Optimal. Leaf size=313 \[ -\frac {12 i f^3 \text {Li}_3\left (-i e^{c+d x}\right )}{a d^4}-\frac {6 f^3 \text {Li}_4\left (-e^{c+d x}\right )}{a d^4}+\frac {6 f^3 \text {Li}_4\left (e^{c+d x}\right )}{a d^4}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {6 f^2 (e+f x) \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {6 f^2 (e+f x) \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {3 f (e+f x)^2 \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {3 f (e+f x)^2 \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {i (e+f x)^3}{a d} \]
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Rubi [A] time = 0.48, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {5575, 4182, 2531, 6609, 2282, 6589, 3318, 4184, 3716, 2190} \[ \frac {12 i f^2 (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {6 f^2 (e+f x) \text {PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac {6 f^2 (e+f x) \text {PolyLog}\left (3,e^{c+d x}\right )}{a d^3}-\frac {3 f (e+f x)^2 \text {PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac {3 f (e+f x)^2 \text {PolyLog}\left (2,e^{c+d x}\right )}{a d^2}-\frac {12 i f^3 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac {6 f^3 \text {PolyLog}\left (4,-e^{c+d x}\right )}{a d^4}+\frac {6 f^3 \text {PolyLog}\left (4,e^{c+d x}\right )}{a d^4}+\frac {6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {i (e+f x)^3}{a d} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3318
Rule 3716
Rule 4182
Rule 4184
Rule 5575
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int \frac {(e+f x)^3 \text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\left (i \int \frac {(e+f x)^3}{a+i a \sinh (c+d x)} \, dx\right )+\frac {\int (e+f x)^3 \text {csch}(c+d x) \, dx}{a}\\ &=-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i \int (e+f x)^3 \csc ^2\left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {i d x}{2}\right ) \, dx}{2 a}-\frac {(3 f) \int (e+f x)^2 \log \left (1-e^{c+d x}\right ) \, dx}{a d}+\frac {(3 f) \int (e+f x)^2 \log \left (1+e^{c+d x}\right ) \, dx}{a d}\\ &=-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {3 f (e+f x)^2 \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {3 f (e+f x)^2 \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {(3 i f) \int (e+f x)^2 \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}+\frac {\left (6 f^2\right ) \int (e+f x) \text {Li}_2\left (-e^{c+d x}\right ) \, dx}{a d^2}-\frac {\left (6 f^2\right ) \int (e+f x) \text {Li}_2\left (e^{c+d x}\right ) \, dx}{a d^2}\\ &=-\frac {i (e+f x)^3}{a d}-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {3 f (e+f x)^2 \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {3 f (e+f x)^2 \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {6 f^2 (e+f x) \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {6 f^2 (e+f x) \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {(6 f) \int \frac {e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )} (e+f x)^2}{1+i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}} \, dx}{a d}-\frac {\left (6 f^3\right ) \int \text {Li}_3\left (-e^{c+d x}\right ) \, dx}{a d^3}+\frac {\left (6 f^3\right ) \int \text {Li}_3\left (e^{c+d x}\right ) \, dx}{a d^3}\\ &=-\frac {i (e+f x)^3}{a d}-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {3 f (e+f x)^2 \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {3 f (e+f x)^2 \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {6 f^2 (e+f x) \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {6 f^2 (e+f x) \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (12 i f^2\right ) \int (e+f x) \log \left (1+i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^2}-\frac {\left (6 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{c+d x}\right )}{a d^4}+\frac {\left (6 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{c+d x}\right )}{a d^4}\\ &=-\frac {i (e+f x)^3}{a d}-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {3 f (e+f x)^2 \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {3 f (e+f x)^2 \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {6 f^2 (e+f x) \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {6 f^2 (e+f x) \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {6 f^3 \text {Li}_4\left (-e^{c+d x}\right )}{a d^4}+\frac {6 f^3 \text {Li}_4\left (e^{c+d x}\right )}{a d^4}-\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (12 i f^3\right ) \int \text {Li}_2\left (-i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^3}\\ &=-\frac {i (e+f x)^3}{a d}-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {3 