Integrand size = 14, antiderivative size = 102 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=-\frac {3 i \log \left (3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {3 i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))} \]
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Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2743, 12, 2739, 630, 31} \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=\frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}-\frac {3 i \log \left (3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {3 i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{64 d} \]
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Rule 12
Rule 31
Rule 630
Rule 2739
Rule 2743
Rubi steps \begin{align*} \text {integral}& = \frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}+\frac {1}{16} \int -\frac {3}{3+5 i \sinh (c+d x)} \, dx \\ & = \frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}-\frac {3}{16} \int \frac {1}{3+5 i \sinh (c+d x)} \, dx \\ & = \frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{3+10 x+3 x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{8 d} \\ & = \frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}+\frac {(9 i) \text {Subst}\left (\int \frac {1}{1+3 x} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{64 d}-\frac {(9 i) \text {Subst}\left (\int \frac {1}{9+3 x} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{64 d} \\ & = -\frac {3 i \log \left (3+i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {3 i \log \left (1+3 i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.39 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=\frac {-9 \left (2 \arctan \left (3 \coth \left (\frac {1}{2} (c+d x)\right )\right )+2 \arctan \left (3 \tanh \left (\frac {1}{2} (c+d x)\right )\right )-i \log (4-5 \cosh (c+d x))+i \log (4+5 \cosh (c+d x))\right )+40 \left (\frac {1}{3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )}+\frac {3}{\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )}\right ) \sinh \left (\frac {1}{2} (c+d x)\right )}{384 d} \]
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Time = 1.17 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {-\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{64}+\frac {5}{16 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}+\frac {3 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{64}+\frac {5}{48 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{d}\) | \(74\) |
default | \(\frac {-\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{64}+\frac {5}{16 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}+\frac {3 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{64}+\frac {5}{48 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{d}\) | \(74\) |
risch | \(\frac {i \left (3 \,{\mathrm e}^{d x +c}-5 i\right )}{8 d \left (5 \,{\mathrm e}^{2 d x +2 c}-5-6 i {\mathrm e}^{d x +c}\right )}+\frac {3 i \ln \left ({\mathrm e}^{d x +c}-\frac {4}{5}-\frac {3 i}{5}\right )}{64 d}-\frac {3 i \ln \left ({\mathrm e}^{d x +c}+\frac {4}{5}-\frac {3 i}{5}\right )}{64 d}\) | \(77\) |
parallelrisch | \(\frac {\left (-9 i+15 \sinh \left (d x +c \right )\right ) \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-9 i\right )+\left (9 i-15 \sinh \left (d x +c \right )\right ) \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )+20 i \cosh \left (d x +c \right )}{320 i d \sinh \left (d x +c \right )+192 d}\) | \(82\) |
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Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=-\frac {3 \, {\left (5 i \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} - 5 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i + \frac {4}{5}\right ) + 3 \, {\left (-5 i \, e^{\left (2 \, d x + 2 \, c\right )} - 6 \, e^{\left (d x + c\right )} + 5 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i - \frac {4}{5}\right ) - 24 i \, e^{\left (d x + c\right )} - 40}{64 \, {\left (5 \, d e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, d e^{\left (d x + c\right )} - 5 \, d\right )}} \]
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Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=\frac {3 i e^{c} e^{d x} + 5}{40 d e^{2 c} e^{2 d x} - 48 i d e^{c} e^{d x} - 40 d} + \frac {\operatorname {RootSum} {\left (4096 z^{2} + 9, \left ( i \mapsto i \log {\left (\frac {\left (256 i i - 9 i\right ) e^{- c}}{15} + e^{d x} \right )} \right )\right )}}{d} \]
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Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=\frac {3 i \, \log \left (\frac {5 \, e^{\left (-d x - c\right )} + 3 i - 4}{5 \, e^{\left (-d x - c\right )} + 3 i + 4}\right )}{64 \, d} + \frac {3 i \, e^{\left (-d x - c\right )} - 5}{-8 \, d {\left (-6 i \, e^{\left (-d x - c\right )} - 5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=-\frac {\frac {8 \, {\left (-3 i \, e^{\left (d x + c\right )} - 5\right )}}{5 \, e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, e^{\left (d x + c\right )} - 5} + 3 i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right ) - 3 i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{64 \, d} \]
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Time = 1.78 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=-\frac {5}{8\,\left (5\,d-5\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,6{}\mathrm {i}\right )}-\frac {\ln \left (-\frac {15}{4}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (-3-\frac {9}{4}{}\mathrm {i}\right )\right )\,3{}\mathrm {i}}{64\,d}+\frac {\ln \left (\frac {15}{4}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (-3+\frac {9}{4}{}\mathrm {i}\right )\right )\,3{}\mathrm {i}}{64\,d}-\frac {{\mathrm {e}}^{c+d\,x}\,3{}\mathrm {i}}{8\,\left (5\,d-5\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,6{}\mathrm {i}\right )} \]
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