Optimal. Leaf size=251 \[ \frac {3 i 2^{-m-4} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{a^3 f}+\frac {3 i 2^{-2 m-5} e^{-4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 i f (c+d x)}{d}\right )}{a^3 f}+\frac {i 2^{-m-4} 3^{-m-1} e^{-6 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {6 i f (c+d x)}{d}\right )}{a^3 f}+\frac {(c+d x)^{m+1}}{8 a^3 d (m+1)} \]
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Rubi [A] time = 0.24, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3729, 2181} \[ \frac {3 i 2^{-m-4} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 i f (c+d x)}{d}\right )}{a^3 f}+\frac {3 i 2^{-2 m-5} e^{-4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {4 i f (c+d x)}{d}\right )}{a^3 f}+\frac {i 2^{-m-4} 3^{-m-1} e^{-6 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {6 i f (c+d x)}{d}\right )}{a^3 f}+\frac {(c+d x)^{m+1}}{8 a^3 d (m+1)} \]
Antiderivative was successfully verified.
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Rule 2181
Rule 3729
Rubi steps
\begin {align*} \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx &=\int \left (\frac {(c+d x)^m}{8 a^3}+\frac {3 e^{-2 i e-2 i f x} (c+d x)^m}{8 a^3}+\frac {3 e^{-4 i e-4 i f x} (c+d x)^m}{8 a^3}+\frac {e^{-6 i e-6 i f x} (c+d x)^m}{8 a^3}\right ) \, dx\\ &=\frac {(c+d x)^{1+m}}{8 a^3 d (1+m)}+\frac {\int e^{-6 i e-6 i f x} (c+d x)^m \, dx}{8 a^3}+\frac {3 \int e^{-2 i e-2 i f x} (c+d x)^m \, dx}{8 a^3}+\frac {3 \int e^{-4 i e-4 i f x} (c+d x)^m \, dx}{8 a^3}\\ &=\frac {(c+d x)^{1+m}}{8 a^3 d (1+m)}+\frac {3 i 2^{-4-m} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{a^3 f}+\frac {3 i 2^{-5-2 m} e^{-4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i f (c+d x)}{d}\right )}{a^3 f}+\frac {i 2^{-4-m} 3^{-1-m} e^{-6 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {6 i f (c+d x)}{d}\right )}{a^3 f}\\ \end {align*}
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Mathematica [A] time = 68.70, size = 269, normalized size = 1.07 \[ \frac {e^{-3 i e} 2^{-2 m-5} 3^{-m-1} (c+d x)^m \sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac {i f (c+d x)}{d}\right )^{-m} \left (e^{6 i e} f 12^{m+1} (c+d x) \left (\frac {i f (c+d x)}{d}\right )^m+i d 2^{m+1} 3^{m+2} (m+1) e^{2 i \left (\frac {c f}{d}+2 e\right )} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )+i d 3^{m+2} (m+1) e^{\frac {4 i c f}{d}+2 i e} \Gamma \left (m+1,\frac {4 i f (c+d x)}{d}\right )+i d 2^{m+1} (m+1) e^{\frac {6 i c f}{d}} \Gamma \left (m+1,\frac {6 i f (c+d x)}{d}\right )\right )}{d f (m+1) (a+i a \tan (e+f x))^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 192, normalized size = 0.76 \[ \frac {{\left (2 i \, d m + 2 i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {6 i \, f}{d}\right ) + 6 i \, d e - 6 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {6 i \, d f x + 6 i \, c f}{d}\right ) + {\left (9 i \, d m + 9 i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {4 i \, f}{d}\right ) + 4 i \, d e - 4 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {4 i \, d f x + 4 i \, c f}{d}\right ) + {\left (18 i \, d m + 18 i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {2 i \, d f x + 2 i \, c f}{d}\right ) + 12 \, {\left (d f x + c f\right )} {\left (d x + c\right )}^{m}}{96 \, {\left (a^{3} d f m + a^{3} d f\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{m}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{m}}{\left (a +i a \tan \left (f x +e \right )\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (6 \, f x + 6 \, e\right )\,{d x} + 3 \, {\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (4 \, f x + 4 \, e\right )\,{d x} + 3 \, {\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (2 \, f x + 2 \, e\right )\,{d x} - {\left (i \, d m + i \, d\right )} \int {\left (d x + c\right )}^{m} \sin \left (6 \, f x + 6 \, e\right )\,{d x} - {\left (3 i \, d m + 3 i \, d\right )} \int {\left (d x + c\right )}^{m} \sin \left (4 \, f x + 4 \, e\right )\,{d x} - {\left (3 i \, d m + 3 i \, d\right )} \int {\left (d x + c\right )}^{m} \sin \left (2 \, f x + 2 \, e\right )\,{d x} + e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )}}{8 \, {\left (a^{3} d m + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \int \frac {\left (c + d x\right )^{m}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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