Optimal. Leaf size=55 \[ \frac {8 \tanh (x)}{15 a^2 \sqrt {a \text {sech}^2(x)}}+\frac {4 \tanh (x)}{15 a \left (a \text {sech}^2(x)\right )^{3/2}}+\frac {\tanh (x)}{5 \left (a \text {sech}^2(x)\right )^{5/2}} \]
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Rubi [A] time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4122, 192, 191} \[ \frac {8 \tanh (x)}{15 a^2 \sqrt {a \text {sech}^2(x)}}+\frac {4 \tanh (x)}{15 a \left (a \text {sech}^2(x)\right )^{3/2}}+\frac {\tanh (x)}{5 \left (a \text {sech}^2(x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 191
Rule 192
Rule 4122
Rubi steps
\begin {align*} \int \frac {1}{\left (a \text {sech}^2(x)\right )^{5/2}} \, dx &=a \operatorname {Subst}\left (\int \frac {1}{\left (a-a x^2\right )^{7/2}} \, dx,x,\tanh (x)\right )\\ &=\frac {\tanh (x)}{5 \left (a \text {sech}^2(x)\right )^{5/2}}+\frac {4}{5} \operatorname {Subst}\left (\int \frac {1}{\left (a-a x^2\right )^{5/2}} \, dx,x,\tanh (x)\right )\\ &=\frac {\tanh (x)}{5 \left (a \text {sech}^2(x)\right )^{5/2}}+\frac {4 \tanh (x)}{15 a \left (a \text {sech}^2(x)\right )^{3/2}}+\frac {8 \operatorname {Subst}\left (\int \frac {1}{\left (a-a x^2\right )^{3/2}} \, dx,x,\tanh (x)\right )}{15 a}\\ &=\frac {\tanh (x)}{5 \left (a \text {sech}^2(x)\right )^{5/2}}+\frac {4 \tanh (x)}{15 a \left (a \text {sech}^2(x)\right )^{3/2}}+\frac {8 \tanh (x)}{15 a^2 \sqrt {a \text {sech}^2(x)}}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 36, normalized size = 0.65 \[ \frac {(150 \sinh (x)+25 \sinh (3 x)+3 \sinh (5 x)) \cosh (x) \sqrt {a \text {sech}^2(x)}}{240 a^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.58, size = 580, normalized size = 10.55 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 41, normalized size = 0.75 \[ -\frac {{\left (150 \, e^{\left (4 \, x\right )} + 25 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-5 \, x\right )} - 3 \, e^{\left (5 \, x\right )} - 25 \, e^{\left (3 \, x\right )} - 150 \, e^{x}}{480 \, a^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.19, size = 196, normalized size = 3.56 \[ \frac {{\mathrm e}^{6 x}}{160 a^{2} \left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{4 x}}{96 a^{2} \left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{2 x}}{16 a^{2} \left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}}-\frac {5}{16 \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right ) a^{2}}-\frac {5 \,{\mathrm e}^{-2 x}}{96 a^{2} \left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}}-\frac {{\mathrm e}^{-4 x}}{160 a^{2} \left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 53, normalized size = 0.96 \[ \frac {e^{\left (5 \, x\right )}}{160 \, a^{\frac {5}{2}}} + \frac {5 \, e^{\left (3 \, x\right )}}{96 \, a^{\frac {5}{2}}} - \frac {5 \, e^{\left (-x\right )}}{16 \, a^{\frac {5}{2}}} - \frac {5 \, e^{\left (-3 \, x\right )}}{96 \, a^{\frac {5}{2}}} - \frac {e^{\left (-5 \, x\right )}}{160 \, a^{\frac {5}{2}}} + \frac {5 \, e^{x}}{16 \, a^{\frac {5}{2}}} \]
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 \[ \int \frac {1}{{\left (\frac {a}{{\mathrm {cosh}\relax (x)}^2}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 10.17, size = 60, normalized size = 1.09 \[ \frac {8 \tanh ^{5}{\relax (x )}}{15 a^{\frac {5}{2}} \left (\operatorname {sech}^{2}{\relax (x )}\right )^{\frac {5}{2}}} - \frac {4 \tanh ^{3}{\relax (x )}}{3 a^{\frac {5}{2}} \left (\operatorname {sech}^{2}{\relax (x )}\right )^{\frac {5}{2}}} + \frac {\tanh {\relax (x )}}{a^{\frac {5}{2}} \left (\operatorname {sech}^{2}{\relax (x )}\right )^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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