Integrand size = 12, antiderivative size = 76 \[ \int \frac {1}{(1-\cosh (c+d x))^3} \, dx=-\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))} \]
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Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2729, 2727} \[ \int \frac {1}{(1-\cosh (c+d x))^3} \, dx=-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3} \]
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Rule 2727
Rule 2729
Rubi steps \begin{align*} \text {integral}& = -\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}+\frac {2}{5} \int \frac {1}{(1-\cosh (c+d x))^2} \, dx \\ & = -\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}+\frac {2}{15} \int \frac {1}{1-\cosh (c+d x)} \, dx \\ & = -\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(1-\cosh (c+d x))^3} \, dx=\frac {(8-6 \cosh (c+d x)+\cosh (2 (c+d x))) \sinh (c+d x)}{15 d (-1+\cosh (c+d x))^3} \]
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Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.49
method | result | size |
risch | \(\frac {\frac {8 \,{\mathrm e}^{2 d x +2 c}}{3}-\frac {4 \,{\mathrm e}^{d x +c}}{3}+\frac {4}{15}}{d \left ({\mathrm e}^{d x +c}-1\right )^{5}}\) | \(37\) |
parallelrisch | \(\frac {3 \coth \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-10 \coth \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+15 \coth \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 d}\) | \(44\) |
derivativedivides | \(\frac {-\frac {1}{6 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {1}{20 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}}{d}\) | \(45\) |
default | \(\frac {-\frac {1}{6 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {1}{20 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}}{d}\) | \(45\) |
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Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (64) = 128\).
Time = 0.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.29 \[ \int \frac {1}{(1-\cosh (c+d x))^3} \, dx=\frac {4 \, {\left (11 \, \cosh \left (d x + c\right ) + 9 \, \sinh \left (d x + c\right ) - 5\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{4} + d \sinh \left (d x + c\right )^{4} - 5 \, d \cosh \left (d x + c\right )^{3} + {\left (4 \, d \cosh \left (d x + c\right ) - 5 \, d\right )} \sinh \left (d x + c\right )^{3} + 10 \, d \cosh \left (d x + c\right )^{2} + {\left (6 \, d \cosh \left (d x + c\right )^{2} - 15 \, d \cosh \left (d x + c\right ) + 10 \, d\right )} \sinh \left (d x + c\right )^{2} - 11 \, d \cosh \left (d x + c\right ) + {\left (4 \, d \cosh \left (d x + c\right )^{3} - 15 \, d \cosh \left (d x + c\right )^{2} + 20 \, d \cosh \left (d x + c\right ) - 9 \, d\right )} \sinh \left (d x + c\right ) + 5 \, d\right )}} \]
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Time = 1.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(1-\cosh (c+d x))^3} \, dx=\begin {cases} \frac {1}{4 d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}} - \frac {1}{6 d \tanh ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}} + \frac {1}{20 d \tanh ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}} & \text {for}\: d \neq 0 \\\frac {x}{\left (1 - \cosh {\left (c \right )}\right )^{3}} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (64) = 128\).
Time = 0.19 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.70 \[ \int \frac {1}{(1-\cosh (c+d x))^3} \, dx=\frac {4 \, e^{\left (-d x - c\right )}}{3 \, d {\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{3 \, d {\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} - \frac {4}{15 \, d {\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.47 \[ \int \frac {1}{(1-\cosh (c+d x))^3} \, dx=\frac {4 \, {\left (10 \, e^{\left (2 \, d x + 2 \, c\right )} - 5 \, e^{\left (d x + c\right )} + 1\right )}}{15 \, d {\left (e^{\left (d x + c\right )} - 1\right )}^{5}} \]
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Time = 1.72 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.47 \[ \int \frac {1}{(1-\cosh (c+d x))^3} \, dx=\frac {4\,\left (10\,{\mathrm {e}}^{2\,c+2\,d\,x}-5\,{\mathrm {e}}^{c+d\,x}+1\right )}{15\,d\,{\left ({\mathrm {e}}^{c+d\,x}-1\right )}^5} \]
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