Optimal. Leaf size=85 \[ \frac {2 \tan (e+f x)}{5 a c^3 f}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2} \]
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Rubi [A] time = 0.16, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2736, 2672, 3767, 8} \[ \frac {2 \tan (e+f x)}{5 a c^3 f}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2672
Rule 2736
Rule 3767
Rubi steps
\begin {align*} \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx &=\frac {\int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{a c}\\ &=\frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {3 \int \frac {\sec ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{5 a c^2}\\ &=\frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {2 \int \sec ^2(e+f x) \, dx}{5 a c^3}\\ &=\frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {2 \operatorname {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{5 a c^3 f}\\ &=\frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {2 \tan (e+f x)}{5 a c^3 f}\\ \end {align*}
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Mathematica [A] time = 0.67, size = 111, normalized size = 1.31 \[ -\frac {12 \sin (e+f x)+32 \sin (2 (e+f x))+12 \sin (3 (e+f x))-8 \sin (4 (e+f x))+32 \cos (e+f x)-12 \cos (2 (e+f x))+32 \cos (3 (e+f x))+3 \cos (4 (e+f x))-15}{160 a c^3 f (\sin (e+f x)-1)^3 (\sin (e+f x)+1)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 83, normalized size = 0.98 \[ -\frac {4 \, \cos \left (f x + e\right )^{2} - {\left (2 \, \cos \left (f x + e\right )^{2} - 3\right )} \sin \left (f x + e\right ) - 2}{5 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.30, size = 105, normalized size = 1.24 \[ -\frac {\frac {5}{a c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {35 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 90 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 120 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 70 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 21}{a c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{20 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 103, normalized size = 1.21 \[ \frac {-\frac {4}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {5}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {7}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{f a \,c^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.58, size = 211, normalized size = 2.48 \[ -\frac {2 \, {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - 2\right )}}{5 \, {\left (a c^{3} - \frac {4 \, a c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {4 \, a c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.16, size = 89, normalized size = 1.05 \[ -\frac {2\,\left (5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+2\right )}{5\,a\,c^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^5\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 8.43, size = 614, normalized size = 7.22 \[ \begin {cases} - \frac {10 \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} + \frac {20 \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} - \frac {20 \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} + \frac {6 \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} - \frac {4}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} & \text {for}\: f \neq 0 \\\frac {x}{\left (a \sin {\relax (e )} + a\right ) \left (- c \sin {\relax (e )} + c\right )^{3}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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