Optimal. Leaf size=50 \[ \frac {2}{5} \sinh ^2(x) \tanh (x) \sqrt {\sinh (x) \tanh (x)}+\frac {16}{15} \tanh (x) \sqrt {\sinh (x) \tanh (x)}-\frac {64}{15} \coth (x) \sqrt {\sinh (x) \tanh (x)} \]
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Rubi [A] time = 0.12, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4398, 4400, 2598, 2594, 2589} \[ \frac {2}{5} \sinh ^2(x) \tanh (x) \sqrt {\sinh (x) \tanh (x)}+\frac {16}{15} \tanh (x) \sqrt {\sinh (x) \tanh (x)}-\frac {64}{15} \coth (x) \sqrt {\sinh (x) \tanh (x)} \]
Antiderivative was successfully verified.
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Rule 2589
Rule 2594
Rule 2598
Rule 4398
Rule 4400
Rubi steps
\begin {align*} \int (\sinh (x) \tanh (x))^{5/2} \, dx &=\frac {\sqrt {\sinh (x) \tanh (x)} \int (-\sinh (x) \tanh (x))^{5/2} \, dx}{\sqrt {-\sinh (x) \tanh (x)}}\\ &=\frac {\sqrt {\sinh (x) \tanh (x)} \int (i \sinh (x))^{5/2} (i \tanh (x))^{5/2} \, dx}{\sqrt {i \sinh (x)} \sqrt {i \tanh (x)}}\\ &=\frac {2}{5} \sinh ^2(x) \tanh (x) \sqrt {\sinh (x) \tanh (x)}+\frac {\left (8 \sqrt {\sinh (x) \tanh (x)}\right ) \int \sqrt {i \sinh (x)} (i \tanh (x))^{5/2} \, dx}{5 \sqrt {i \sinh (x)} \sqrt {i \tanh (x)}}\\ &=\frac {16}{15} \tanh (x) \sqrt {\sinh (x) \tanh (x)}+\frac {2}{5} \sinh ^2(x) \tanh (x) \sqrt {\sinh (x) \tanh (x)}-\frac {\left (32 \sqrt {\sinh (x) \tanh (x)}\right ) \int \sqrt {i \sinh (x)} \sqrt {i \tanh (x)} \, dx}{15 \sqrt {i \sinh (x)} \sqrt {i \tanh (x)}}\\ &=-\frac {64}{15} \coth (x) \sqrt {\sinh (x) \tanh (x)}+\frac {16}{15} \tanh (x) \sqrt {\sinh (x) \tanh (x)}+\frac {2}{5} \sinh ^2(x) \tanh (x) \sqrt {\sinh (x) \tanh (x)}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 29, normalized size = 0.58 \[ -\frac {2}{15} \tanh (x) \sqrt {\sinh (x) \tanh (x)} \left (-3 \cosh ^2(x)+32 \coth ^2(x)-5\right ) \]
Antiderivative was successfully verified.
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fricas [B] time = 0.42, size = 253, normalized size = 5.06 \[ \frac {\sqrt {\frac {1}{2}} {\left (3 \, \cosh \relax (x)^{8} + 24 \, \cosh \relax (x) \sinh \relax (x)^{7} + 3 \, \sinh \relax (x)^{8} + 12 \, {\left (7 \, \cosh \relax (x)^{2} - 9\right )} \sinh \relax (x)^{6} - 108 \, \cosh \relax (x)^{6} + 24 \, {\left (7 \, \cosh \relax (x)^{3} - 27 \, \cosh \relax (x)\right )} \sinh \relax (x)^{5} + 2 \, {\left (105 \, \cosh \relax (x)^{4} - 810 \, \cosh \relax (x)^{2} - 151\right )} \sinh \relax (x)^{4} - 302 \, \cosh \relax (x)^{4} + 8 \, {\left (21 \, \cosh \relax (x)^{5} - 270 \, \cosh \relax (x)^{3} - 151 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 12 \, {\left (7 \, \cosh \relax (x)^{6} - 135 \, \cosh \relax (x)^{4} - 151 \, \cosh \relax (x)^{2} - 9\right )} \sinh \relax (x)^{2} - 108 \, \cosh \relax (x)^{2} + 8 \, {\left (3 \, \cosh \relax (x)^{7} - 81 \, \cosh \relax (x)^{5} - 151 \, \cosh \relax (x)^{3} - 27 \, \cosh \relax (x)\right )} \sinh \relax (x) + 3\right )}}{30 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + {\left (6 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + \cosh \relax (x)^{2} + 2 \, {\left (2 \, \cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x)\right )} \sqrt {\cosh \relax (x)^{3} + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} + \sinh \relax (x)^{3} + {\left (3 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x) + \cosh \relax (x)}} \]
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 \[ \int \left (\sinh \relax (x) \tanh \relax (x)\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.44, size = 0, normalized size = 0.00 \[ \int \left (\sinh \relax (x ) \tanh \relax (x )\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.69, size = 103, normalized size = 2.06 \[ -\frac {\sqrt {2} e^{\left (\frac {5}{2} \, x\right )}}{20 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} + \frac {7 \, \sqrt {2} e^{\left (\frac {1}{2} \, x\right )}}{4 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} + \frac {41 \, \sqrt {2} e^{\left (-\frac {3}{2} \, x\right )}}{6 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} + \frac {41 \, \sqrt {2} e^{\left (-\frac {7}{2} \, x\right )}}{6 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} + \frac {7 \, \sqrt {2} e^{\left (-\frac {11}{2} \, x\right )}}{4 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} - \frac {\sqrt {2} e^{\left (-\frac {15}{2} \, x\right )}}{20 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (\mathrm {sinh}\relax (x)\,\mathrm {tanh}\relax (x)\right )}^{5/2} \,d x \]
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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