Integrand size = 41, antiderivative size = 181 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\frac {(5 A+3 i B) x}{32 a c^4}-\frac {A+i B}{32 a c^4 f (i-\tan (e+f x))}-\frac {i A+B}{16 a c^4 f (i+\tan (e+f x))^4}-\frac {A}{12 a c^4 f (i+\tan (e+f x))^3}+\frac {3 i A-B}{32 a c^4 f (i+\tan (e+f x))^2}+\frac {2 A+i B}{16 a c^4 f (i+\tan (e+f x))} \]
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Time = 0.28 (sec) , antiderivative size = 181, 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} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=-\frac {A+i B}{32 a c^4 f (-\tan (e+f x)+i)}+\frac {2 A+i B}{16 a c^4 f (\tan (e+f x)+i)}+\frac {-B+3 i A}{32 a c^4 f (\tan (e+f x)+i)^2}-\frac {B+i A}{16 a c^4 f (\tan (e+f x)+i)^4}+\frac {x (5 A+3 i B)}{32 a c^4}-\frac {A}{12 a c^4 f (\tan (e+f x)+i)^3} \]
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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)^2 (c-i c x)^5} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {-A-i B}{32 a^2 c^5 (-i+x)^2}+\frac {i A+B}{4 a^2 c^5 (i+x)^5}+\frac {A}{4 a^2 c^5 (i+x)^4}+\frac {-3 i A+B}{16 a^2 c^5 (i+x)^3}+\frac {-2 A-i B}{16 a^2 c^5 (i+x)^2}+\frac {5 A+3 i B}{32 a^2 c^5 \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {A+i B}{32 a c^4 f (i-\tan (e+f x))}-\frac {i A+B}{16 a c^4 f (i+\tan (e+f x))^4}-\frac {A}{12 a c^4 f (i+\tan (e+f x))^3}+\frac {3 i A-B}{32 a c^4 f (i+\tan (e+f x))^2}+\frac {2 A+i B}{16 a c^4 f (i+\tan (e+f x))}+\frac {(5 A+3 i B) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{32 a c^4 f} \\ & = \frac {(5 A+3 i B) x}{32 a c^4}-\frac {A+i B}{32 a c^4 f (i-\tan (e+f x))}-\frac {i A+B}{16 a c^4 f (i+\tan (e+f x))^4}-\frac {A}{12 a c^4 f (i+\tan (e+f x))^3}+\frac {3 i A-B}{32 a c^4 f (i+\tan (e+f x))^2}+\frac {2 A+i B}{16 a c^4 f (i+\tan (e+f x))} \\ \end{align*}
Time = 3.78 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.78 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\frac {i \left (3 (-5 i A+3 B) \arctan (\tan (e+f x))+\frac {32 i A+3 (5 A+3 i B) \tan (e+f x)+7 i (5 A+3 i B) \tan ^2(e+f x)+9 (5 A+3 i B) \tan ^3(e+f x)+3 (-5 i A+3 B) \tan ^4(e+f x)}{(-i+\tan (e+f x)) (i+\tan (e+f x))^4}\right )}{96 a c^4 f} \]
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Time = 0.21 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\frac {3 i x B}{32 a \,c^{4}}+\frac {5 x A}{32 a \,c^{4}}-\frac {{\mathrm e}^{8 i \left (f x +e \right )} B}{256 a \,c^{4} f}-\frac {i {\mathrm e}^{8 i \left (f x +e \right )} A}{256 a \,c^{4} f}-\frac {{\mathrm e}^{6 i \left (f x +e \right )} B}{64 a \,c^{4} f}-\frac {5 i {\mathrm e}^{6 i \left (f x +e \right )} A}{192 a \,c^{4} f}-\frac {{\mathrm e}^{4 i \left (f x +e \right )} B}{64 a \,c^{4} f}-\frac {5 i {\mathrm e}^{4 i \left (f x +e \right )} A}{64 a \,c^{4} f}+\frac {\cos \left (2 f x +2 e \right ) B}{64 a \,c^{4} f}-\frac {9 i \cos \left (2 f x +2 e \right ) A}{64 a \,c^{4} f}+\frac {3 i \sin \left (2 f x +2 e \right ) B}{64 a \,c^{4} f}+\frac {11 \sin \left (2 f x +2 e \right ) A}{64 a \,c^{4} f}\) | \(238\) |
derivativedivides | \(-\frac {A}{12 a \,c^{4} f \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 i B \arctan \left (\tan \left (f x +e \right )\right )}{32 f a \,c^{4}}+\frac {A}{8 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )}+\frac {i B}{32 f a \,c^{4} \left (-i+\tan \left (f x +e \right )\right )}+\frac {5 A \arctan \left (\tan \left (f x +e \right )\right )}{32 f a \,c^{4}}+\frac {3 i A}{32 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{16 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {B}{32 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {B}{16 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {i B}{16 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )}+\frac {A}{32 f a \,c^{4} \left (-i+\tan \left (f x +e \right )\right )}\) | \(252\) |
default | \(-\frac {A}{12 a \,c^{4} f \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 i B \arctan \left (\tan \left (f x +e \right )\right )}{32 f a \,c^{4}}+\frac {A}{8 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )}+\frac {i B}{32 f a \,c^{4} \left (-i+\tan \left (f x +e \right )\right )}+\frac {5 A \arctan \left (\tan \left (f x +e \right )\right )}{32 f a \,c^{4}}+\frac {3 i A}{32 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{16 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {B}{32 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {B}{16 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {i B}{16 f a \,c^{4} \left (i+\tan \left (f x +e \right )\right )}+\frac {A}{32 f a \,c^{4} \left (-i+\tan \left (f x +e \right )\right )}\) | \(252\) |
