Optimal. Leaf size=64 \[ \frac{2 i a^2 (c-i c \tan (e+f x))^n}{f n}-\frac{i a^2 (c-i c \tan (e+f x))^{n+1}}{c f (n+1)} \]
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Rubi [A] time = 0.127287, antiderivative size = 64, 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} \[ \frac{2 i a^2 (c-i c \tan (e+f x))^n}{f n}-\frac{i a^2 (c-i c \tan (e+f x))^{n+1}}{c f (n+1)} \]
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))^2 (c-i c \tan (e+f x))^n \, dx &=\left (a^2 c^2\right ) \int \sec ^4(e+f x) (c-i c \tan (e+f x))^{-2+n} \, dx\\ &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int (c-x) (c+x)^{-1+n} \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \left (2 c (c+x)^{-1+n}-(c+x)^n\right ) \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac{2 i a^2 (c-i c \tan (e+f x))^n}{f n}-\frac{i a^2 (c-i c \tan (e+f x))^{1+n}}{c f (1+n)}\\ \end{align*}
Mathematica [A] time = 2.46127, size = 72, normalized size = 1.12 \[ -\frac{a^2 (n \tan (e+f x)-i (n+2)) (c \sec (e+f x))^n \exp (n (-\log (c \sec (e+f x))+\log (c-i c \tan (e+f x))))}{f n (n+1)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.278, size = 100, normalized size = 1.6 \begin{align*}{\frac{i{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}{a}^{2}}{f \left ( 1+n \right ) }}+{\frac{2\,i{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}{a}^{2}}{fn \left ( 1+n \right ) }}-{\frac{{a}^{2}\tan \left ( fx+e \right ){{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+n \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.8301, size = 367, normalized size = 5.73 \begin{align*} \frac{2^{n + 1} a^{2} c^{n} \cos \left (n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) - i \cdot 2^{n + 1} a^{2} c^{n} \sin \left (n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + 2 \,{\left (a^{2} c^{n} n + a^{2} c^{n}\right )} 2^{n} \cos \left (-2 \, f x + n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 2 \, e\right ) -{\left (2 i \, a^{2} c^{n} n + 2 i \, a^{2} c^{n}\right )} 2^{n} \sin \left (-2 \, f x + n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 2 \, e\right )}{{\left (-i \, n^{2} +{\left (-i \, n^{2} - i \, n\right )} \cos \left (2 \, f x + 2 \, e\right ) +{\left (n^{2} + n\right )} \sin \left (2 \, f x + 2 \, e\right ) - i \, n\right )}{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac{1}{2} \, n} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.64454, size = 188, normalized size = 2.94 \begin{align*} \frac{{\left (2 i \, a^{2} +{\left (2 i \, a^{2} n + 2 i \, a^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \left (\frac{2 \, c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}}{f n^{2} + f n +{\left (f n^{2} + f n\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 \, a \tan \left (f x + e\right ) + a\right )}^{2}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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