Integrand size = 41, antiderivative size = 183 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {(5 A+i B) x}{16 a^2 c^3}-\frac {i A-B}{32 a^2 c^3 f (i-\tan (e+f x))^2}-\frac {2 A+i B}{16 a^2 c^3 f (i-\tan (e+f x))}-\frac {A-i B}{24 a^2 c^3 f (i+\tan (e+f x))^3}+\frac {3 i A+B}{32 a^2 c^3 f (i+\tan (e+f x))^2}+\frac {3 A}{16 a^2 c^3 f (i+\tan (e+f x))} \]
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Time = 0.37 (sec) , antiderivative size = 183, 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} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=-\frac {2 A+i B}{16 a^2 c^3 f (-\tan (e+f x)+i)}-\frac {-B+i A}{32 a^2 c^3 f (-\tan (e+f x)+i)^2}+\frac {B+3 i A}{32 a^2 c^3 f (\tan (e+f x)+i)^2}-\frac {A-i B}{24 a^2 c^3 f (\tan (e+f x)+i)^3}+\frac {x (5 A+i B)}{16 a^2 c^3}+\frac {3 A}{16 a^2 c^3 f (\tan (e+f x)+i)} \]
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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)^3 (c-i c x)^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {i (A+i B)}{16 a^3 c^4 (-i+x)^3}+\frac {-2 A-i B}{16 a^3 c^4 (-i+x)^2}+\frac {A-i B}{8 a^3 c^4 (i+x)^4}-\frac {i (3 A-i B)}{16 a^3 c^4 (i+x)^3}-\frac {3 A}{16 a^3 c^4 (i+x)^2}+\frac {5 A+i B}{16 a^3 c^4 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i A-B}{32 a^2 c^3 f (i-\tan (e+f x))^2}-\frac {2 A+i B}{16 a^2 c^3 f (i-\tan (e+f x))}-\frac {A-i B}{24 a^2 c^3 f (i+\tan (e+f x))^3}+\frac {3 i A+B}{32 a^2 c^3 f (i+\tan (e+f x))^2}+\frac {3 A}{16 a^2 c^3 f (i+\tan (e+f x))}+\frac {(5 A+i B) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^2 c^3 f} \\ & = \frac {(5 A+i B) x}{16 a^2 c^3}-\frac {i A-B}{32 a^2 c^3 f (i-\tan (e+f x))^2}-\frac {2 A+i B}{16 a^2 c^3 f (i-\tan (e+f x))}-\frac {A-i B}{24 a^2 c^3 f (i+\tan (e+f x))^3}+\frac {3 i A+B}{32 a^2 c^3 f (i+\tan (e+f x))^2}+\frac {3 A}{16 a^2 c^3 f (i+\tan (e+f x))} \\ \end{align*}
Time = 6.00 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {\sec ^4(e+f x) (47 A-5 i B-2 (7 A+11 i B) \cos (2 (e+f x))-A \cos (4 (e+f x))-5 i B \cos (4 (e+f x))+40 i A \sin (2 (e+f x))-8 B \sin (2 (e+f x))+5 i A \sin (4 (e+f x))-B \sin (4 (e+f x))+12 (5 A+i B) \arctan (\tan (e+f x)) (i+\tan (e+f x)))}{192 a^2 c^3 f (-i+\tan (e+f x))^2 (i+\tan (e+f x))^3} \]
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Time = 0.28 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.14
method | result | size |
norman | \(\frac {\frac {\left (i B +5 A \right ) x}{16 a c}-\frac {i A +B}{6 a c f}+\frac {\left (-i B +11 A \right ) \tan \left (f x +e \right )}{16 a c f}+\frac {\left (i B +5 A \right ) \tan \left (f x +e \right )^{3}}{6 a c f}+\frac {\left (i B +5 A \right ) \tan \left (f x +e \right )^{5}}{16 a c f}+\frac {3 \left (i B +5 A \right ) x \tan \left (f x +e \right )^{2}}{16 a c}+\frac {3 \left (i B +5 A \right ) x \tan \left (f x +e \right )^{4}}{16 a c}+\frac {\left (i B +5 A \right ) x \tan \left (f x +e \right )^{6}}{16 a c}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} a \,c^{2}}\) | \(209\) |
risch | \(\frac {i x B}{16 a^{2} c^{3}}+\frac {5 x A}{16 a^{2} c^{3}}-\frac {{\mathrm e}^{6 i \left (f x +e \right )} B}{192 a^{2} c^{3} f}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )} A}{192 a^{2} c^{3} f}-\frac {\cos \left (4 f x +4 e \right ) B}{32 a^{2} c^{3} f}-\frac {i \cos \left (4 f x +4 e \right ) A}{32 a^{2} c^{3} f}-\frac {i \sin \left (4 f x +4 e \right ) B}{64 a^{2} c^{3} f}+\frac {3 \sin \left (4 f x +4 e \right ) A}{64 a^{2} c^{3} f}-\frac {5 \cos \left (2 f x +2 e \right ) B}{64 a^{2} c^{3} f}-\frac {5 i \cos \left (2 f x +2 e \right ) A}{64 a^{2} c^{3} f}+\frac {i \sin \left (2 f x +2 e \right ) B}{64 a^{2} c^{3} f}+\frac {15 \sin \left (2 f x +2 e \right ) A}{64 a^{2} c^{3} f}\) | \(238\) |
derivativedivides | \(\frac {i B}{16 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {A}{8 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}+\frac {5 A \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}+\frac {i B}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {B}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {A}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 i A}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {3 A}{16 a^{2} c^{3} f \left (i+\tan \left (f x +e \right )\right )}+\frac {B}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(252\) |
