Integrand size = 31, antiderivative size = 71 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {4 c^3 x}{a}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}+\frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))} \]
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Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 45} \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}-\frac {4 c^3 x}{a} \]
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Rule 45
Rule 3568
Rule 3603
Rubi steps \begin{align*} \text {integral}& = \left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(a+i a \tan (e+f x))^4} \, dx \\ & = -\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {(a-x)^2}{(a+x)^2} \, dx,x,i a \tan (e+f x)\right )}{a^2 f} \\ & = -\frac {\left (i c^3\right ) \text {Subst}\left (\int \left (1+\frac {4 a^2}{(a+x)^2}-\frac {4 a}{a+x}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^2 f} \\ & = -\frac {4 c^3 x}{a}-\frac {4 i c^3 \log (\cos (e+f x))}{a f}+\frac {c^3 \tan (e+f x)}{a f}+\frac {4 i c^3}{f (a+i a \tan (e+f x))} \\ \end{align*}
Time = 2.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {i c^3 \left (-4 \log (i-\tan (e+f x))+i \tan (e+f x)+\frac {4 i}{-i+\tan (e+f x)}\right )}{a f} \]
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Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {c^{3} \tan \left (f x +e \right )}{a f}-\frac {4 c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f a}+\frac {2 i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f a}+\frac {4 c^{3}}{f a \left (\tan \left (f x +e \right )-i\right )}\) | \(81\) |
default | \(\frac {c^{3} \tan \left (f x +e \right )}{a f}-\frac {4 c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f a}+\frac {2 i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f a}+\frac {4 c^{3}}{f a \left (\tan \left (f x +e \right )-i\right )}\) | \(81\) |
risch | \(\frac {2 i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )}}{a f}-\frac {8 c^{3} x}{a}-\frac {8 c^{3} e}{a f}+\frac {2 i c^{3}}{f a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {4 i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a f}\) | \(93\) |
norman | \(\frac {\frac {4 i c^{3}}{a f}+\frac {c^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{a f}-\frac {4 c^{3} x}{a}-\frac {4 c^{3} x \left (\tan ^{2}\left (f x +e \right )\right )}{a}+\frac {5 c^{3} \tan \left (f x +e \right )}{a f}}{1+\tan ^{2}\left (f x +e \right )}+\frac {2 i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f a}\) | \(112\) |
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none
Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.66 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {2 \, {\left (4 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} - i \, c^{3} + 2 \, {\left (2 \, c^{3} f x - i \, c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}} \]
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Time = 0.23 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.87 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {2 i c^{3}}{a f e^{2 i e} e^{2 i f x} + a f} + \begin {cases} \frac {2 i c^{3} e^{- 2 i e} e^{- 2 i f x}}{a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (\frac {8 c^{3}}{a} + \frac {\left (- 8 c^{3} e^{2 i e} + 4 c^{3}\right ) e^{- 2 i e}}{a}\right ) & \text {otherwise} \end {cases} - \frac {8 c^{3} x}{a} - \frac {4 i c^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} \]
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Exception generated. \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (65) = 130\).
Time = 0.47 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.46 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {2 \, {\left (-\frac {2 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a} + \frac {4 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a} - \frac {2 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a} + \frac {2 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 i \, c^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a} - \frac {2 \, {\left (3 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 8 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 i \, c^{3}\right )}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{2}}\right )}}{f} \]
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Time = 6.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {(c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {c^3\,\mathrm {tan}\left (e+f\,x\right )}{a\,f}+\frac {c^3\,4{}\mathrm {i}}{a\,f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {c^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,4{}\mathrm {i}}{a\,f} \]
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