Optimal. Leaf size=53 \[ \frac{8}{15} a^2 \tanh (x) \sqrt{a \cosh ^2(x)}+\frac{1}{5} \tanh (x) \left (a \cosh ^2(x)\right )^{5/2}+\frac{4}{15} a \tanh (x) \left (a \cosh ^2(x)\right )^{3/2} \]
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Rubi [A] time = 0.0356787, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3203, 3207, 2637} \[ \frac{8}{15} a^2 \tanh (x) \sqrt{a \cosh ^2(x)}+\frac{1}{5} \tanh (x) \left (a \cosh ^2(x)\right )^{5/2}+\frac{4}{15} a \tanh (x) \left (a \cosh ^2(x)\right )^{3/2} \]
Antiderivative was successfully verified.
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Rule 3203
Rule 3207
Rule 2637
Rubi steps
\begin{align*} \int \left (a \cosh ^2(x)\right )^{5/2} \, dx &=\frac{1}{5} \left (a \cosh ^2(x)\right )^{5/2} \tanh (x)+\frac{1}{5} (4 a) \int \left (a \cosh ^2(x)\right )^{3/2} \, dx\\ &=\frac{4}{15} a \left (a \cosh ^2(x)\right )^{3/2} \tanh (x)+\frac{1}{5} \left (a \cosh ^2(x)\right )^{5/2} \tanh (x)+\frac{1}{15} \left (8 a^2\right ) \int \sqrt{a \cosh ^2(x)} \, dx\\ &=\frac{4}{15} a \left (a \cosh ^2(x)\right )^{3/2} \tanh (x)+\frac{1}{5} \left (a \cosh ^2(x)\right )^{5/2} \tanh (x)+\frac{1}{15} \left (8 a^2 \sqrt{a \cosh ^2(x)} \text{sech}(x)\right ) \int \cosh (x) \, dx\\ &=\frac{8}{15} a^2 \sqrt{a \cosh ^2(x)} \tanh (x)+\frac{4}{15} a \left (a \cosh ^2(x)\right )^{3/2} \tanh (x)+\frac{1}{5} \left (a \cosh ^2(x)\right )^{5/2} \tanh (x)\\ \end{align*}
Mathematica [A] time = 0.0162901, size = 36, normalized size = 0.68 \[ \frac{1}{240} a^2 (150 \sinh (x)+25 \sinh (3 x)+3 \sinh (5 x)) \text{sech}(x) \sqrt{a \cosh ^2(x)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.037, size = 32, normalized size = 0.6 \begin{align*}{\frac{{a}^{3}\cosh \left ( x \right ) \sinh \left ( x \right ) \left ( 3\, \left ( \cosh \left ( x \right ) \right ) ^{4}+4\, \left ( \cosh \left ( x \right ) \right ) ^{2}+8 \right ) }{15}{\frac{1}{\sqrt{a \left ( \cosh \left ( x \right ) \right ) ^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.56399, size = 72, normalized size = 1.36 \begin{align*} \frac{1}{160} \, a^{\frac{5}{2}} e^{\left (5 \, x\right )} + \frac{5}{96} \, a^{\frac{5}{2}} e^{\left (3 \, x\right )} - \frac{5}{16} \, a^{\frac{5}{2}} e^{\left (-x\right )} - \frac{5}{96} \, a^{\frac{5}{2}} e^{\left (-3 \, x\right )} - \frac{1}{160} \, a^{\frac{5}{2}} e^{\left (-5 \, x\right )} + \frac{5}{16} \, a^{\frac{5}{2}} e^{x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.20837, size = 1462, normalized size = 27.58 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.1672, size = 82, normalized size = 1.55 \begin{align*} \frac{1}{480} \,{\left (3 \, a^{2} e^{\left (5 \, x\right )} + 25 \, a^{2} e^{\left (3 \, x\right )} + 150 \, a^{2} e^{x} -{\left (150 \, a^{2} e^{\left (4 \, x\right )} + 25 \, a^{2} e^{\left (2 \, x\right )} + 3 \, a^{2}\right )} e^{\left (-5 \, x\right )}\right )} \sqrt{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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