Optimal. Leaf size=193 \[ -\frac{5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac{5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )}-\frac{3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}+\frac{15 x}{64 a^2 c^4}-\frac{i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac{i}{32 a^2 f (c-i c \tan (e+f x))^4} \]
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Rubi [A] time = 0.190532, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3522, 3487, 44, 206} \[ -\frac{5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac{5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )}-\frac{3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}+\frac{15 x}{64 a^2 c^4}-\frac{i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac{i}{32 a^2 f (c-i c \tan (e+f x))^4} \]
Antiderivative was successfully verified.
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Rule 3522
Rule 3487
Rule 44
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx &=\frac{\int \frac{\cos ^4(e+f x)}{(c-i c \tan (e+f x))^2} \, dx}{a^2 c^2}\\ &=\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^3 (c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{32 c^5 (c-x)^3}+\frac{5}{64 c^6 (c-x)^2}+\frac{1}{8 c^3 (c+x)^5}+\frac{3}{16 c^4 (c+x)^4}+\frac{3}{16 c^5 (c+x)^3}+\frac{5}{32 c^6 (c+x)^2}+\frac{15}{64 c^6 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=-\frac{i}{32 a^2 f (c-i c \tan (e+f x))^4}-\frac{i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac{3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac{5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac{5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )}+\frac{(15 i) \operatorname{Subst}\left (\int \frac{1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{64 a^2 c^3 f}\\ &=\frac{15 x}{64 a^2 c^4}-\frac{i}{32 a^2 f (c-i c \tan (e+f x))^4}-\frac{i}{16 a^2 c f (c-i c \tan (e+f x))^3}-\frac{3 i}{32 a^2 f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{i}{64 a^2 f \left (c^2+i c^2 \tan (e+f x)\right )^2}-\frac{5 i}{32 a^2 f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac{5 i}{64 a^2 f \left (c^4+i c^4 \tan (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 1.18212, size = 139, normalized size = 0.72 \[ \frac{\sec ^2(e+f x) (\sin (4 (e+f x))-i \cos (4 (e+f x))) (-120 f x \sin (2 (e+f x))-30 i \sin (2 (e+f x))-32 i \sin (4 (e+f x))-3 i \sin (6 (e+f x))+(-30-120 i f x) \cos (2 (e+f x))+16 \cos (4 (e+f x))+\cos (6 (e+f x))-80)}{512 a^2 c^4 f (\tan (e+f x)-i)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.048, size = 181, normalized size = 0.9 \begin{align*}{\frac{-{\frac{15\,i}{128}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{2}{c}^{4}}}-{\frac{{\frac{i}{64}}}{f{a}^{2}{c}^{4} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{5}{64\,f{a}^{2}{c}^{4} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{3\,i}{32}}}{f{a}^{2}{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{{\frac{i}{32}}}{f{a}^{2}{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}}+{\frac{{\frac{15\,i}{128}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{a}^{2}{c}^{4}}}-{\frac{1}{16\,f{a}^{2}{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{5}{32\,f{a}^{2}{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.35322, size = 285, normalized size = 1.48 \begin{align*} \frac{{\left (120 \, f x e^{\left (4 i \, f x + 4 i \, e\right )} - i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 8 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 30 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 80 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 24 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{512 \, a^{2} c^{4} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 1.38695, size = 298, normalized size = 1.54 \begin{align*} \begin{cases} \frac{\left (- 8589934592 i a^{10} c^{20} f^{5} e^{14 i e} e^{8 i f x} - 68719476736 i a^{10} c^{20} f^{5} e^{12 i e} e^{6 i f x} - 257698037760 i a^{10} c^{20} f^{5} e^{10 i e} e^{4 i f x} - 687194767360 i a^{10} c^{20} f^{5} e^{8 i e} e^{2 i f x} + 206158430208 i a^{10} c^{20} f^{5} e^{4 i e} e^{- 2 i f x} + 17179869184 i a^{10} c^{20} f^{5} e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{4398046511104 a^{12} c^{24} f^{6}} & \text{for}\: 4398046511104 a^{12} c^{24} f^{6} e^{6 i e} \neq 0 \\x \left (\frac{\left (e^{12 i e} + 6 e^{10 i e} + 15 e^{8 i e} + 20 e^{6 i e} + 15 e^{4 i e} + 6 e^{2 i e} + 1\right ) e^{- 4 i e}}{64 a^{2} c^{4}} - \frac{15}{64 a^{2} c^{4}}\right ) & \text{otherwise} \end{cases} + \frac{15 x}{64 a^{2} c^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.33375, size = 201, normalized size = 1.04 \begin{align*} -\frac{-\frac{60 i \, \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{a^{2} c^{4}} + \frac{60 i \, \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a^{2} c^{4}} + \frac{2 \,{\left (-45 i \, \tan \left (f x + e\right )^{2} - 110 \, \tan \left (f x + e\right ) + 69 i\right )}}{a^{2} c^{4}{\left (\tan \left (f x + e\right ) - i\right )}^{2}} + \frac{125 i \, \tan \left (f x + e\right )^{4} - 580 \, \tan \left (f x + e\right )^{3} - 1038 i \, \tan \left (f x + e\right )^{2} + 868 \, \tan \left (f x + e\right ) + 301 i}{a^{2} c^{4}{\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{512 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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