Integrand size = 41, antiderivative size = 153 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {(2 A-i B) x}{8 a^3 c}+\frac {A+i B}{12 a^3 c f (i-\tan (e+f x))^3}-\frac {i A}{8 a^3 c f (i-\tan (e+f x))^2}-\frac {3 A-i B}{16 a^3 c f (i-\tan (e+f x))}+\frac {A-i B}{16 a^3 c f (i+\tan (e+f x))} \]
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Time = 0.24 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {3669, 78, 209} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=-\frac {3 A-i B}{16 a^3 c f (-\tan (e+f x)+i)}+\frac {A-i B}{16 a^3 c f (\tan (e+f x)+i)}+\frac {A+i B}{12 a^3 c f (-\tan (e+f x)+i)^3}+\frac {x (2 A-i B)}{8 a^3 c}-\frac {i A}{8 a^3 c f (-\tan (e+f x)+i)^2} \]
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Rule 78
Rule 209
Rule 3669
Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^4 (c-i c x)^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {A+i B}{4 a^4 c^2 (-i+x)^4}+\frac {i A}{4 a^4 c^2 (-i+x)^3}+\frac {-3 A+i B}{16 a^4 c^2 (-i+x)^2}+\frac {-A+i B}{16 a^4 c^2 (i+x)^2}+\frac {2 A-i B}{8 a^4 c^2 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {A+i B}{12 a^3 c f (i-\tan (e+f x))^3}-\frac {i A}{8 a^3 c f (i-\tan (e+f x))^2}-\frac {3 A-i B}{16 a^3 c f (i-\tan (e+f x))}+\frac {A-i B}{16 a^3 c f (i+\tan (e+f x))}+\frac {(2 A-i B) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 c f} \\ & = \frac {(2 A-i B) x}{8 a^3 c}+\frac {A+i B}{12 a^3 c f (i-\tan (e+f x))^3}-\frac {i A}{8 a^3 c f (i-\tan (e+f x))^2}-\frac {3 A-i B}{16 a^3 c f (i-\tan (e+f x))}+\frac {A-i B}{16 a^3 c f (i+\tan (e+f x))} \\ \end{align*}
Time = 5.83 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.97 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {\sec ^3(e+f x) (-9 i A \cos (e+f x)+(-1+2 \cos (2 (e+f x))) ((i A+2 B) \cos (e+f x)+(-2 A+i B) \sin (e+f x))-3 (2 A-i B) \arctan (\tan (e+f x)) \sec (e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))))}{24 a^3 c f (-i+\tan (e+f x))^3 (i+\tan (e+f x))} \]
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Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.27
method | result | size |
risch | \(-\frac {i x B}{8 a^{3} c}+\frac {x A}{4 a^{3} c}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} B}{32 a^{3} c f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} A}{16 a^{3} c f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} B}{96 a^{3} c f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} A}{96 a^{3} c f}-\frac {\cos \left (2 f x +2 e \right ) B}{32 a^{3} c f}+\frac {5 i \cos \left (2 f x +2 e \right ) A}{32 a^{3} c f}-\frac {i \sin \left (2 f x +2 e \right ) B}{32 a^{3} c f}+\frac {7 \sin \left (2 f x +2 e \right ) A}{32 a^{3} c f}\) | \(195\) |
derivativedivides | \(-\frac {i A}{8 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {3 A}{16 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )}-\frac {i B}{16 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )}-\frac {A}{12 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B}{12 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3} c}+\frac {A \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3} c}+\frac {A}{16 f \,a^{3} c \left (i+\tan \left (f x +e \right )\right )}-\frac {i B}{16 f \,a^{3} c \left (i+\tan \left (f x +e \right )\right )}\) | \(206\) |
default | \(-\frac {i A}{8 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {3 A}{16 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )}-\frac {i B}{16 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )}-\frac {A}{12 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B}{12 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3} c}+\frac {A \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3} c}+\frac {A}{16 f \,a^{3} c \left (i+\tan \left (f x +e \right )\right )}-\frac {i B}{16 f \,a^{3} c \left (i+\tan \left (f x +e \right )\right )}\) | \(206\) |
