Integrand size = 36, antiderivative size = 110 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {B c^2 x}{a^3}-\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac {2 B c^2 \cos (e+f x)}{f \left (a^3+a^3 \sin (e+f x)\right )} \]
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Time = 0.18 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3046, 2938, 2759, 8} \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {2 B c^2 \cos (e+f x)}{f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {B c^2 x}{a^3}-\frac {a^2 c^2 (A-B) \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]
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Rule 8
Rule 2759
Rule 2938
Rule 3046
Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(a+a \sin (e+f x))^5} \, dx \\ & = -\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}+\left (a B c^2\right ) \int \frac {\cos ^4(e+f x)}{(a+a \sin (e+f x))^4} \, dx \\ & = -\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}-\frac {\left (B c^2\right ) \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^2} \, dx}{a} \\ & = -\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac {2 B c^2 \cos (e+f x)}{f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\left (B c^2\right ) \int 1 \, dx}{a^3} \\ & = \frac {B c^2 x}{a^3}-\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac {2 B c^2 \cos (e+f x)}{f \left (a^3+a^3 \sin (e+f x)\right )} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(272\) vs. \(2(110)=220\).
Time = 11.25 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.47 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (24 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )-12 (A-B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-8 (3 A-8 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+4 (3 A-8 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+2 (3 A-43 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+15 B (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right ) (c-c \sin (e+f x))^2}{15 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (1+\sin (e+f x))^3} \]
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Time = 0.78 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.15
method | result | size |
derivativedivides | \(\frac {2 c^{2} \left (B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {-32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {A -B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {16 A -16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {24 A -16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}\right )}{f \,a^{3}}\) | \(127\) |
default | \(\frac {2 c^{2} \left (B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {-32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {A -B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {16 A -16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {24 A -16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}\right )}{f \,a^{3}}\) | \(127\) |
parallelrisch | \(-\frac {2 c^{2} \left (-\frac {B \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) f x}{2}+\left (-\frac {5}{2} f x B +A -B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-5 f x -4\right ) B \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-5 f x B +2 A -\frac {34}{3} B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+B \left (-\frac {5 f x}{2}-\frac {20}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {f x B}{2}+\frac {A}{5}-\frac {23 B}{15}\right )}{f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(135\) |
risch | \(\frac {B \,c^{2} x}{a^{3}}-\frac {2 \left (-30 A \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+15 A \,c^{2} {\mathrm e}^{4 i \left (f x +e \right )}+250 B \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-180 i B \,c^{2} {\mathrm e}^{3 i \left (f x +e \right )}+140 i B \,c^{2} {\mathrm e}^{i \left (f x +e \right )}-75 B \,c^{2} {\mathrm e}^{4 i \left (f x +e \right )}+3 A \,c^{2}-43 B \,c^{2}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(138\) |
norman | \(\frac {\frac {B \,c^{2} x}{a}+\frac {8 B \,c^{2} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {48 B \,c^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {B \,c^{2} x \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {6 A \,c^{2}-46 B \,c^{2}}{15 f a}+\frac {40 B \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f a}+\frac {64 B \,c^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {112 B \,c^{2} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {\left (2 A \,c^{2}-2 B \,c^{2}\right ) \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {\left (30 A \,c^{2}-86 B \,c^{2}\right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {2 \left (38 A \,c^{2}-198 B \,c^{2}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 f a}-\frac {\left (78 A \,c^{2}-478 B \,c^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 f a}-\frac {2 \left (138 A \,c^{2}-578 B \,c^{2}\right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 f a}+\frac {5 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}+\frac {13 B \,c^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {25 B \,c^{2} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {38 B \,c^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {46 B \,c^{2} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {46 B \,c^{2} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {38 B \,c^{2} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {25 B \,c^{2} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {13 B \,c^{2} x \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {5 B \,c^{2} x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(565\) |
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Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (106) = 212\).
Time = 0.28 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.54 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {60 \, B c^{2} f x - {\left (15 \, B c^{2} f x - {\left (3 \, A - 43 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (A - B\right )} c^{2} - {\left (45 \, B c^{2} f x - {\left (9 \, A + 11 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B c^{2} f x - {\left (A - 11 \, B\right )} c^{2}\right )} \cos \left (f x + e\right ) + {\left (60 \, B c^{2} f x + 12 \, {\left (A - B\right )} c^{2} - {\left (15 \, B c^{2} f x + {\left (3 \, A - 43 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B c^{2} f x + {\left (A + 9 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1647 vs. \(2 (102) = 204\).
Time = 7.72 (sec) , antiderivative size = 1647, normalized size of antiderivative = 14.97 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1134 vs. \(2 (106) = 212\).
Time = 0.31 (sec) , antiderivative size = 1134, normalized size of antiderivative = 10.31 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \]
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Time = 0.44 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} B c^{2}}{a^{3}} - \frac {2 \, {\left (15 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 60 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 170 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 100 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A c^{2} - 23 \, B c^{2}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \]
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Time = 16.20 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.09 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {c^2\,\left (120\,B+150\,B\,\left (e+f\,x\right )\right )}{15}-10\,B\,c^2\,\left (e+f\,x\right )\right )+\frac {c^2\,\left (46\,B-6\,A+15\,B\,\left (e+f\,x\right )\right )}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {c^2\,\left (30\,B-30\,A+75\,B\,\left (e+f\,x\right )\right )}{15}-5\,B\,c^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {c^2\,\left (340\,B-60\,A+150\,B\,\left (e+f\,x\right )\right )}{15}-10\,B\,c^2\,\left (e+f\,x\right )\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {c^2\,\left (200\,B+75\,B\,\left (e+f\,x\right )\right )}{15}-5\,B\,c^2\,\left (e+f\,x\right )\right )-B\,c^2\,\left (e+f\,x\right )}{a^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^5}+\frac {B\,c^2\,x}{a^3} \]
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