Optimal. Leaf size=86 \[ -\frac {2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {11 c \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac {4 c \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \]
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Rubi [A]
time = 0.13, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {4093, 4085,
3879} \begin {gather*} -\frac {4 c \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac {11 c \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 3879
Rule 4085
Rule 4093
Rubi steps
\begin {align*} \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx &=-\frac {2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {\int \frac {\sec (e+f x) (-6 a c+5 a c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {11 c \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac {(4 c) \int \frac {\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{15 a^2}\\ &=-\frac {2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {11 c \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac {4 c \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.17, size = 43, normalized size = 0.50 \begin {gather*} -\frac {c (4+\cos (e+f x)) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \tan ^3\left (\frac {1}{2} (e+f x)\right )}{30 a^3 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.20, size = 37, normalized size = 0.43
method | result | size |
derivativedivides | \(\frac {c \left (-\frac {\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}\right )}{2 f \,a^{3}}\) | \(37\) |
default | \(\frac {c \left (-\frac {\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}\right )}{2 f \,a^{3}}\) | \(37\) |
risch | \(\frac {2 i c \left (15 \,{\mathrm e}^{3 i \left (f x +e \right )}-5 \,{\mathrm e}^{2 i \left (f x +e \right )}+5 \,{\mathrm e}^{i \left (f x +e \right )}+1\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\) | \(59\) |
norman | \(\frac {-\frac {c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{6 a f}+\frac {7 c \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{30 a f}+\frac {c \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{30 a f}-\frac {c \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{10 a f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2} a^{2}}\) | \(101\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 125, normalized size = 1.45 \begin {gather*} -\frac {\frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, c {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.55, size = 84, normalized size = 0.98 \begin {gather*} \frac {{\left (c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) - 4 \, c\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \end {gather*}
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 \begin {gather*} - \frac {c \left (\int \left (- \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.51, size = 37, normalized size = 0.43 \begin {gather*} -\frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}{30 \, a^{3} f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.70, size = 35, normalized size = 0.41 \begin {gather*} -\frac {c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\right )}{30\,a^3\,f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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