Integrand size = 21, antiderivative size = 209 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=-\frac {11 i d x}{96 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {11 d}{96 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
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Time = 0.28 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3560, 8, 3811} \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {x (c+d x)}{8 a^3}+\frac {11 d}{96 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {11 i d x}{96 a^3 f}-\frac {d x^2}{16 a^3}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3} \]
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Rule 8
Rule 3560
Rule 3811
Rubi steps \begin{align*} \text {integral}& = \frac {x (c+d x)}{8 a^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-d \int \left (\frac {x}{8 a^3}+\frac {i}{6 f (a+i a \tan (e+f x))^3}+\frac {i}{8 a f (a+i a \tan (e+f x))^2}+\frac {i}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}\right ) \, dx \\ & = -\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {(i d) \int \frac {1}{a^3+i a^3 \tan (e+f x)} \, dx}{8 f}-\frac {(i d) \int \frac {1}{(a+i a \tan (e+f x))^3} \, dx}{6 f}-\frac {(i d) \int \frac {1}{(a+i a \tan (e+f x))^2} \, dx}{8 a f} \\ & = -\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {d}{32 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {d}{16 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {(i d) \int 1 \, dx}{16 a^3 f}-\frac {(i d) \int \frac {1}{a+i a \tan (e+f x)} \, dx}{16 a^2 f}-\frac {(i d) \int \frac {1}{(a+i a \tan (e+f x))^2} \, dx}{12 a f} \\ & = -\frac {i d x}{16 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {3 d}{32 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {(i d) \int 1 \, dx}{32 a^3 f}-\frac {(i d) \int \frac {1}{a+i a \tan (e+f x)} \, dx}{24 a^2 f} \\ & = -\frac {3 i d x}{32 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {11 d}{96 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {(i d) \int 1 \, dx}{48 a^3 f} \\ & = -\frac {11 i d x}{96 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {11 d}{96 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \\ \end{align*}
Time = 1.52 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \sec ^3(e+f x) \left (27 (12 i c f+d (5+12 i f x)) \cos (e+f x)+4 \left (6 c f (i+6 f x)+d \left (1+6 i f x+18 f^2 x^2\right )\right ) \cos (3 (e+f x))+81 i d \sin (e+f x)-108 c f \sin (e+f x)-108 d f x \sin (e+f x)-4 i d \sin (3 (e+f x))+24 c f \sin (3 (e+f x))+24 d f x \sin (3 (e+f x))+144 i c f^2 x \sin (3 (e+f x))+72 i d f^2 x^2 \sin (3 (e+f x))\right )}{1152 a^3 f^2 (-i+\tan (e+f x))^3} \]
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Time = 0.88 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.55
method | result | size |
risch | \(\frac {d \,x^{2}}{16 a^{3}}+\frac {c x}{8 a^{3}}+\frac {3 i \left (2 d f x +2 c f -i d \right ) {\mathrm e}^{-2 i \left (f x +e \right )}}{32 a^{3} f^{2}}+\frac {3 i \left (4 d f x +4 c f -i d \right ) {\mathrm e}^{-4 i \left (f x +e \right )}}{128 a^{3} f^{2}}+\frac {i \left (6 d f x +6 c f -i d \right ) {\mathrm e}^{-6 i \left (f x +e \right )}}{288 a^{3} f^{2}}\) | \(114\) |
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Time = 0.25 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.50 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (24 i \, d f x + 24 i \, c f + 72 \, {\left (d f^{2} x^{2} + 2 \, c f^{2} x\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 108 \, {\left (-2 i \, d f x - 2 i \, c f - d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 27 \, {\left (-4 i \, d f x - 4 i \, c f - d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, d\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1152 \, a^{3} f^{2}} \]
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Time = 0.34 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.49 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (24576 i a^{6} c f^{5} e^{6 i e} + 24576 i a^{6} d f^{5} x e^{6 i e} + 4096 a^{6} d f^{4} e^{6 i e}\right ) e^{- 6 i f x} + \left (110592 i a^{6} c f^{5} e^{8 i e} + 110592 i a^{6} d f^{5} x e^{8 i e} + 27648 a^{6} d f^{4} e^{8 i e}\right ) e^{- 4 i f x} + \left (221184 i a^{6} c f^{5} e^{10 i e} + 221184 i a^{6} d f^{5} x e^{10 i e} + 110592 a^{6} d f^{4} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{1179648 a^{9} f^{6}} & \text {for}\: a^{9} f^{6} e^{12 i e} \neq 0 \\\frac {x^{2} \cdot \left (3 d e^{4 i e} + 3 d e^{2 i e} + d\right ) e^{- 6 i e}}{16 a^{3}} + \frac {x \left (3 c e^{4 i e} + 3 c e^{2 i e} + c\right ) e^{- 6 i e}}{8 a^{3}} & \text {otherwise} \end {cases} + \frac {c x}{8 a^{3}} + \frac {d x^{2}}{16 a^{3}} \]
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Exception generated. \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.56 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.68 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (72 \, d f^{2} x^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 144 \, c f^{2} x e^{\left (6 i \, f x + 6 i \, e\right )} + 216 i \, d f x e^{\left (4 i \, f x + 4 i \, e\right )} + 108 i \, d f x e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, d f x + 216 i \, c f e^{\left (4 i \, f x + 4 i \, e\right )} + 108 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, c f + 108 \, d e^{\left (4 i \, f x + 4 i \, e\right )} + 27 \, d e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, d\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1152 \, a^{3} f^{2}} \]
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Time = 3.94 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.70 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {d\,x^2}{16\,a^3}-{\mathrm {e}}^{-e\,4{}\mathrm {i}-f\,x\,4{}\mathrm {i}}\,\left (\frac {\left (-12\,c\,f+d\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{128\,a^3\,f^2}-\frac {d\,x\,3{}\mathrm {i}}{32\,a^3\,f}\right )-{\mathrm {e}}^{-e\,6{}\mathrm {i}-f\,x\,6{}\mathrm {i}}\,\left (\frac {\left (-6\,c\,f+d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{288\,a^3\,f^2}-\frac {d\,x\,1{}\mathrm {i}}{48\,a^3\,f}\right )-{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,\left (\frac {\left (-6\,c\,f+d\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^3\,f^2}-\frac {d\,x\,3{}\mathrm {i}}{16\,a^3\,f}\right )+\frac {c\,x}{8\,a^3} \]
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