Integrand size = 24, antiderivative size = 46 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x)) \, dx=a (c-i d) x-\frac {a (i c+d) \log (\cos (e+f x))}{f}+\frac {i a d \tan (e+f x)}{f} \]
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Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3606, 3556} \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x)) \, dx=-\frac {a (d+i c) \log (\cos (e+f x))}{f}+a x (c-i d)+\frac {i a d \tan (e+f x)}{f} \]
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Rule 3556
Rule 3606
Rubi steps \begin{align*} \text {integral}& = a (c-i d) x+\frac {i a d \tan (e+f x)}{f}+(a (i c+d)) \int \tan (e+f x) \, dx \\ & = a (c-i d) x-\frac {a (i c+d) \log (\cos (e+f x))}{f}+\frac {i a d \tan (e+f x)}{f} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.43 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x)) \, dx=a c x-\frac {i a d \arctan (\tan (e+f x))}{f}-\frac {i a c \log (\cos (e+f x))}{f}-\frac {a d \log (\cos (e+f x))}{f}+\frac {i a d \tan (e+f x)}{f} \]
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Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {a \left (i \tan \left (f x +e \right ) d +\frac {\left (i c +d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (-i d +c \right ) \arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\) | \(50\) |
default | \(\frac {a \left (i \tan \left (f x +e \right ) d +\frac {\left (i c +d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (-i d +c \right ) \arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\) | \(50\) |
norman | \(\left (-i a d +a c \right ) x +\frac {i a d \tan \left (f x +e \right )}{f}+\frac {\left (i a c +a d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f}\) | \(52\) |
parts | \(a c x +\frac {\left (i a c +a d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {i a d \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\) | \(55\) |
parallelrisch | \(\frac {-2 i x a d f +i \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a c +2 i a d \tan \left (f x +e \right )+2 x a c f +\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a d}{2 f}\) | \(61\) |
risch | \(\frac {2 i a d e}{f}-\frac {2 a c e}{f}-\frac {2 a d}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) d}{f}-\frac {i a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) c}{f}\) | \(78\) |
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Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.39 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x)) \, dx=-\frac {2 \, a d - {\left (-i \, a c - a d + {\left (-i \, a c - a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \]
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Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.15 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x)) \, dx=- \frac {2 a d}{f e^{2 i e} e^{2 i f x} + f} - \frac {i a \left (c - i d\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{f} \]
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Time = 0.47 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.13 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x)) \, dx=-\frac {-2 i \, a d \tan \left (f x + e\right ) - 2 \, {\left (a c - i \, a d\right )} {\left (f x + e\right )} + {\left (-i \, a c - a d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{2 \, f} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (40) = 80\).
Time = 0.34 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.24 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x)) \, dx=\frac {-i \, a c e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - a d e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - i \, a c \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - a d \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 2 \, a d}{f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \]
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Time = 5.38 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (a\,d+a\,c\,1{}\mathrm {i}\right )}{f}+\frac {a\,d\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{f} \]
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