Optimal. Leaf size=162 \[ \frac {5 x}{32 a c^4}-\frac {i}{16 a f (c-i c \tan (e+f x))^4}-\frac {i}{12 a c f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}-\frac {i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )} \]
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Rubi [A]
time = 0.12, antiderivative size = 162, normalized size of antiderivative = 1.00, 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 = {3603, 3568, 46,
212} \begin {gather*} -\frac {i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )}+\frac {5 x}{32 a c^4}-\frac {3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}-\frac {i}{12 a c f (c-i c \tan (e+f x))^3}-\frac {i}{16 a f (c-i c \tan (e+f x))^4} \end {gather*}
Antiderivative was successfully verified.
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Rule 46
Rule 212
Rule 3568
Rule 3603
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx &=\frac {\int \frac {\cos ^2(e+f x)}{(c-i c \tan (e+f x))^3} \, dx}{a c}\\ &=\frac {\left (i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=\frac {\left (i c^2\right ) \text {Subst}\left (\int \left (\frac {1}{32 c^5 (c-x)^2}+\frac {1}{4 c^2 (c+x)^5}+\frac {1}{4 c^3 (c+x)^4}+\frac {3}{16 c^4 (c+x)^3}+\frac {1}{8 c^5 (c+x)^2}+\frac {5}{32 c^5 \left (c^2-x^2\right )}\right ) \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=-\frac {i}{16 a f (c-i c \tan (e+f x))^4}-\frac {i}{12 a c f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}-\frac {i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )}+\frac {(5 i) \text {Subst}\left (\int \frac {1}{c^2-x^2} \, dx,x,-i c \tan (e+f x)\right )}{32 a c^3 f}\\ &=\frac {5 x}{32 a c^4}-\frac {i}{16 a f (c-i c \tan (e+f x))^4}-\frac {i}{12 a c f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}-\frac {i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.78, size = 134, normalized size = 0.83 \begin {gather*} \frac {\sec (e+f x) (\cos (4 (e+f x))+i \sin (4 (e+f x))) (-180 \cos (e+f x)+(-20-120 i f x) \cos (3 (e+f x))+9 \cos (5 (e+f x))+60 i \sin (e+f x)-20 i \sin (3 (e+f x))-120 f x \sin (3 (e+f x))-15 i \sin (5 (e+f x)))}{768 a c^4 f (-i+\tan (e+f x))} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.22, size = 105, normalized size = 0.65
method | result | size |
derivativedivides | \(\frac {\frac {3 i}{32 \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {i}{16 \left (\tan \left (f x +e \right )+i\right )^{4}}+\frac {5 i \ln \left (\tan \left (f x +e \right )+i\right )}{64}-\frac {1}{12 \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {1}{8 \tan \left (f x +e \right )+8 i}-\frac {5 i \ln \left (\tan \left (f x +e \right )-i\right )}{64}+\frac {1}{32 \tan \left (f x +e \right )-32 i}}{f a \,c^{4}}\) | \(105\) |
default | \(\frac {\frac {3 i}{32 \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {i}{16 \left (\tan \left (f x +e \right )+i\right )^{4}}+\frac {5 i \ln \left (\tan \left (f x +e \right )+i\right )}{64}-\frac {1}{12 \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {1}{8 \tan \left (f x +e \right )+8 i}-\frac {5 i \ln \left (\tan \left (f x +e \right )-i\right )}{64}+\frac {1}{32 \tan \left (f x +e \right )-32 i}}{f a \,c^{4}}\) | \(105\) |
risch | \(\frac {5 x}{32 a \,c^{4}}-\frac {i {\mathrm e}^{8 i \left (f x +e \right )}}{256 a \,c^{4} f}-\frac {5 i {\mathrm e}^{6 i \left (f x +e \right )}}{192 a \,c^{4} f}-\frac {5 i {\mathrm e}^{4 i \left (f x +e \right )}}{64 a \,c^{4} f}-\frac {9 i \cos \left (2 f x +2 e \right )}{64 a \,c^{4} f}+\frac {11 \sin \left (2 f x +2 e \right )}{64 a \,c^{4} f}\) | \(115\) |
norman | \(\frac {\frac {5 x}{32 a c}+\frac {27 \tan \left (f x +e \right )}{32 a c f}+\frac {73 \left (\tan ^{3}\left (f x +e \right )\right )}{96 a c f}+\frac {55 \left (\tan ^{5}\left (f x +e \right )\right )}{96 a c f}+\frac {5 \left (\tan ^{7}\left (f x +e \right )\right )}{32 a c f}+\frac {5 x \left (\tan ^{2}\left (f x +e \right )\right )}{8 a c}+\frac {15 x \left (\tan ^{4}\left (f x +e \right )\right )}{16 a c}+\frac {5 x \left (\tan ^{6}\left (f x +e \right )\right )}{8 a c}+\frac {5 x \left (\tan ^{8}\left (f x +e \right )\right )}{32 a c}-\frac {i}{3 a c f}+\frac {i \left (\tan ^{2}\left (f x +e \right )\right )}{6 a c f}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{4} c^{3}}\) | \(201\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.63, size = 85, normalized size = 0.52 \begin {gather*} \frac {{\left (120 \, f x e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 20 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 60 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 120 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{768 \, a c^{4} f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.28, size = 246, normalized size = 1.52 \begin {gather*} \begin {cases} \frac {\left (- 25165824 i a^{4} c^{16} f^{4} e^{10 i e} e^{8 i f x} - 167772160 i a^{4} c^{16} f^{4} e^{8 i e} e^{6 i f x} - 503316480 i a^{4} c^{16} f^{4} e^{6 i e} e^{4 i f x} - 1006632960 i a^{4} c^{16} f^{4} e^{4 i e} e^{2 i f x} + 100663296 i a^{4} c^{16} f^{4} e^{- 2 i f x}\right ) e^{- 2 i e}}{6442450944 a^{5} c^{20} f^{5}} & \text {for}\: a^{5} c^{20} f^{5} e^{2 i e} \neq 0 \\x \left (\frac {\left (e^{10 i e} + 5 e^{8 i e} + 10 e^{6 i e} + 10 e^{4 i e} + 5 e^{2 i e} + 1\right ) e^{- 2 i e}}{32 a c^{4}} - \frac {5}{32 a c^{4}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{32 a c^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.64, size = 140, normalized size = 0.86 \begin {gather*} -\frac {\frac {60 i \, \log \left (i \, \tan \left (f x + e\right ) + 1\right )}{a c^{4}} - \frac {60 i \, \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a c^{4}} - \frac {12 \, {\left (5 \, \tan \left (f x + e\right ) - 7 i\right )}}{a c^{4} {\left (-i \, \tan \left (f x + e\right ) - 1\right )}} + \frac {125 i \, \tan \left (f x + e\right )^{4} - 596 \, \tan \left (f x + e\right )^{3} - 1110 i \, \tan \left (f x + e\right )^{2} + 996 \, \tan \left (f x + e\right ) + 405 i}{a c^{4} {\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{768 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.17, size = 88, normalized size = 0.54 \begin {gather*} \frac {5\,x}{32\,a\,c^4}-\frac {-\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,5{}\mathrm {i}}{32}+\frac {15\,{\mathrm {tan}\left (e+f\,x\right )}^3}{32}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,35{}\mathrm {i}}{96}+\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{32}+\frac {1}{3}{}\mathrm {i}}{a\,c^4\,f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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