norman | \(\frac {\frac {\left (-i B +2 A \right ) x}{8 a c}-\frac {-4 i A +B}{12 a c f}+\frac {B \tan \left (f x +e \right )^{2}}{4 a c f}+\frac {\left (-i B +2 A \right ) \tan \left (f x +e \right )^{3}}{3 a c f}+\frac {\left (-i B +2 A \right ) \tan \left (f x +e \right )^{5}}{8 a c f}+\frac {3 \left (-i B +2 A \right ) x \tan \left (f x +e \right )^{2}}{8 a c}+\frac {3 \left (-i B +2 A \right ) x \tan \left (f x +e \right )^{4}}{8 a c}+\frac {\left (-i B +2 A \right ) x \tan \left (f x +e \right )^{6}}{8 a c}+\frac {\left (i B +6 A \right ) \tan \left (f x +e \right )}{8 a c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} a^{2}}\) | \(226\) |
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Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.59 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {{\left (12 \, {\left (2 \, A - i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 18 i \, A e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-2 i \, A + B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} c f} \]
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Time = 0.35 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.22 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\begin {cases} \frac {\left (294912 i A a^{9} c^{3} f^{3} e^{10 i e} e^{- 2 i f x} + \left (16384 i A a^{9} c^{3} f^{3} e^{6 i e} - 16384 B a^{9} c^{3} f^{3} e^{6 i e}\right ) e^{- 6 i f x} + \left (98304 i A a^{9} c^{3} f^{3} e^{8 i e} - 49152 B a^{9} c^{3} f^{3} e^{8 i e}\right ) e^{- 4 i f x} + \left (- 49152 i A a^{9} c^{3} f^{3} e^{14 i e} - 49152 B a^{9} c^{3} f^{3} e^{14 i e}\right ) e^{2 i f x}\right ) e^{- 12 i e}}{1572864 a^{12} c^{4} f^{4}} & \text {for}\: a^{12} c^{4} f^{4} e^{12 i e} \neq 0 \\x \left (- \frac {2 A - i B}{8 a^{3} c} + \frac {\left (A e^{8 i e} + 4 A e^{6 i e} + 6 A e^{4 i e} + 4 A e^{2 i e} + A - i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{16 a^{3} c}\right ) & \text {otherwise} \end {cases} + \frac {x \left (2 A - i B\right )}{8 a^{3} c} \]
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Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.63 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=-\frac {\frac {6 \, {\left (-2 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c} + \frac {6 \, {\left (2 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c} + \frac {6 \, {\left (2 i \, A \tan \left (f x + e\right ) + B \tan \left (f x + e\right ) - 3 \, A + 2 i \, B\right )}}{a^{3} c {\left (\tan \left (f x + e\right ) + i\right )}} + \frac {-22 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 84 \, A \tan \left (f x + e\right )^{2} + 39 i \, B \tan \left (f x + e\right )^{2} + 114 i \, A \tan \left (f x + e\right ) + 45 \, B \tan \left (f x + e\right ) + 60 \, A - 9 i \, B}{a^{3} c {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \]
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Time = 8.73 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {\frac {A}{3\,a^3\,c}+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {A}{2\,a^3\,c}-\frac {B\,1{}\mathrm {i}}{4\,a^3\,c}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {B}{8\,a^3\,c}+\frac {A\,1{}\mathrm {i}}{4\,a^3\,c}\right )-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {B}{24\,a^3\,c}+\frac {A\,1{}\mathrm {i}}{12\,a^3\,c}\right )+\frac {B\,1{}\mathrm {i}}{12\,a^3\,c}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (e+f\,x\right )}^3+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {x\,\left (B+A\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3\,c} \]
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