Integrand size = 13, antiderivative size = 44 \[ \int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx=-\frac {x}{8 a}-\frac {\cosh (x) \sinh (x)}{8 a}+\frac {\cosh ^3(x) \sinh (x)}{4 a}-\frac {\sinh ^3(x)}{3 a} \]
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Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3957, 2918, 2644, 30, 2648, 2715, 8} \[ \int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx=-\frac {x}{8 a}-\frac {\sinh ^3(x)}{3 a}+\frac {\sinh (x) \cosh ^3(x)}{4 a}-\frac {\sinh (x) \cosh (x)}{8 a} \]
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Rule 8
Rule 30
Rule 2644
Rule 2648
Rule 2715
Rule 2918
Rule 3957
Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cosh (x) \sinh ^4(x)}{-a-a \cosh (x)} \, dx \\ & = -\frac {\int \cosh (x) \sinh ^2(x) \, dx}{a}+\frac {\int \cosh ^2(x) \sinh ^2(x) \, dx}{a} \\ & = \frac {\cosh ^3(x) \sinh (x)}{4 a}-\frac {i \text {Subst}\left (\int x^2 \, dx,x,i \sinh (x)\right )}{a}-\frac {\int \cosh ^2(x) \, dx}{4 a} \\ & = -\frac {\cosh (x) \sinh (x)}{8 a}+\frac {\cosh ^3(x) \sinh (x)}{4 a}-\frac {\sinh ^3(x)}{3 a}-\frac {\int 1 \, dx}{8 a} \\ & = -\frac {x}{8 a}-\frac {\cosh (x) \sinh (x)}{8 a}+\frac {\cosh ^3(x) \sinh (x)}{4 a}-\frac {\sinh ^3(x)}{3 a} \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.64 \[ \int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx=\frac {24 \sinh (x)-8 \sinh (3 x)+3 (-4 x+\sinh (4 x))}{96 a} \]
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Time = 8.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.36
method | result | size |
risch | \(-\frac {x}{8 a}+\frac {{\mathrm e}^{4 x}}{64 a}-\frac {{\mathrm e}^{3 x}}{24 a}+\frac {{\mathrm e}^{x}}{8 a}-\frac {{\mathrm e}^{-x}}{8 a}+\frac {{\mathrm e}^{-3 x}}{24 a}-\frac {{\mathrm e}^{-4 x}}{64 a}\) | \(60\) |
default | \(\frac {\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {5}{6 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {7}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {16}{128 \tanh \left (\frac {x}{2}\right )-128}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8}-\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {5}{6 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {7}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {16}{128 \tanh \left (\frac {x}{2}\right )+128}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8}}{a}\) | \(105\) |
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Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx=\frac {{\left (3 \, \cosh \left (x\right ) - 2\right )} \sinh \left (x\right )^{3} + 3 \, {\left (\cosh \left (x\right )^{3} - 2 \, \cosh \left (x\right )^{2} + 2\right )} \sinh \left (x\right ) - 3 \, x}{24 \, a} \]
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\[ \int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx=\frac {\int \frac {\sinh ^{4}{\left (x \right )}}{\operatorname {sech}{\left (x \right )} + 1}\, dx}{a} \]
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Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.23 \[ \int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx=-\frac {{\left (8 \, e^{\left (-x\right )} - 24 \, e^{\left (-3 \, x\right )} - 3\right )} e^{\left (4 \, x\right )}}{192 \, a} - \frac {x}{8 \, a} - \frac {24 \, e^{\left (-x\right )} - 8 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )}}{192 \, a} \]
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Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx=-\frac {{\left (24 \, e^{\left (3 \, x\right )} - 8 \, e^{x} + 3\right )} e^{\left (-4 \, x\right )} + 24 \, x - 3 \, e^{\left (4 \, x\right )} + 8 \, e^{\left (3 \, x\right )} - 24 \, e^{x}}{192 \, a} \]
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Time = 2.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.34 \[ \int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx=\frac {{\mathrm {e}}^{-3\,x}}{24\,a}-\frac {{\mathrm {e}}^{-x}}{8\,a}-\frac {{\mathrm {e}}^{3\,x}}{24\,a}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a}+\frac {{\mathrm {e}}^{4\,x}}{64\,a}-\frac {x}{8\,a}+\frac {{\mathrm {e}}^x}{8\,a} \]
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