Integrand size = 14, antiderivative size = 88 \[ \int \frac {1}{(1-i \sinh (c+d x))^3} \, dx=-\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))^2}-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))} \]
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Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2729, 2727} \[ \int \frac {1}{(1-i \sinh (c+d x))^3} \, dx=-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))}-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))^2}-\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3} \]
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Rule 2727
Rule 2729
Rubi steps \begin{align*} \text {integral}& = -\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}+\frac {2}{5} \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx \\ & = -\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))^2}+\frac {2}{15} \int \frac {1}{1-i \sinh (c+d x)} \, dx \\ & = -\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))^2}-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))} \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(1-i \sinh (c+d x))^3} \, dx=\frac {10-15 \cosh (c+d x)-6 \cosh (2 (c+d x))+\cosh (3 (c+d x))-15 i \sinh (c+d x)+6 i \sinh (2 (c+d x))+i \sinh (3 (c+d x))}{30 d (i+\sinh (c+d x))^3} \]
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Time = 0.96 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45
method | result | size |
risch | \(\frac {4 i \left (5 i {\mathrm e}^{d x +c}+10 \,{\mathrm e}^{2 d x +2 c}-1\right )}{15 d \left ({\mathrm e}^{d x +c}+i\right )^{5}}\) | \(40\) |
derivativedivides | \(\frac {\frac {2}{i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {8}{5 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {16}{3 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {4 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}}{d}\) | \(88\) |
default | \(\frac {\frac {2}{i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {8}{5 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {16}{3 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {4 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}}{d}\) | \(88\) |
parallelrisch | \(\frac {\frac {14}{15}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {16 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}-\frac {8 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}}{d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+5 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-10 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-10 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}\) | \(128\) |
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Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(1-i \sinh (c+d x))^3} \, dx=-\frac {4 \, {\left (-10 i \, e^{\left (2 \, d x + 2 \, c\right )} + 5 \, e^{\left (d x + c\right )} + i\right )}}{15 \, {\left (d e^{\left (5 \, d x + 5 \, c\right )} + 5 i \, d e^{\left (4 \, d x + 4 \, c\right )} - 10 \, d e^{\left (3 \, d x + 3 \, c\right )} - 10 i \, d e^{\left (2 \, d x + 2 \, c\right )} + 5 \, d e^{\left (d x + c\right )} + i \, d\right )}} \]
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Time = 0.16 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(1-i \sinh (c+d x))^3} \, dx=\frac {40 i e^{2 c} e^{2 d x} - 20 e^{c} e^{d x} - 4 i}{15 d e^{5 c} e^{5 d x} + 75 i d e^{4 c} e^{4 d x} - 150 d e^{3 c} e^{3 d x} - 150 i d e^{2 c} e^{2 d x} + 75 d e^{c} e^{d x} + 15 i d} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (70) = 140\).
Time = 0.22 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.40 \[ \int \frac {1}{(1-i \sinh (c+d x))^3} \, dx=\frac {20 i \, e^{\left (-d x - c\right )}}{-15 \, d {\left (-5 i \, e^{\left (-d x - c\right )} + 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - i \, e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} - \frac {40 \, e^{\left (-2 \, d x - 2 \, c\right )}}{-15 \, d {\left (-5 i \, e^{\left (-d x - c\right )} + 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - i \, e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} + \frac {4}{-15 \, d {\left (-5 i \, e^{\left (-d x - c\right )} + 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - i \, e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.41 \[ \int \frac {1}{(1-i \sinh (c+d x))^3} \, dx=\frac {4 i \, {\left (10 \, e^{\left (2 \, d x + 2 \, c\right )} + 5 i \, e^{\left (d x + c\right )} - 1\right )}}{15 \, d {\left (e^{\left (d x + c\right )} + i\right )}^{5}} \]
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Time = 1.42 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45 \[ \int \frac {1}{(1-i \sinh (c+d x))^3} \, dx=-\frac {4\,\left (10\,{\mathrm {e}}^{2\,c+2\,d\,x}-1+{\mathrm {e}}^{c+d\,x}\,5{}\mathrm {i}\right )}{15\,d\,{\left (-1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^5} \]
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