Integrand size = 41, antiderivative size = 97 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {i B c^2 x}{a^2}-\frac {B c^2 \log (\cos (e+f x))}{a^2 f}-\frac {(i A-B) c^2}{a^2 f (i-\tan (e+f x))^2}+\frac {(A+3 i B) c^2}{a^2 f (i-\tan (e+f x))} \]
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Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3669, 78} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^2 (A+3 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac {c^2 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}-\frac {B c^2 \log (\cos (e+f x))}{a^2 f}+\frac {i B c^2 x}{a^2} \]
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Rule 78
Rule 3669
Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {(A+B x) (c-i c x)}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {2 i (A+i B) c}{a^3 (-i+x)^3}+\frac {(A+3 i B) c}{a^3 (-i+x)^2}+\frac {B c}{a^3 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i B c^2 x}{a^2}-\frac {B c^2 \log (\cos (e+f x))}{a^2 f}-\frac {(i A-B) c^2}{a^2 f (i-\tan (e+f x))^2}+\frac {(A+3 i B) c^2}{a^2 f (i-\tan (e+f x))} \\ \end{align*}
Time = 5.64 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^2 \left (B \log (i-\tan (e+f x))-\frac {2 B+(A+3 i B) \tan (e+f x)}{(-i+\tan (e+f x))^2}\right )}{a^2 f} \]
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Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.18
method | result | size |
risch | \(\frac {c^{2} B \,{\mathrm e}^{-2 i \left (f x +e \right )}}{a^{2} f}-\frac {c^{2} {\mathrm e}^{-4 i \left (f x +e \right )} B}{4 a^{2} f}+\frac {i c^{2} {\mathrm e}^{-4 i \left (f x +e \right )} A}{4 a^{2} f}+\frac {2 i B \,c^{2} x}{a^{2}}+\frac {2 i c^{2} B e}{a^{2} f}-\frac {c^{2} B \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a^{2} f}\) | \(114\) |
derivativedivides | \(-\frac {3 i c^{2} B}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}-\frac {c^{2} A}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}+\frac {c^{2} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}+\frac {i c^{2} B \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}-\frac {i c^{2} A}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {c^{2} B}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}\) | \(138\) |
default | \(-\frac {3 i c^{2} B}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}-\frac {c^{2} A}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}+\frac {c^{2} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}+\frac {i c^{2} B \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}-\frac {i c^{2} A}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {c^{2} B}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}\) | \(138\) |
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Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (8 i \, B c^{2} f x e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, B c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + 4 \, B c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A - B\right )} c^{2}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, a^{2} f} \]
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Time = 0.29 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.12 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {2 i B c^{2} x}{a^{2}} - \frac {B c^{2} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} + \begin {cases} \frac {\left (4 B a^{2} c^{2} f e^{4 i e} e^{- 2 i f x} + \left (i A a^{2} c^{2} f e^{2 i e} - B a^{2} c^{2} f e^{2 i e}\right ) e^{- 4 i f x}\right ) e^{- 6 i e}}{4 a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {2 i B c^{2}}{a^{2}} + \frac {\left (A c^{2} + 2 i B c^{2} e^{4 i e} - 2 i B c^{2} e^{2 i e} + i B c^{2}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} \]
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Exception generated. \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (84) = 168\).
Time = 0.54 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.97 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {6 \, B c^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2}} - \frac {12 \, B c^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{2}} + \frac {6 \, B c^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{2}} + \frac {25 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 12 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 112 i \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 198 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 112 i \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 25 \, B c^{2}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{4}}}{6 \, f} \]
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Time = 8.84 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^2\,\left (2\,B+A\,\mathrm {tan}\left (e+f\,x\right )+B\,\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+B\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+B\,\mathrm {tan}\left (e+f\,x\right )\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,2{}\mathrm {i}\right )}{a^2\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \]
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