Optimal. Leaf size=123 \[ \frac {\text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {\text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b} \]
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Rubi [A] time = 0.24, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5457, 4182, 2531, 2282, 6589, 4186, 3770} \[ -\frac {x \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {x \text {PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac {\text {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac {\text {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 3770
Rule 4182
Rule 4186
Rule 5457
Rule 6589
Rubi steps
\begin {align*} \int x^2 \coth ^2(a+b x) \text {csch}(a+b x) \, dx &=\int x^2 \text {csch}(a+b x) \, dx+\int x^2 \text {csch}^3(a+b x) \, dx\\ &=-\frac {2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {1}{2} \int x^2 \text {csch}(a+b x) \, dx+\frac {\int \text {csch}(a+b x) \, dx}{b^2}-\frac {2 \int x \log \left (1-e^{a+b x}\right ) \, dx}{b}+\frac {2 \int x \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {2 x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {2 x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {2 \int \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}-\frac {2 \int \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}+\frac {\int x \log \left (1-e^{a+b x}\right ) \, dx}{b}-\frac {\int x \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}-\frac {2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}-\frac {\int \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}+\frac {\int \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {2 \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {2 \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {\text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {\text {Li}_3\left (e^{a+b x}\right )}{b^3}\\ \end {align*}
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Mathematica [A] time = 4.15, size = 222, normalized size = 1.80 \[ -\frac {-4 b^2 x^2 \log \left (1-e^{a+b x}\right )+4 b^2 x^2 \log \left (e^{a+b x}+1\right )+b^2 x^2 \text {csch}^2\left (\frac {1}{2} (a+b x)\right )+b^2 x^2 \text {sech}^2\left (\frac {1}{2} (a+b x)\right )+8 b x \text {Li}_2\left (-e^{a+b x}\right )-8 b x \text {Li}_2\left (e^{a+b x}\right )-8 \text {Li}_3\left (-e^{a+b x}\right )+8 \text {Li}_3\left (e^{a+b x}\right )-8 \log \left (1-e^{a+b x}\right )+8 \log \left (e^{a+b x}+1\right )+8 b x \text {csch}(a)-4 b x \text {csch}\left (\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}\right ) \text {csch}\left (\frac {1}{2} (a+b x)\right )-4 b x \text {sech}\left (\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}\right ) \text {sech}\left (\frac {1}{2} (a+b x)\right )}{8 b^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.45, size = 1311, normalized size = 10.66 \[ \text {result too large to display} \]
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 \[ \int x^{2} \cosh \left (b x + a\right )^{2} \operatorname {csch}\left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.72, size = 210, normalized size = 1.71 \[ -\frac {x \,{\mathrm e}^{b x +a} \left (b x \,{\mathrm e}^{2 b x +2 a}+b x +2 \,{\mathrm e}^{2 b x +2 a}-2\right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}-\frac {a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{2 b}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}-\frac {\polylog \left (2, -{\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {\polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{2 b}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}+\frac {\polylog \left (2, {\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {\polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 197, normalized size = 1.60 \[ -\frac {{\left (b x^{2} e^{\left (3 \, a\right )} + 2 \, x e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} + {\left (b x^{2} e^{a} - 2 \, x e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} - \frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{2 \, b^{3}} + \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{2 \, b^{3}} - \frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{b^{3}} + \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{b^{3}} \]
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 \[ \int \frac {x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^2}{{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \]
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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