Optimal. Leaf size=240 \[ -\frac {3 i \text {Li}_2\left (-i e^{a+b x}\right )}{b^4}+\frac {3 i \text {Li}_2\left (i e^{a+b x}\right )}{b^4}-\frac {3 i \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {3 i \text {Li}_4\left (i e^{a+b x}\right )}{b^4}+\frac {3 i x \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {3 i x \text {Li}_3\left (i e^{a+b x}\right )}{b^3}+\frac {6 x \tan ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac {3 i x^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {3 i x^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}-\frac {3 x^2 \text {sech}(a+b x)}{2 b^2}+\frac {x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b} \]
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Rubi [A] time = 0.30, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5455, 4180, 2531, 6609, 2282, 6589, 4186, 2279, 2391} \[ -\frac {3 i x^2 \text {PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}+\frac {3 i x^2 \text {PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}+\frac {3 i x \text {PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}-\frac {3 i x \text {PolyLog}\left (3,i e^{a+b x}\right )}{b^3}-\frac {3 i \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^4}+\frac {3 i \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^4}-\frac {3 i \text {PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}+\frac {3 i \text {PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac {3 x^2 \text {sech}(a+b x)}{2 b^2}+\frac {6 x \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x^3 \tanh (a+b x) \text {sech}(a+b x)}{2 b} \]
Antiderivative was successfully verified.
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Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 4180
Rule 4186
Rule 5455
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \text {sech}(a+b x) \tanh ^2(a+b x) \, dx &=\int x^3 \text {sech}(a+b x) \, dx-\int x^3 \text {sech}^3(a+b x) \, dx\\ &=\frac {2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 x^2 \text {sech}(a+b x)}{2 b^2}-\frac {x^3 \text {sech}(a+b x) \tanh (a+b x)}{2 b}-\frac {1}{2} \int x^3 \text {sech}(a+b x) \, dx+\frac {3 \int x \text {sech}(a+b x) \, dx}{b^2}-\frac {(3 i) \int x^2 \log \left (1-i e^{a+b x}\right ) \, dx}{b}+\frac {(3 i) \int x^2 \log \left (1+i e^{a+b x}\right ) \, dx}{b}\\ &=\frac {6 x \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i x^2 \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {3 i x^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac {3 x^2 \text {sech}(a+b x)}{2 b^2}-\frac {x^3 \text {sech}(a+b x) \tanh (a+b x)}{2 b}-\frac {(3 i) \int \log \left (1-i e^{a+b x}\right ) \, dx}{b^3}+\frac {(3 i) \int \log \left (1+i e^{a+b x}\right ) \, dx}{b^3}+\frac {(6 i) \int x \text {Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}-\frac {(6 i) \int x \text {Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}+\frac {(3 i) \int x^2 \log \left (1-i e^{a+b x}\right ) \, dx}{2 b}-\frac {(3 i) \int x^2 \log \left (1+i e^{a+b x}\right ) \, dx}{2 b}\\ &=\frac {6 x \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i x^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {3 i x^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac {6 i x \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {3 x^2 \text {sech}(a+b x)}{2 b^2}-\frac {x^3 \text {sech}(a+b x) \tanh (a+b x)}{2 b}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac {(6 i) \int \text {Li}_3\left (-i e^{a+b x}\right ) \, dx}{b^3}+\frac {(6 i) \int \text {Li}_3\left (i e^{a+b x}\right ) \, dx}{b^3}-\frac {(3 i) \int x \text {Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}+\frac {(3 i) \int x \text {Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}\\ &=\frac {6 x \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i \text {Li}_2\left (-i e^{a+b x}\right )}{b^4}-\frac {3 i x^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {3 i \text {Li}_2\left (i e^{a+b x}\right )}{b^4}+\frac {3 i x^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac {3 i x \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {3 i x \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {3 x^2 \text {sech}(a+b x)}{2 b^2}-\frac {x^3 \text {sech}(a+b x) \tanh (a+b x)}{2 b}-\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {(3 i) \int \text {Li}_3\left (-i e^{a+b x}\right ) \, dx}{b^3}-\frac {(3 i) \int \text {Li}_3\left (i e^{a+b x}\right ) \, dx}{b^3}\\ &=\frac {6 x \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i \text {Li}_2\left (-i e^{a+b x}\right )}{b^4}-\frac {3 i x^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {3 i \text {Li}_2\left (i e^{a+b x}\right )}{b^4}+\frac {3 i x^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac {3 i x \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {3 i x \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {6 i \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {6 i \text {Li}_4\left (i e^{a+b x}\right )}{b^4}-\frac {3 x^2 \text {sech}(a+b x)}{2 b^2}-\frac {x^3 \text {sech}(a+b x) \tanh (a+b x)}{2 b}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=\frac {6 x \tan ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 i \text {Li}_2\left (-i e^{a+b x}\right )}{b^4}-\frac {3 i x^2 \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {3 i \text {Li}_2\left (i e^{a+b x}\right )}{b^4}+\frac {3 i x^2 \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}+\frac {3 i x \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}-\frac {3 i x \text {Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac {3 i \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}+\frac {3 i \text {Li}_4\left (i e^{a+b x}\right )}{b^4}-\frac {3 x^2 \text {sech}(a+b x)}{2 b^2}-\frac {x^3 \text {sech}(a+b x) \tanh (a+b x)}{2 b}\\ \end {align*}
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Mathematica [A] time = 3.42, size = 245, normalized size = 1.02 \[ -\frac {b^3 x^3 \text {sech}(a) \sinh (b x) \text {sech}^2(a+b x)+b^2 x^2 (b x \tanh (a)+3) \text {sech}(a+b x)-i \left (b^3 x^3 \log \left (1-i e^{a+b x}\right )-b^3 x^3 \log \left (1+i e^{a+b x}\right )-3 \left (b^2 x^2+2\right ) \text {Li}_2\left (-i e^{a+b x}\right )+3 \left (b^2 x^2+2\right ) \text {Li}_2\left (i e^{a+b x}\right )+6 b x \text {Li}_3\left (-i e^{a+b x}\right )-6 b x \text {Li}_3\left (i e^{a+b x}\right )-6 \text {Li}_4\left (-i e^{a+b x}\right )+6 \text {Li}_4\left (i e^{a+b x}\right )+6 b x \log \left (1-i e^{a+b x}\right )-6 b x \log \left (1+i e^{a+b x}\right )\right )}{2 b^4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.90, size = 2125, normalized size = 8.85 \[ \text {result too large to display} \]
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 \[ \int x^{3} \operatorname {sech}\left (b x + a\right )^{3} \sinh \left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.37, size = 0, normalized size = 0.00 \[ \int x^{3} \mathrm {sech}\left (b x +a \right )^{3} \left (\sinh ^{2}\left (b x +a \right )\right )\, dx \]
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 \[ -\frac {{\left (b x^{3} e^{\left (3 \, a\right )} + 3 \, x^{2} e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} - {\left (b x^{3} e^{a} - 3 \, x^{2} e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + 2 \, \int \frac {{\left (b^{2} x^{3} e^{a} + 6 \, x e^{a}\right )} e^{\left (b x\right )}}{2 \, {\left (b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}}\,{d x} \]
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 \[ \int \frac {x^3\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\mathrm {cosh}\left (a+b\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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