rule 3475
Rule 4182
Rule 4184
Rule 5575
Rubi steps
\begin {align*} \int \frac {(e+f x) \text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\left (i \int \frac {e+f x}{a+i a \sinh (c+d x)} \, dx\right )+\frac {\int (e+f x) \text {csch}(c+d x) \, dx}{a}\\ &=-\frac {2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i \int (e+f x) \csc ^2\left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {i d x}{2}\right ) \, dx}{2 a}-\frac {f \int \log \left (1-e^{c+d x}\right ) \, dx}{a d}+\frac {f \int \log \left (1+e^{c+d x}\right ) \, dx}{a d}\\ &=-\frac {2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {f \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{c+d x}\right )}{a d^2}+\frac {f \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{c+d x}\right )}{a d^2}+\frac {(i f) \int \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}\\ &=-\frac {2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {2 i f \log \left (\cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )\right )}{a d^2}-\frac {f \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {f \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}\\ \end {align*}
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Mathematica [B] time = 1.22, size = 345, normalized size = 2.74 \[ \frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (-2 i d (e+f x) \sinh \left (\frac {1}{2} (c+d x)\right )+d e \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )+f \left (\text {Li}_2\left (-e^{-c-d x}\right )-\text {Li}_2\left (e^{-c-d x}\right )+(c+d x) \left (\log \left (1-e^{-c-d x}\right )-\log \left (e^{-c-d x}+1\right )\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )+f (c+d x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )+i f \log (\cosh (c+d x)) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )-c f \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )-2 f \tan ^{-1}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )\right )}{d^2 (a+i a \sinh (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.48, size = 211, normalized size = 1.67 \[ \frac {-2 i \, d f x e^{\left (d x + c\right )} + 2 \, d e - {\left (f e^{\left (d x + c\right )} - i \, f\right )} {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) + {\left (f e^{\left (d x + c\right )} - i \, f\right )} {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) + {\left (i \, d f x + i \, d e - {\left (d f x + d e\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) - 2 \, {\left (-i \, f e^{\left (d x + c\right )} - f\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + {\left (-i \, d e + i \, c f + {\left (d e - c f\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) + {\left (-i \, d f x - i \, c f + {\left (d f x + c f\right )} e^{\left (d x + c\right )}\right )} \log \left (-e^{\left (d x + c\right )} + 1\right )}{a d^{2} e^{\left (d x + c\right )} - i \, a d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )} \operatorname {csch}\left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 211, normalized size = 1.67 \[ \frac {2 f x +2 e}{d a \left ({\mathrm e}^{d x +c}-i\right )}-\frac {2 i f \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}-\frac {e \ln \left ({\mathrm e}^{d x +c}+1\right )}{a d}+\frac {e \ln \left ({\mathrm e}^{d x +c}-1\right )}{a d}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right ) f x}{a d}+\frac {\ln \left (1-{\mathrm e}^{d x +c}\right ) f x}{a d}+\frac {\ln \left (1-{\mathrm e}^{d x +c}\right ) c f}{a \,d^{2}}-\frac {f c \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{2}}-\frac {f \polylog \left (2, -{\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {f \polylog \left (2, {\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {2 i f \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ 2 \, f {\left (\frac {x e^{\left (d x + c\right )}}{i \, a d e^{\left (d x + c\right )} + a d} + \frac {i \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}} + \int \frac {x}{2 \, {\left (a e^{\left (d x + c\right )} + a\right )}}\,{d x} + \int \frac {x}{2 \, {\left (a e^{\left (d x + c\right )} - a\right )}}\,{d x}\right )} - e {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} - \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {e+f\,x}{\mathrm {sinh}\left (c+d\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \left (\int \frac {e \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f x \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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