f (e+f x)^2 \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {3 f (e+f x)^2 \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {6 f^2 (e+f x) \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {6 f^2 (e+f x) \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {6 f^3 \text {Li}_4\left (-e^{c+d x}\right )}{a d^4}+\frac {6 f^3 \text {Li}_4\left (e^{c+d x}\right )}{a d^4}-\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (12 i f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{a d^4}\\ &=-\frac {i (e+f x)^3}{a d}-\frac {2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {3 f (e+f x)^2 \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {3 f (e+f x)^2 \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {6 f^2 (e+f x) \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {12 i f^3 \text {Li}_3\left (-i e^{c+d x}\right )}{a d^4}-\frac {6 f^2 (e+f x) \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {6 f^3 \text {Li}_4\left (-e^{c+d x}\right )}{a d^4}+\frac {6 f^3 \text {Li}_4\left (e^{c+d x}\right )}{a d^4}-\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}\\ \end {align*}
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Mathematica [A] time = 5.37, size = 363, normalized size = 1.16 \[ \frac {\frac {2 d^3 (e+f x)^3}{e^c-i}-\frac {2 i d^3 \sinh \left (\frac {d x}{2}\right ) (e+f x)^3}{\left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}-2 d^3 (e+f x)^3 \tanh ^{-1}(\sinh (c+d x)+\cosh (c+d x))-3 f \left (d^2 (e+f x)^2 \text {Li}_2(-\cosh (c+d x)-\sinh (c+d x))-2 d f (e+f x) \text {Li}_3(-\cosh (c+d x)-\sinh (c+d x))+2 f^2 \text {Li}_4(-\cosh (c+d x)-\sinh (c+d x))\right )+3 f \left (d^2 (e+f x)^2 \text {Li}_2(\cosh (c+d x)+\sinh (c+d x))-2 d f (e+f x) \text {Li}_3(\cosh (c+d x)+\sinh (c+d x))+2 f^2 \text {Li}_4(\cosh (c+d x)+\sinh (c+d x))\right )+6 i d^2 f (e+f x)^2 \log \left (1-i e^{-c-d x}\right )-12 i f^2 \left (d (e+f x) \text {Li}_2\left (i e^{-c-d x}\right )+f \text {Li}_3\left (i e^{-c-d x}\right )\right )}{a d^4} \]
Warning: Unable to verify antiderivative.
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fricas [C] time = 0.86, size = 997, normalized size = 3.19 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (f x + e\right )}^{3} \operatorname {csch}\left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.36, size = 1034, normalized size = 3.30 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.70, size = 580, normalized size = 1.85 \[ -e^{3} {\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 )} - \frac {6 i \, e^{2} f x}{a d} - \frac {3 \, {\left (d x \log \left (e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (d x + c\right )}\right )\right )} e^{2} f}{a d^{2}} + \frac {3 \, {\left (d x \log \left (-e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (d x + c\right )}\right )\right )} e^{2} f}{a d^{2}} + \frac {6 i \, e^{2} f \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{a d^{2}} + \frac {2 \, {\left (f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac {3 \, {\left (d^{2} x^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (d x + c\right )})\right )} e f^{2}}{a d^{3}} + \frac {3 \, {\left (d^{2} x^{2} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (d x + c\right )})\right )} e f^{2}}{a d^{3}} + \frac {12 i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e f^{2}}{a d^{3}} - \frac {{\left (d^{3} x^{3} \log \left (e^{\left (d x + c\right )} + 1\right ) + 3 \, d^{2} x^{2} {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 6 \, d x {\rm Li}_{3}(-e^{\left (d x + c\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} + \frac {{\left (d^{3} x^{3} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 3 \, d^{2} x^{2} {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 6 \, d x {\rm Li}_{3}(e^{\left (d x + c\right )}) + 6 \, {\rm Li}_{4}(e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} + \frac {6 i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} - \frac {2 i \, d^{3} f^{3} x^{3} + 6 i \, d^{3} e f^{2} x^{2}}{a d^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e+f\,x\right )}^3}{\mathrm {sinh}\left (c+d\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]
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 \[ - \frac {i \left (\int \frac {e^{3} \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f^{3} x^{3} \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e f^{2} x^{2} \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e^{2} f x \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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