Integrand size = 14, antiderivative size = 66 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {5 x}{64}-\frac {5 i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))} \]
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Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2743, 12, 2736} \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=-\frac {5 i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}+\frac {5 x}{64} \]
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Rule 12
Rule 2736
Rule 2743
Rubi steps \begin{align*} \text {integral}& = -\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}-\frac {1}{16} \int -\frac {5}{5+3 i \sinh (c+d x)} \, dx \\ & = -\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}+\frac {5}{16} \int \frac {1}{5+3 i \sinh (c+d x)} \, dx \\ & = \frac {5 x}{64}-\frac {5 i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))} \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(183\) vs. \(2(66)=132\).
Time = 0.37 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.77 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {24 i-50 i \arctan \left (\frac {2 \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-2 \sinh \left (\frac {1}{2} (c+d x)\right )}\right )+50 i \arctan \left (\frac {\cosh \left (\frac {1}{2} (c+d x)\right )+2 \sinh \left (\frac {1}{2} (c+d x)\right )}{2 \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )}\right )-25 \log (5 \cosh (c+d x)-4 \sinh (c+d x))+25 \log (5 \cosh (c+d x)+4 \sinh (c+d x))-\frac {120 \cosh (c+d x)}{-5 i+3 \sinh (c+d x)}}{640 d} \]
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Time = 1.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11
method | result | size |
risch | \(-\frac {i \left (5 \,{\mathrm e}^{d x +c}-3 i\right )}{8 d \left (3 \,{\mathrm e}^{2 d x +2 c}-3-10 i {\mathrm e}^{d x +c}\right )}-\frac {5 \ln \left ({\mathrm e}^{d x +c}-3 i\right )}{64 d}+\frac {5 \ln \left (-\frac {i}{3}+{\mathrm e}^{d x +c}\right )}{64 d}\) | \(73\) |
derivativedivides | \(\frac {\frac {-\frac {9}{80}-\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i}+\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{64}+\frac {-\frac {9}{80}+\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i}-\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{64}}{d}\) | \(84\) |
default | \(\frac {\frac {-\frac {9}{80}-\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i}+\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{64}+\frac {-\frac {9}{80}+\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i}-\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{64}}{d}\) | \(84\) |
parallelrisch | \(\frac {125 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )-125 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )-60 i+75 i \sinh \left (d x +c \right ) \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )-75 i \sinh \left (d x +c \right ) \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )-60 i \cosh \left (d x +c \right )+36 \sinh \left (d x +c \right )}{960 i d \sinh \left (d x +c \right )+1600 d}\) | \(124\) |
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Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.56 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {5 \, {\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )} \log \left (e^{\left (d x + c\right )} - \frac {1}{3} i\right ) - 5 \, {\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )} \log \left (e^{\left (d x + c\right )} - 3 i\right ) - 40 i \, e^{\left (d x + c\right )} - 24}{64 \, {\left (3 \, d e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, d e^{\left (d x + c\right )} - 3 \, d\right )}} \]
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Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {- 5 i e^{c} e^{d x} - 3}{24 d e^{2 c} e^{2 d x} - 80 i d e^{c} e^{d x} - 24 d} + \frac {- \frac {5 \log {\left (e^{d x} - 3 i e^{- c} \right )}}{64} + \frac {5 \log {\left (e^{d x} - \frac {i e^{- c}}{3} \right )}}{64}}{d} \]
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Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=-\frac {5 i \, \arctan \left (\frac {3}{4} \, e^{\left (-d x - c\right )} + \frac {5}{4} i\right )}{32 \, d} - \frac {5 i \, e^{\left (-d x - c\right )} - 3}{-8 \, d {\left (-10 i \, e^{\left (-d x - c\right )} - 3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=-\frac {\frac {8 \, {\left (5 i \, e^{\left (d x + c\right )} + 3\right )}}{3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3} - 5 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right ) + 5 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{64 \, d} \]
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Time = 1.78 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.55 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {3}{8\,\left (3\,d-3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}\right )}-\frac {5\,\ln \left (-\frac {5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4}+\frac {15}{4}{}\mathrm {i}\right )}{64\,d}+\frac {5\,\ln \left (\frac {45\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4}-\frac {15}{4}{}\mathrm {i}\right )}{64\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,5{}\mathrm {i}}{8\,\left (3\,d-3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}\right )} \]
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