Integrand size = 14, antiderivative size = 37 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {x}{4}-\frac {i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{2 d} \]
[Out]
Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2736} \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {x}{4}-\frac {i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{2 d} \]
[In]
[Out]
Rule 2736
Rubi steps \begin{align*} \text {integral}& = \frac {x}{4}-\frac {i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{2 d} \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(171\) vs. \(2(37)=74\).
Time = 0.13 (sec) , antiderivative size = 171, normalized size of antiderivative = 4.62 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=-\frac {i \arctan \left (\frac {2 \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-2 \sinh \left (\frac {1}{2} (c+d x)\right )}\right )}{4 d}+\frac {i \arctan \left (\frac {\cosh \left (\frac {1}{2} (c+d x)\right )+2 \sinh \left (\frac {1}{2} (c+d x)\right )}{2 \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )}\right )}{4 d}-\frac {\log (5 \cosh (c+d x)-4 \sinh (c+d x))}{8 d}+\frac {\log (5 \cosh (c+d x)+4 \sinh (c+d x))}{8 d} \]
[In]
[Out]
Time = 1.75 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {\ln \left (-\frac {i}{3}+{\mathrm e}^{d x +c}\right )}{4 d}-\frac {\ln \left ({\mathrm e}^{d x +c}-3 i\right )}{4 d}\) | \(32\) |
parallelrisch | \(\frac {-\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )+\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{4 d}\) | \(41\) |
derivativedivides | \(\frac {-\frac {\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{4}+\frac {\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{4}}{d}\) | \(42\) |
default | \(\frac {-\frac {\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{4}+\frac {\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{4}}{d}\) | \(42\) |
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {\log \left (e^{\left (d x + c\right )} - \frac {1}{3} i\right ) - \log \left (e^{\left (d x + c\right )} - 3 i\right )}{4 \, d} \]
[In]
[Out]
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {- \frac {\log {\left (e^{d x} - 3 i e^{- c} \right )}}{4} + \frac {\log {\left (e^{d x} - \frac {i e^{- c}}{3} \right )}}{4}}{d} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {\log \left (-\frac {6 \, {\left (-i \, e^{\left (-d x - c\right )} + 3\right )}}{6 i \, e^{\left (-d x - c\right )} - 2}\right )}{4 \, d} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {\log \left (3 \, e^{\left (d x + c\right )} - i\right ) - \log \left (e^{\left (d x + c\right )} - 3 i\right )}{4 \, d} \]
[In]
[Out]
Time = 1.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=-\frac {\ln \left (-\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{2}+\frac {3}{2}{}\mathrm {i}\right )-\ln \left (\frac {9\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{2}-\frac {3}{2}{}\mathrm {i}\right )}{4\,d} \]
[In]
[Out]