Optimal. Leaf size=99 \[ -\frac{i c^3 (a+i a \tan (e+f x))^{m+2}}{a^2 f (m+2)}-\frac{4 i c^3 (a+i a \tan (e+f x))^m}{f m}+\frac{4 i c^3 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)} \]
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Rubi [A] time = 0.145773, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ -\frac{i c^3 (a+i a \tan (e+f x))^{m+2}}{a^2 f (m+2)}-\frac{4 i c^3 (a+i a \tan (e+f x))^m}{f m}+\frac{4 i c^3 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)} \]
Antiderivative was successfully verified.
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Rule 3522
Rule 3487
Rule 43
Rubi steps
\begin{align*} \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^3 \, dx &=\left (a^3 c^3\right ) \int \sec ^6(e+f x) (a+i a \tan (e+f x))^{-3+m} \, dx\\ &=-\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int (a-x)^2 (a+x)^{-1+m} \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=-\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \left (4 a^2 (a+x)^{-1+m}-4 a (a+x)^m+(a+x)^{1+m}\right ) \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=-\frac{4 i c^3 (a+i a \tan (e+f x))^m}{f m}+\frac{4 i c^3 (a+i a \tan (e+f x))^{1+m}}{a f (1+m)}-\frac{i c^3 (a+i a \tan (e+f x))^{2+m}}{a^2 f (2+m)}\\ \end{align*}
Mathematica [F] time = 180.002, size = 0, normalized size = 0. \[ \text{\$Aborted} \]
Verification is Not applicable to the result.
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Maple [B] time = 0.357, size = 192, normalized size = 1.9 \begin{align*}{\frac{i{c}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{2}{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 2+m \right ) }}-2\,{\frac{{c}^{3} \left ( 3+m \right ) \tan \left ( fx+e \right ){{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+m \right ) \left ( 2+m \right ) }}-{\frac{i{c}^{3}m{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+m \right ) \left ( 2+m \right ) }}-{\frac{5\,i{c}^{3}{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+m \right ) \left ( 2+m \right ) }}-{\frac{8\,i{c}^{3}{{\rm e}^{m\ln \left ( a+ia\tan \left ( fx+e \right ) \right ) }}}{fm \left ( 1+m \right ) \left ( 2+m \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.63905, size = 396, normalized size = 4. \begin{align*} \frac{{\left (-4 i \, c^{3} m^{2} - 12 i \, c^{3} m - 8 i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 8 i \, c^{3} +{\left (-8 i \, c^{3} m - 16 i \, c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \left (\frac{2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m}}{f m^{3} + 3 \, f m^{2} + 2 \, f m +{\left (f m^{3} + 3 \, f m^{2} + 2 \, f m\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \,{\left (f m^{3} + 3 \, f m^{2} + 2 \, f m\right )} e^{\left (2 i \, f x + 2 i \, e\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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