Optimal. Leaf size=92 \[ -\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}+\frac {5 a^3 x}{c^2}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \]
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Rubi [A] time = 0.19, antiderivative size = 92, 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, 2680, 2682, 8} \[ -\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}+\frac {5 a^3 x}{c^2}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2680
Rule 2682
Rule 2736
Rubi steps
\begin {align*} \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {1}{3} \left (5 a^3 c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^3} \, dx\\ &=\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac {\left (5 a^3\right ) \int \frac {\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{c}\\ &=-\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac {\left (5 a^3\right ) \int 1 \, dx}{c^2}\\ &=\frac {5 a^3 x}{c^2}-\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}\\ \end {align*}
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Mathematica [A] time = 1.00, size = 149, normalized size = 1.62 \[ \frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 (15 e+15 f x+23) \cos \left (\frac {1}{2} (e+f x)\right )-(30 e+30 f x+121) \cos \left (\frac {3}{2} (e+f x)\right )+3 \cos \left (\frac {5}{2} (e+f x)\right )-6 \sin \left (\frac {1}{2} (e+f x)\right ) (2 (5 e+5 f x-2) \cos (e+f x)-\cos (2 (e+f x))+20 e+20 f x+31)\right )}{12 c^2 f (\sin (e+f x)-1)^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.45, size = 184, normalized size = 2.00 \[ -\frac {3 \, a^{3} \cos \left (f x + e\right )^{3} + 30 \, a^{3} f x + 8 \, a^{3} - {\left (15 \, a^{3} f x + 31 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (15 \, a^{3} f x - 26 \, a^{3}\right )} \cos \left (f x + e\right ) - {\left (30 \, a^{3} f x - 3 \, a^{3} \cos \left (f x + e\right )^{2} - 8 \, a^{3} + {\left (15 \, a^{3} f x - 34 \, a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.38, size = 101, normalized size = 1.10 \[ \frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c^{2}} - \frac {6 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} c^{2}} + \frac {8 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, a^{3}\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 121, normalized size = 1.32 \[ -\frac {32 a^{3}}{3 c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {16 a^{3}}{c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {8 a^{3}}{c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {2 a^{3}}{c^{2} f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}+\frac {10 a^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c^{2} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.50, size = 594, normalized size = 6.46 \[ \frac {2 \, {\left (2 \, a^{3} {\left (\frac {\frac {12 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {11 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {9 \, \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 )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 5}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {4 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {4 \, c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{2} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}}\right )} + 3 \, a^{3} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 4}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}}\right )} - \frac {a^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, a^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.75, size = 218, normalized size = 2.37 \[ \frac {5\,a^3\,x}{c^2}+\frac {5\,a^3\,\left (e+f\,x\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (45\,e+45\,f\,x-114\right )}{3}\right )-\frac {a^3\,\left (15\,e+15\,f\,x-46\right )}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (45\,e+45\,f\,x-24\right )}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (20\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (60\,e+60\,f\,x-82\right )}{3}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (20\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (60\,e+60\,f\,x-102\right )}{3}\right )}{c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 13.65, size = 1282, normalized size = 13.93 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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