Integrand size = 15, antiderivative size = 38 \[ \int (a+i a \tan (c+d x))^2 \, dx=2 a^2 x-\frac {2 i a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \tan (c+d x)}{d} \]
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Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3558, 3556} \[ \int (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \tan (c+d x)}{d}-\frac {2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x \]
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Rule 3556
Rule 3558
Rubi steps \begin{align*} \text {integral}& = 2 a^2 x-\frac {a^2 \tan (c+d x)}{d}+\left (2 i a^2\right ) \int \tan (c+d x) \, dx \\ & = 2 a^2 x-\frac {2 i a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \tan (c+d x)}{d} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int (a+i a \tan (c+d x))^2 \, dx=-\frac {i a (-2 a \log (i+\tan (c+d x))-i a \tan (c+d x))}{d} \]
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Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\tan \left (d x +c \right )+i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(40\) |
default | \(\frac {a^{2} \left (-\tan \left (d x +c \right )+i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(40\) |
parallelrisch | \(\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 a^{2} d x -a^{2} \tan \left (d x +c \right )}{d}\) | \(41\) |
norman | \(2 a^{2} x -\frac {a^{2} \tan \left (d x +c \right )}{d}+\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(42\) |
parts | \(a^{2} x +\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(51\) |
risch | \(-\frac {4 a^{2} c}{d}-\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(54\) |
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Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.47 \[ \int (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (i \, a^{2} + {\left (i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]
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Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int (a+i a \tan (c+d x))^2 \, dx=- \frac {2 i a^{2}}{d e^{2 i c} e^{2 i d x} + d} - \frac {2 i a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} \]
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Time = 0.37 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int (a+i a \tan (c+d x))^2 \, dx=a^{2} x + \frac {{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2}}{d} + \frac {2 i \, a^{2} \log \left (\sec \left (d x + c\right )\right )}{d} \]
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Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.74 \[ \int (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + i \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + i \, a^{2}\right )}}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]
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Time = 3.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,\left (-\mathrm {tan}\left (c+d\,x\right )+\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}\right )}{d} \]
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