Optimal. Leaf size=67 \[ \frac {15 x}{8 a}-\frac {4 \sinh ^3(x)}{3 a}-\frac {4 \sinh (x)}{a}+\frac {5 \sinh (x) \cosh ^3(x)}{4 a}+\frac {15 \sinh (x) \cosh (x)}{8 a}-\frac {\sinh (x) \cosh ^3(x)}{a \text {sech}(x)+a} \]
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Rubi [A] time = 0.10, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3819, 3787, 2635, 8, 2633} \[ \frac {15 x}{8 a}-\frac {4 \sinh ^3(x)}{3 a}-\frac {4 \sinh (x)}{a}+\frac {5 \sinh (x) \cosh ^3(x)}{4 a}+\frac {15 \sinh (x) \cosh (x)}{8 a}-\frac {\sinh (x) \cosh ^3(x)}{a \text {sech}(x)+a} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 3787
Rule 3819
Rubi steps
\begin {align*} \int \frac {\cosh ^4(x)}{a+a \text {sech}(x)} \, dx &=-\frac {\cosh ^3(x) \sinh (x)}{a+a \text {sech}(x)}-\frac {\int \cosh ^4(x) (-5 a+4 a \text {sech}(x)) \, dx}{a^2}\\ &=-\frac {\cosh ^3(x) \sinh (x)}{a+a \text {sech}(x)}-\frac {4 \int \cosh ^3(x) \, dx}{a}+\frac {5 \int \cosh ^4(x) \, dx}{a}\\ &=\frac {5 \cosh ^3(x) \sinh (x)}{4 a}-\frac {\cosh ^3(x) \sinh (x)}{a+a \text {sech}(x)}-\frac {(4 i) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (x)\right )}{a}+\frac {15 \int \cosh ^2(x) \, dx}{4 a}\\ &=-\frac {4 \sinh (x)}{a}+\frac {15 \cosh (x) \sinh (x)}{8 a}+\frac {5 \cosh ^3(x) \sinh (x)}{4 a}-\frac {\cosh ^3(x) \sinh (x)}{a+a \text {sech}(x)}-\frac {4 \sinh ^3(x)}{3 a}+\frac {15 \int 1 \, dx}{8 a}\\ &=\frac {15 x}{8 a}-\frac {4 \sinh (x)}{a}+\frac {15 \cosh (x) \sinh (x)}{8 a}+\frac {5 \cosh ^3(x) \sinh (x)}{4 a}-\frac {\cosh ^3(x) \sinh (x)}{a+a \text {sech}(x)}-\frac {4 \sinh ^3(x)}{3 a}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 63, normalized size = 0.94 \[ \frac {\text {sech}\left (\frac {x}{2}\right ) \left (-360 \sinh \left (\frac {x}{2}\right )-120 \sinh \left (\frac {3 x}{2}\right )+40 \sinh \left (\frac {5 x}{2}\right )-5 \sinh \left (\frac {7 x}{2}\right )+3 \sinh \left (\frac {9 x}{2}\right )+360 x \cosh \left (\frac {x}{2}\right )\right )}{192 a} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.39, size = 139, normalized size = 2.07 \[ \frac {3 \, \cosh \relax (x)^{5} + {\left (15 \, \cosh \relax (x) - 8\right )} \sinh \relax (x)^{4} + 3 \, \sinh \relax (x)^{5} - 8 \, \cosh \relax (x)^{4} + {\left (30 \, \cosh \relax (x)^{2} - 8 \, \cosh \relax (x) + 35\right )} \sinh \relax (x)^{3} + 45 \, \cosh \relax (x)^{3} + {\left (30 \, \cosh \relax (x)^{3} - 48 \, \cosh \relax (x)^{2} + 135 \, \cosh \relax (x) - 160\right )} \sinh \relax (x)^{2} + 24 \, {\left (15 \, x - 2\right )} \cosh \relax (x) - 160 \, \cosh \relax (x)^{2} + {\left (15 \, \cosh \relax (x)^{4} - 8 \, \cosh \relax (x)^{3} + 105 \, \cosh \relax (x)^{2} + 360 \, x - 160 \, \cosh \relax (x) - 288\right )} \sinh \relax (x) + 360 \, x + 552}{192 \, {\left (a \cosh \relax (x) + a \sinh \relax (x) + a\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.11, size = 86, normalized size = 1.28 \[ \frac {15 \, x}{8 \, a} + \frac {{\left (552 \, e^{\left (4 \, x\right )} + 120 \, e^{\left (3 \, x\right )} - 40 \, e^{\left (2 \, x\right )} + 5 \, e^{x} - 3\right )} e^{\left (-4 \, x\right )}}{192 \, a {\left (e^{x} + 1\right )}} + \frac {3 \, a^{3} e^{\left (4 \, x\right )} - 8 \, a^{3} e^{\left (3 \, x\right )} + 48 \, a^{3} e^{\left (2 \, x\right )} - 168 \, a^{3} e^{x}}{192 \, a^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.14, size = 139, normalized size = 2.07 \[ -\frac {\tanh \left (\frac {x}{2}\right )}{a}+\frac {1}{4 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {5}{6 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {15}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {25}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {15 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 a}-\frac {1}{4 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {5}{6 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {15}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {25}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {15 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 80, normalized size = 1.19 \[ \frac {15 \, x}{8 \, a} + \frac {168 \, e^{\left (-x\right )} - 48 \, e^{\left (-2 \, x\right )} + 8 \, e^{\left (-3 \, x\right )} - 3 \, e^{\left (-4 \, x\right )}}{192 \, a} - \frac {5 \, e^{\left (-x\right )} - 40 \, e^{\left (-2 \, x\right )} + 120 \, e^{\left (-3 \, x\right )} + 552 \, e^{\left (-4 \, x\right )} - 3}{192 \, {\left (a e^{\left (-4 \, x\right )} + a e^{\left (-5 \, x\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.45, size = 88, normalized size = 1.31 \[ \frac {7\,{\mathrm {e}}^{-x}}{8\,a}-\frac {{\mathrm {e}}^{-2\,x}}{4\,a}+\frac {{\mathrm {e}}^{2\,x}}{4\,a}+\frac {{\mathrm {e}}^{-3\,x}}{24\,a}-\frac {{\mathrm {e}}^{3\,x}}{24\,a}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a}+\frac {{\mathrm {e}}^{4\,x}}{64\,a}+\frac {15\,x}{8\,a}+\frac {2}{a\,\left ({\mathrm {e}}^x+1\right )}-\frac {7\,{\mathrm {e}}^x}{8\,a} \]
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 \[ \frac {\int \frac {\cosh ^{4}{\relax (x )}}{\operatorname {sech}{\relax (x )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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