Optimal. Leaf size=146 \[ \frac {2 i \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {2 i \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {2 \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {2 \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {2 x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {2 x \text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac {x^2 \text {sech}(a+b x)}{b} \]
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Rubi [A] time = 0.23, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {2622, 321, 207, 5462, 14, 6273, 12, 4182, 2531, 2282, 6589, 4180, 2279, 2391} \[ -\frac {2 x \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {2 x \text {PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac {2 i \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac {2 i \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac {2 \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac {2 \text {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac {x^2 \text {sech}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 207
Rule 321
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 2622
Rule 4180
Rule 4182
Rule 5462
Rule 6273
Rule 6589
Rubi steps
\begin {align*} \int x^2 \text {csch}(a+b x) \text {sech}^2(a+b x) \, dx &=-\frac {x^2 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac {x^2 \text {sech}(a+b x)}{b}-2 \int x \left (-\frac {\tanh ^{-1}(\cosh (a+b x))}{b}+\frac {\text {sech}(a+b x)}{b}\right ) \, dx\\ &=-\frac {x^2 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac {x^2 \text {sech}(a+b x)}{b}-2 \int \left (-\frac {x \tanh ^{-1}(\cosh (a+b x))}{b}+\frac {x \text {sech}(a+b x)}{b}\right ) \, dx\\ &=-\frac {x^2 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac {x^2 \text {sech}(a+b x)}{b}+\frac {2 \int x \tanh ^{-1}(\cosh (a+b x)) \, dx}{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 {\int b x^2 \text {csch}(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) \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}+\int x^2 \text {csch}(a+b x) \, dx\\ &=-\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\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}-\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 {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {2 x \text {Li}_2\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 {2 x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {x^2 \text {sech}(a+b x)}{b}+\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 {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {2 x \text {Li}_2\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 {2 x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {x^2 \text {sech}(a+b x)}{b}+\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 {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {2 x \text {Li}_2\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 {2 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 {x^2 \text {sech}(a+b x)}{b}\\ \end {align*}
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Mathematica [A] time = 0.65, size = 225, normalized size = 1.54 \[ \frac {-2 \left (b^2 x^2 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))+b x \text {Li}_2(-\cosh (a+b x)-\sinh (a+b x))-b x \text {Li}_2(\cosh (a+b x)+\sinh (a+b x))-\text {Li}_3(-\cosh (a+b x)-\sinh (a+b x))+\text {Li}_3(\cosh (a+b x)+\sinh (a+b x))\right )+b^2 x^2 \text {sech}(a+b x)+2 i \left (\text {Li}_2\left (-i e^{a+b x}\right )-\text {Li}_2\left (i e^{a+b x}\right )\right )+(-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-(\pi -2 i a) \log \left (\cot \left (\frac {1}{4} (2 i a+2 i b x+\pi )\right )\right )}{b^3} \]
Warning: Unable to verify antiderivative.
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fricas [C] time = 0.45, size = 929, normalized size = 6.36 \[ \text {result too large to display} \]
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 = 1.75, size = 0, normalized size = 0.00 \[ \int x^{2} \mathrm {csch}\left (b x +a \right ) \mathrm {sech}\left (b x +a \right )^{2}\, dx \]
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 \[ \frac {2 \, x^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{b^{3}} + \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{b^{3}} - 8 \, \int \frac {x e^{\left (b x + a\right )}}{2 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )} \,d x \]
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 \[ \int x^{2} \operatorname {csch}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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