Optimal. Leaf size=237 \[ \frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac {6 \text {Li}_3\left (e^{a+b x}\right )}{b^4}+\frac {6 i \text {Li}_4\left (-i e^{a+b x}\right )}{b^4}-\frac {6 i \text {Li}_4\left (i e^{a+b x}\right )}{b^4}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}-\frac {6 i x \text {Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac {6 i x \text {Li}_3\left (i e^{a+b x}\right )}{b^3}+\frac {3 i x^2 \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac {3 i x^2 \text {Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x^3 \text {csch}(a+b x)}{b} \]
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Rubi [A] time = 0.34, antiderivative size = 237, 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 = {2621, 321, 207, 5462, 14, 5205, 12, 4180, 2531, 6609, 2282, 6589, 4182} \[ \frac {3 i x^2 \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac {3 i x^2 \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac {6 x \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac {6 x \text {PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac {6 i x \text {PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac {6 i x \text {PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac {6 \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^4}-\frac {6 \text {PolyLog}\left (3,e^{a+b x}\right )}{b^4}+\frac {6 i \text {PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac {6 i \text {PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x^3 \text {csch}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 207
Rule 321
Rule 2282
Rule 2531
Rule 2621
Rule 4180
Rule 4182
Rule 5205
Rule 5462
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx &=-\frac {x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac {x^3 \text {csch}(a+b x)}{b}-3 \int x^2 \left (-\frac {\tan ^{-1}(\sinh (a+b x))}{b}-\frac {\text {csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac {x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac {x^3 \text {csch}(a+b x)}{b}-3 \int \left (-\frac {x^2 \tan ^{-1}(\sinh (a+b x))}{b}-\frac {x^2 \text {csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac {x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac {x^3 \text {csch}(a+b x)}{b}+\frac {3 \int x^2 \tan ^{-1}(\sinh (a+b x)) \, dx}{b}+\frac {3 \int x^2 \text {csch}(a+b x) \, dx}{b}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 \int x \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac {6 \int x \log \left (1+e^{a+b x}\right ) \, dx}{b^2}-\frac {\int b x^3 \text {sech}(a+b x) \, dx}{b}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \int \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^3}-\frac {6 \int \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^3}-\int x^3 \text {sech}(a+b x) \, dx\\ &=-\frac {2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\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 {2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\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 {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac {6 \text {Li}_3\left (e^{a+b x}\right )}{b^4}-\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 {2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\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 {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\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 {6 \text {Li}_3\left (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 {2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\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 {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\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 {6 \text {Li}_3\left (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 {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac {2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\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 {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\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 {6 \text {Li}_3\left (e^{a+b x}\right )}{b^4}+\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}\\ \end {align*}
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Mathematica [A] time = 1.81, size = 333, normalized size = 1.41 \[ \frac {-2 b^3 x^3 \text {csch}(a)+b^3 x^3 \text {csch}\left (\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}\right ) \text {csch}\left (\frac {1}{2} (a+b x)\right )+b^3 x^3 \text {sech}\left (\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}\right ) \text {sech}\left (\frac {1}{2} (a+b x)\right )-12 \left (b^2 x^2 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))+b x \text {Li}_2(-\cosh (a+b x)-\sinh (a+b x))-b x \text {Li}_2(\cosh (a+b x)+\sinh (a+b x))-\text {Li}_3(-\cosh (a+b x)-\sinh (a+b x))+\text {Li}_3(\cosh (a+b x)+\sinh (a+b x))\right )-2 i \left (b^3 x^3 \log \left (1-i e^{a+b x}\right )-b^3 x^3 \log \left (1+i e^{a+b x}\right )-3 b^2 x^2 \text {Li}_2\left (-i e^{a+b x}\right )+3 b^2 x^2 \text {Li}_2\left (i e^{a+b x}\right )+6 b x \text {Li}_3\left (-i e^{a+b x}\right )-6 b x \text {Li}_3\left (i e^{a+b x}\right )-6 \text {Li}_4\left (-i e^{a+b x}\right )+6 \text {Li}_4\left (i e^{a+b x}\right )\right )}{2 b^4} \]
Warning: Unable to verify antiderivative.
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fricas [C] time = 0.48, size = 1307, normalized size = 5.51 \[ \text {result too large to display} \]
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 \[ \int x^{3} \operatorname {csch}\left (b x + a\right )^{2} \operatorname {sech}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.55, size = 0, normalized size = 0.00 \[ \int x^{3} \mathrm {csch}\left (b x +a \right )^{2} \mathrm {sech}\left (b x +a \right )\, dx \]
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 \[ -\frac {2 \, x^{3} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac {3 \, {\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}} - 8 \, \int \frac {x^{3} e^{\left (b x + a\right )}}{4 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \]
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 \[ \int \frac {x^3}{\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int x^{3} \operatorname {csch}^{2}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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