norman | \(\frac {\frac {\left (3 i B +5 A \right ) x}{32 a c}-\frac {i A}{3 a c f}+\frac {\left (i A +3 B \right ) \tan \left (f x +e \right )^{2}}{6 a c f}+\frac {11 \left (3 i B +5 A \right ) \tan \left (f x +e \right )^{5}}{96 a c f}+\frac {\left (3 i B +5 A \right ) \tan \left (f x +e \right )^{7}}{32 a c f}+\frac {\left (3 i B +5 A \right ) x \tan \left (f x +e \right )^{2}}{8 a c}+\frac {3 \left (3 i B +5 A \right ) x \tan \left (f x +e \right )^{4}}{16 a c}+\frac {\left (3 i B +5 A \right ) x \tan \left (f x +e \right )^{6}}{8 a c}+\frac {\left (3 i B +5 A \right ) x \tan \left (f x +e \right )^{8}}{32 a c}+\frac {3 \left (-i B +9 A \right ) \tan \left (f x +e \right )}{32 a c f}+\frac {\left (63 i B +73 A \right ) \tan \left (f x +e \right )^{3}}{96 a c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{4} c^{3}}\) | \(281\) |
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Time = 0.25 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.64 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\frac {{\left (24 \, {\left (5 \, A + 3 i \, B\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} - 3 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} - 4 \, {\left (5 i \, A + 3 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 12 \, {\left (5 i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 24 \, {\left (5 i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, A - 12 \, B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{768 \, a c^{4} f} \]
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Time = 0.41 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.43 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\begin {cases} \frac {\left (\left (100663296 i A a^{4} c^{16} f^{4} - 100663296 B a^{4} c^{16} f^{4}\right ) e^{- 2 i f x} + \left (- 1006632960 i A a^{4} c^{16} f^{4} e^{4 i e} + 201326592 B a^{4} c^{16} f^{4} e^{4 i e}\right ) e^{2 i f x} + \left (- 503316480 i A a^{4} c^{16} f^{4} e^{6 i e} - 100663296 B a^{4} c^{16} f^{4} e^{6 i e}\right ) e^{4 i f x} + \left (- 167772160 i A a^{4} c^{16} f^{4} e^{8 i e} - 100663296 B a^{4} c^{16} f^{4} e^{8 i e}\right ) e^{6 i f x} + \left (- 25165824 i A a^{4} c^{16} f^{4} e^{10 i e} - 25165824 B a^{4} c^{16} f^{4} e^{10 i e}\right ) e^{8 i f x}\right ) e^{- 2 i e}}{6442450944 a^{5} c^{20} f^{5}} & \text {for}\: a^{5} c^{20} f^{5} e^{2 i e} \neq 0 \\x \left (- \frac {5 A + 3 i B}{32 a c^{4}} + \frac {\left (A e^{10 i e} + 5 A e^{8 i e} + 10 A e^{6 i e} + 10 A e^{4 i e} + 5 A e^{2 i e} + A - i B e^{10 i e} - 3 i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{4 i e} + 3 i B e^{2 i e} + i B\right ) e^{- 2 i e}}{32 a c^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (5 A + 3 i B\right )}{32 a c^{4}} \]
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Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.82 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.14 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\frac {\frac {12 \, {\left (5 i \, A - 3 \, B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{4}} + \frac {12 \, {\left (-5 i \, A + 3 \, B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{4}} + \frac {12 \, {\left (5 \, A \tan \left (f x + e\right ) + 3 i \, B \tan \left (f x + e\right ) - 7 i \, A + 5 \, B\right )}}{a c^{4} {\left (-i \, \tan \left (f x + e\right ) - 1\right )}} + \frac {-125 i \, A \tan \left (f x + e\right )^{4} + 75 \, B \tan \left (f x + e\right )^{4} + 596 \, A \tan \left (f x + e\right )^{3} + 348 i \, B \tan \left (f x + e\right )^{3} + 1110 i \, A \tan \left (f x + e\right )^{2} - 618 \, B \tan \left (f x + e\right )^{2} - 996 \, A \tan \left (f x + e\right ) - 492 i \, B \tan \left (f x + e\right ) - 405 i \, A + 99 \, B}{a c^{4} {\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{768 \, f} \]
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Time = 9.83 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-\frac {3\,B}{32\,a\,c^4}+\frac {A\,5{}\mathrm {i}}{32\,a\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {5\,A}{32\,a\,c^4}+\frac {B\,3{}\mathrm {i}}{32\,a\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (-\frac {9\,B}{32\,a\,c^4}+\frac {A\,15{}\mathrm {i}}{32\,a\,c^4}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {35\,A}{96\,a\,c^4}+\frac {B\,7{}\mathrm {i}}{32\,a\,c^4}\right )-\frac {A}{3\,a\,c^4}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^5-{\mathrm {tan}\left (e+f\,x\right )}^4\,3{}\mathrm {i}+2\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,2{}\mathrm {i}+3\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {x\,\left (-3\,B+A\,5{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a\,c^4} \]
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