Optimal. Leaf size=69 \[ -\frac {2 i \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac {2 i \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^2 \text {sech}(a+b x)}{b} \]
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Rubi [A] time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5418, 4180, 2279, 2391} \[ -\frac {2 i \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac {2 i \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^2 \text {sech}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 2279
Rule 2391
Rule 4180
Rule 5418
Rubi steps
\begin {align*} \int x^2 \text {sech}(a+b x) \tanh (a+b x) \, dx &=-\frac {x^2 \text {sech}(a+b x)}{b}+\frac {2 \int x \text {sech}(a+b x) \, dx}{b}\\ &=\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^2 \text {sech}(a+b x)}{b}-\frac {(2 i) \int \log \left (1-i e^{a+b x}\right ) \, dx}{b^2}+\frac {(2 i) \int \log \left (1+i e^{a+b x}\right ) \, dx}{b^2}\\ &=\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^2 \text {sech}(a+b x)}{b}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 i \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac {2 i \text {Li}_2\left (i e^{a+b x}\right )}{b^3}-\frac {x^2 \text {sech}(a+b x)}{b}\\ \end {align*}
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Mathematica [A] time = 0.45, size = 125, normalized size = 1.81 \[ -\frac {b^2 x^2 \text {sech}(a+b x)+2 i \left (\text {Li}_2\left (-i e^{a+b x}\right )-\text {Li}_2\left (i e^{a+b x}\right )\right )+(-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-(\pi -2 i a) \log \left (\cot \left (\frac {1}{4} (2 i a+2 i b x+\pi )\right )\right )}{b^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.61, size = 474, normalized size = 6.87 \[ -\frac {2 \, b^{2} x^{2} \cosh \left (b x + a\right ) + 2 \, b^{2} x^{2} \sinh \left (b x + a\right ) - {\left (2 i \, \cosh \left (b x + a\right )^{2} + 4 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 i \, \sinh \left (b x + a\right )^{2} + 2 i\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - {\left (-2 i \, \cosh \left (b x + a\right )^{2} - 4 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 2 i \, \sinh \left (b x + a\right )^{2} - 2 i\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - {\left (-2 i \, a \cosh \left (b x + a\right )^{2} - 4 i \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 2 i \, a \sinh \left (b x + a\right )^{2} - 2 i \, a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) - {\left (2 i \, a \cosh \left (b x + a\right )^{2} + 4 i \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 i \, a \sinh \left (b x + a\right )^{2} + 2 i \, a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - {\left ({\left (-2 i \, b x - 2 i \, a\right )} \cosh \left (b x + a\right )^{2} + {\left (-4 i \, b x - 4 i \, a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (-2 i \, b x - 2 i \, a\right )} \sinh \left (b x + a\right )^{2} - 2 i \, b x - 2 i \, a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - {\left ({\left (2 i \, b x + 2 i \, a\right )} \cosh \left (b x + a\right )^{2} + {\left (4 i \, b x + 4 i \, a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (2 i \, b x + 2 i \, a\right )} \sinh \left (b x + a\right )^{2} + 2 i \, b x + 2 i \, a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} + b^{3}} \]
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 \[ \int x^{2} \operatorname {sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.33, size = 154, normalized size = 2.23 \[ -\frac {2 x^{2} {\mathrm e}^{b x +a}}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {2 i \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {2 i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{b^{3}}+\frac {2 i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {2 i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{b^{3}}-\frac {2 i \dilog \left (1+i {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {2 i \dilog \left (1-i {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {4 a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{3}} \]
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 \[ -\frac {2 \, x^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} + 4 \, \int \frac {x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\mathrm {sinh}\left (a+b\,x\right )}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int x^{2} \sinh {\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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