Integrand size = 14, antiderivative size = 73 \[ \int \frac {1}{3+5 i \sinh (c+d x)} \, dx=\frac {i \log \left (3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]
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Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2739, 630, 31} \[ \int \frac {1}{3+5 i \sinh (c+d x)} \, dx=\frac {i \log \left (3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]
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Rule 31
Rule 630
Rule 2739
Rubi steps \begin{align*} \text {integral}& = -\frac {(2 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 )}{d} \\ & = -\frac {(3 i) \text {Subst}\left (\int \frac {1}{1+3 x} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{4 d}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{9+3 x} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{4 d} \\ & = \frac {i \log \left (3+i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {i \log \left (1+3 i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.11 \[ \int \frac {1}{3+5 i \sinh (c+d x)} \, dx=\frac {\arctan \left (3 \coth \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}+\frac {\arctan \left (3 \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {i \log (4-5 \cosh (c+d x))}{8 d}+\frac {i \log (4+5 \cosh (c+d x))}{8 d} \]
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Time = 2.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.49
method | result | size |
risch | \(-\frac {i \ln \left ({\mathrm e}^{d x +c}-\frac {4}{5}-\frac {3 i}{5}\right )}{4 d}+\frac {i \ln \left ({\mathrm e}^{d x +c}+\frac {4}{5}-\frac {3 i}{5}\right )}{4 d}\) | \(36\) |
derivativedivides | \(\frac {-\frac {i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{4}+\frac {i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{4}}{d}\) | \(40\) |
default | \(\frac {-\frac {i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{4}+\frac {i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{4}}{d}\) | \(40\) |
parallelrisch | \(-\frac {i \left (-\ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-9 i\right )+\ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )\right )}{4 d}\) | \(40\) |
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Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.38 \[ \int \frac {1}{3+5 i \sinh (c+d x)} \, dx=\frac {i \, \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i + \frac {4}{5}\right ) - i \, \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i - \frac {4}{5}\right )}{4 \, d} \]
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Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.42 \[ \int \frac {1}{3+5 i \sinh (c+d x)} \, dx=\frac {\operatorname {RootSum} {\left (16 z^{2} + 1, \left ( i \mapsto i \log {\left (\frac {\left (- 16 i i - 3 i\right ) e^{- c}}{5} + e^{d x} \right )} \right )\right )}}{d} \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.26 \[ \int \frac {1}{3+5 i \sinh (c+d x)} \, dx=\frac {\arctan \left (\frac {5}{4} i \, e^{\left (-d x - c\right )} - \frac {3}{4}\right )}{2 \, d} \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.44 \[ \int \frac {1}{3+5 i \sinh (c+d x)} \, dx=-\frac {-i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right ) + i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{4 \, d} \]
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Time = 0.35 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.53 \[ \int \frac {1}{3+5 i \sinh (c+d x)} \, dx=-\frac {\ln \left (-\frac {5}{2}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (2-\frac {3}{2}{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{4\,d}+\frac {\ln \left (\frac {5}{2}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (2+\frac {3}{2}{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{4\,d} \]
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