Integrand size = 28, antiderivative size = 129 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {(c-i d)^2 x}{8 a^3}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}+\frac {(c+i d) (i c+3 d)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c-i d)^2}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
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Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3621, 3607, 3560, 8} \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {i (c-i d)^2}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {x (c-i d)^2}{8 a^3}+\frac {(c+i d) (3 d+i c)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3} \]
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Rule 8
Rule 3560
Rule 3607
Rule 3621
Rubi steps \begin{align*} \text {integral}& = \frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}+\frac {\int \frac {a \left (c^2-2 i c d+d^2\right )-2 i a d^2 \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx}{2 a^2} \\ & = \frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}+\frac {(c+i d) (i c+3 d)}{8 a f (a+i a \tan (e+f x))^2}+\frac {(c-i d)^2 \int \frac {1}{a+i a \tan (e+f x)} \, dx}{4 a^2} \\ & = \frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}+\frac {(c+i d) (i c+3 d)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c-i d)^2}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {(c-i d)^2 \int 1 \, dx}{8 a^3} \\ & = \frac {(c-i d)^2 x}{8 a^3}+\frac {i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3}+\frac {(c+i d) (i c+3 d)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c-i d)^2}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \\ \end{align*}
Time = 2.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.38 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \left (\frac {3 \sec ^2(e+f x) \left (2 c^2+2 d^2+2 c^2 \cos (2 (e+f x))+i c^2 \sin (2 (e+f x))+2 c d \sin (2 (e+f x))+i d^2 \sin (2 (e+f x))+2 (c-i d)^2 \arctan (\tan (e+f x)) (-i \cos (2 (e+f x))+\sin (2 (e+f x)))\right )}{a^3 (-i+\tan (e+f x))^2}-\frac {8 (c+d \tan (e+f x))^3}{(c+i d) (a+i a \tan (e+f x))^3}\right )}{48 f} \]
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Time = 0.37 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.64
method | result | size |
risch | \(-\frac {i x c d}{4 a^{3}}+\frac {x \,c^{2}}{8 a^{3}}-\frac {x \,d^{2}}{8 a^{3}}+\frac {{\mathrm e}^{-2 i \left (f x +e \right )} c d}{8 a^{3} f}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} c^{2}}{16 a^{3} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} d^{2}}{16 a^{3} f}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} c d}{16 a^{3} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )} c^{2}}{32 a^{3} f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} d^{2}}{32 a^{3} f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} c d}{24 a^{3} f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} c^{2}}{48 a^{3} f}-\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} d^{2}}{48 a^{3} f}\) | \(212\) |
derivativedivides | \(-\frac {i c d \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3}}+\frac {c^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c d}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {d^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {c d}{4 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {3 i d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i c d}{4 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {c^{2}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {d^{2}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {c^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}\) | \(255\) |
default | \(-\frac {i c d \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3}}+\frac {c^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c d}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {d^{2} \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3}}-\frac {i c^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {c d}{4 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {3 i d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i c d}{4 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {c^{2}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {d^{2}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {c^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}\) | \(255\) |
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Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (12 \, {\left (c^{2} - 2 i \, c d - d^{2}\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 2 i \, c^{2} - 4 \, c d - 2 i \, d^{2} - 6 \, {\left (-3 i \, c^{2} - 2 \, c d - i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-3 i \, c^{2} + 2 \, c d - i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \]
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Time = 0.35 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.11 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (512 i a^{6} c^{2} f^{2} e^{6 i e} - 1024 a^{6} c d f^{2} e^{6 i e} - 512 i a^{6} d^{2} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (2304 i a^{6} c^{2} f^{2} e^{8 i e} - 1536 a^{6} c d f^{2} e^{8 i e} + 768 i a^{6} d^{2} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (4608 i a^{6} c^{2} f^{2} e^{10 i e} + 3072 a^{6} c d f^{2} e^{10 i e} + 1536 i a^{6} d^{2} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text {for}\: a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {c^{2} - 2 i c d - d^{2}}{8 a^{3}} + \frac {\left (c^{2} e^{6 i e} + 3 c^{2} e^{4 i e} + 3 c^{2} e^{2 i e} + c^{2} - 2 i c d e^{6 i e} - 2 i c d e^{4 i e} + 2 i c d e^{2 i e} + 2 i c d - d^{2} e^{6 i e} + d^{2} e^{4 i e} + d^{2} e^{2 i e} - d^{2}\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (c^{2} - 2 i c d - d^{2}\right )}{8 a^{3}} \]
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Exception generated. \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.68 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.56 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=-\frac {\frac {6 \, {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3}} + \frac {6 \, {\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3}} + \frac {-11 i \, c^{2} \tan \left (f x + e\right )^{3} - 22 \, c d \tan \left (f x + e\right )^{3} + 11 i \, d^{2} \tan \left (f x + e\right )^{3} - 45 \, c^{2} \tan \left (f x + e\right )^{2} + 90 i \, c d \tan \left (f x + e\right )^{2} + 45 \, d^{2} \tan \left (f x + e\right )^{2} + 69 i \, c^{2} \tan \left (f x + e\right ) + 138 \, c d \tan \left (f x + e\right ) - 21 i \, d^{2} \tan \left (f x + e\right ) + 51 \, c^{2} - 38 i \, c d - 3 \, d^{2}}{a^{3} {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \]
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Time = 6.00 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.15 \[ \int \frac {(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {c\,d}{6\,a^3}-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,c^2}{8\,a^3}+\frac {d^2}{8\,a^3}-\frac {c\,d\,3{}\mathrm {i}}{4\,a^3}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {c\,d}{4\,a^3}+\frac {c^2\,1{}\mathrm {i}}{8\,a^3}-\frac {d^2\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {c^2\,5{}\mathrm {i}}{12\,a^3}+\frac {d^2\,1{}\mathrm {i}}{12\,a^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {x\,{\left (d+c\,1{}\mathrm {i}\right )}^2}{8\,a^3} \]
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