Optimal. Leaf size=226 \[ -\frac {6 i \text {Li}_3\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i \text {Li}_3\left (i e^{a+b x}\right )}{b^4}-\frac {6 \text {Li}_4\left (-e^{a+b x}\right )}{b^4}+\frac {6 \text {Li}_4\left (e^{a+b x}\right )}{b^4}+\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {6 x \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {3 x^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {3 x^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac {x^3 \text {sech}(a+b x)}{b} \]
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Rubi [A] time = 0.34, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 13, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {2622, 321, 207, 5462, 14, 6273, 12, 4182, 2531, 6609, 2282, 6589, 4180} \[ -\frac {3 x^2 \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {3 x^2 \text {PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac {6 i x \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac {6 x \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac {6 x \text {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {6 i \text {PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac {6 i \text {PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac {6 \text {PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac {6 \text {PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac {x^3 \text {sech}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 207
Rule 321
Rule 2282
Rule 2531
Rule 2622
Rule 4180
Rule 4182
Rule 5462
Rule 6273
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx &=-\frac {x^3 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac {x^3 \text {sech}(a+b x)}{b}-3 \int x^2 \left (-\frac {\tanh ^{-1}(\cosh (a+b x))}{b}+\frac {\text {sech}(a+b x)}{b}\right ) \, dx\\ &=-\frac {x^3 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac {x^3 \text {sech}(a+b x)}{b}-3 \int \left (-\frac {x^2 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac {x^2 \text {sech}(a+b x)}{b}\right ) \, dx\\ &=-\frac {x^3 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac {x^3 \text {sech}(a+b x)}{b}+\frac {3 \int x^2 \tanh ^{-1}(\cosh (a+b x)) \, dx}{b}-\frac {3 \int x^2 \text {sech}(a+b x) \, dx}{b}\\ &=-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac {x^3 \text {sech}(a+b x)}{b}+\frac {(6 i) \int x \log \left (1-i e^{a+b x}\right ) \, dx}{b^2}-\frac {(6 i) \int x \log \left (1+i e^{a+b x}\right ) \, dx}{b^2}+\frac {\int b x^3 \text {csch}(a+b x) \, dx}{b}\\ &=-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {x^3 \text {sech}(a+b x)}{b}-\frac {(6 i) \int \text {Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^3}+\frac {(6 i) \int \text {Li}_2\left (i e^{a+b x}\right ) \, dx}{b^3}+\int x^3 \text {csch}(a+b x) \, dx\\ &=-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {x^3 \text {sech}(a+b x)}{b}-\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac {3 \int x^2 \log \left (1-e^{a+b x}\right ) \, dx}{b}+\frac {3 \int x^2 \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 x^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {3 x^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {6 i \text {Li}_3\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i \text {Li}_3\left (i e^{a+b x}\right )}{b^4}+\frac {x^3 \text {sech}(a+b x)}{b}+\frac {6 \int x \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}-\frac {6 \int x \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 x^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {3 x^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {6 x \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {6 i \text {Li}_3\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i \text {Li}_3\left (i e^{a+b x}\right )}{b^4}-\frac {6 x \text {Li}_3\left (e^{a+b x}\right )}{b^3}+\frac {x^3 \text {sech}(a+b x)}{b}-\frac {6 \int \text {Li}_3\left (-e^{a+b x}\right ) \, dx}{b^3}+\frac {6 \int \text {Li}_3\left (e^{a+b x}\right ) \, dx}{b^3}\\ &=-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 x^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {3 x^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {6 x \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {6 i \text {Li}_3\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i \text {Li}_3\left (i e^{a+b x}\right )}{b^4}-\frac {6 x \text {Li}_3\left (e^{a+b x}\right )}{b^3}+\frac {x^3 \text {sech}(a+b x)}{b}-\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 x^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {3 x^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {6 x \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {6 i \text {Li}_3\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i \text {Li}_3\left (i e^{a+b x}\right )}{b^4}-\frac {6 x \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {6 \text {Li}_4\left (-e^{a+b x}\right )}{b^4}+\frac {6 \text {Li}_4\left (e^{a+b x}\right )}{b^4}+\frac {x^3 \text {sech}(a+b x)}{b}\\ \end {align*}
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Mathematica [A] time = 0.40, size = 282, normalized size = 1.25 \[ \frac {b^3 x^3 \text {sech}(a+b x)-2 b^3 x^3 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))-3 i \left (b^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 x^2 \log \left (1+i e^{a+b x}\right )-2 b x \text {Li}_2\left (-i e^{a+b x}\right )+2 b x \text {Li}_2\left (i e^{a+b x}\right )+2 \text {Li}_3\left (-i e^{a+b x}\right )-2 \text {Li}_3\left (i e^{a+b x}\right )\right )-3 b^2 x^2 \text {Li}_2(-\cosh (a+b x)-\sinh (a+b x))+3 b^2 x^2 \text {Li}_2(\cosh (a+b x)+\sinh (a+b x))+6 b x \text {Li}_3(-\cosh (a+b x)-\sinh (a+b x))-6 b x \text {Li}_3(\cosh (a+b x)+\sinh (a+b x))-6 \text {Li}_4(-\cosh (a+b x)-\sinh (a+b x))+6 \text {Li}_4(\cosh (a+b x)+\sinh (a+b x))}{b^4} \]
Warning: Unable to verify antiderivative.
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fricas [C] time = 0.46, size = 1270, normalized size = 5.62 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.58, size = 0, normalized size = 0.00 \[ \int x^{3} \mathrm {csch}\left (b x +a \right ) \mathrm {sech}\left (b x +a \right )^{2}\, dx \]
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 \[ \frac {2 \, x^{3} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} + \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} - 24 \, \int \frac {x^{2} e^{\left (b x + a\right )}}{4 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}}\,{d x} \]
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 \[ \int \frac {x^3}{{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )} \,d x \]
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 \[ \int x^{3} \operatorname {csch}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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