default | \(\frac {i B}{16 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {A}{8 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}+\frac {5 A \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}+\frac {i B}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {B}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {A}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 i A}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {3 A}{16 a^{2} c^{3} f \left (i+\tan \left (f x +e \right )\right )}+\frac {B}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(252\) |
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Time = 0.25 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.63 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {{\left (24 \, {\left (5 \, A + i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} - 3 \, {\left (5 i \, A + 3 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 12 \, {\left (5 i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 6 \, {\left (-5 i \, A + 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} c^{3} f} \]
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Time = 0.42 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.48 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (50331648 i A a^{8} c^{12} f^{4} e^{2 i e} - 50331648 B a^{8} c^{12} f^{4} e^{2 i e}\right ) e^{- 4 i f x} + \left (503316480 i A a^{8} c^{12} f^{4} e^{4 i e} - 301989888 B a^{8} c^{12} f^{4} e^{4 i e}\right ) e^{- 2 i f x} + \left (- 1006632960 i A a^{8} c^{12} f^{4} e^{8 i e} - 201326592 B a^{8} c^{12} f^{4} e^{8 i e}\right ) e^{2 i f x} + \left (- 251658240 i A a^{8} c^{12} f^{4} e^{10 i e} - 150994944 B a^{8} c^{12} f^{4} e^{10 i e}\right ) e^{4 i f x} + \left (- 33554432 i A a^{8} c^{12} f^{4} e^{12 i e} - 33554432 B a^{8} c^{12} f^{4} e^{12 i e}\right ) e^{6 i f x}\right ) e^{- 6 i e}}{6442450944 a^{10} c^{15} f^{5}} & \text {for}\: a^{10} c^{15} f^{5} e^{6 i e} \neq 0 \\x \left (- \frac {5 A + i B}{16 a^{2} c^{3}} + \frac {\left (A e^{10 i e} + 5 A e^{8 i e} + 10 A e^{6 i e} + 10 A e^{4 i e} + 5 A e^{2 i e} + A - i B e^{10 i e} - 3 i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{4 i e} + 3 i B e^{2 i e} + i B\right ) e^{- 4 i e}}{32 a^{2} c^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (5 A + i B\right )}{16 a^{2} c^{3}} \]
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Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.79 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=-\frac {\frac {6 \, {\left (-5 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2} c^{3}} + \frac {6 \, {\left (5 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} c^{3}} + \frac {3 \, {\left (15 i \, A \tan \left (f x + e\right )^{2} - 3 \, B \tan \left (f x + e\right )^{2} + 38 \, A \tan \left (f x + e\right ) + 10 i \, B \tan \left (f x + e\right ) - 25 i \, A + 9 \, B\right )}}{a^{2} c^{3} {\left (i \, \tan \left (f x + e\right ) + 1\right )}^{2}} + \frac {55 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 201 \, A \tan \left (f x + e\right )^{2} - 33 i \, B \tan \left (f x + e\right )^{2} - 255 i \, A \tan \left (f x + e\right ) + 27 \, B \tan \left (f x + e\right ) + 117 \, A - 3 i \, B}{a^{2} c^{3} {\left (\tan \left (f x + e\right ) + i\right )}^{3}}}{192 \, f} \]
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Time = 9.09 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.14 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-\frac {5\,B}{48\,a^2\,c^3}+\frac {A\,25{}\mathrm {i}}{48\,a^2\,c^3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (-\frac {B}{16\,a^2\,c^3}+\frac {A\,5{}\mathrm {i}}{16\,a^2\,c^3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {5\,A}{16\,a^2\,c^3}+\frac {B\,1{}\mathrm {i}}{16\,a^2\,c^3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {25\,A}{48\,a^2\,c^3}+\frac {B\,5{}\mathrm {i}}{48\,a^2\,c^3}\right )+\frac {A}{6\,a^2\,c^3}-\frac {B\,1{}\mathrm {i}}{6\,a^2\,c^3}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5+{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (e+f\,x\right )}^3+{\mathrm {tan}\left (e+f\,x\right )}^2\,2{}\mathrm {i}+\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {x\,\left (-B+A\,5{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^2\,c^3} \]
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