Integrand size = 36, antiderivative size = 112 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {a^2 B x}{c^3}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )} \]
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Time = 0.19 (sec) , antiderivative size = 112, 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} \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {a^2 c^2 (A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac {2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {a^2 B x}{c^3}-\frac {2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]
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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))}{(c-c \sin (e+f x))^5} \, dx \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\left (a^2 B c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^4} \, dx \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {\left (a^2 B\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{c} \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {\left (a^2 B\right ) \int 1 \, dx}{c^3} \\ & = -\frac {a^2 B x}{c^3}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(278\) vs. \(2(112)=224\).
Time = 11.28 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.48 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (12 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-4 (3 A+8 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-15 B (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+24 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-8 (3 A+8 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+2 (3 A+43 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^3} \]
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Time = 0.77 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12
method | result | size |
derivativedivides | \(\frac {2 a^{2} \left (-B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {A +B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {16 A +16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {24 A +16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}\right )}{f \,c^{3}}\) | \(126\) |
default | \(\frac {2 a^{2} \left (-B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {A +B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {16 A +16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {24 A +16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}\right )}{f \,c^{3}}\) | \(126\) |
parallelrisch | \(-\frac {2 \left (\frac {B \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) f x}{2}+\left (-\frac {5}{2} f x B +A +B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+B \left (5 f x -4\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-5 f x B +2 A +\frac {34}{3} B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+B \left (\frac {5 f x}{2}-\frac {20}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {f x B}{2}+\frac {A}{5}+\frac {23 B}{15}\right ) a^{2}}{f \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(133\) |
risch | \(-\frac {a^{2} B x}{c^{3}}+\frac {-4 A \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+2 A \,a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-\frac {100 B \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{3}-24 i B \,a^{2} {\mathrm e}^{3 i \left (f x +e \right )}+\frac {56 i B \,a^{2} {\mathrm e}^{i \left (f x +e \right )}}{3}+10 B \,a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 A \,a^{2}}{5}+\frac {86 B \,a^{2}}{15}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3}}\) | \(139\) |
norman | \(\frac {\frac {a^{2} x B}{c}+\frac {8 B \,a^{2} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {48 B \,a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {6 A \,a^{2}+46 B \,a^{2}}{15 c f}+\frac {40 B \,a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 c f}+\frac {64 B \,a^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {112 B \,a^{2} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {\left (2 A \,a^{2}+2 B \,a^{2}\right ) \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {\left (30 A \,a^{2}+86 B \,a^{2}\right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {2 \left (38 A \,a^{2}+198 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}-\frac {\left (78 A \,a^{2}+478 B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}-\frac {2 \left (138 A \,a^{2}+578 B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}-\frac {5 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}+\frac {13 a^{2} x B \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {25 a^{2} x B \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {38 a^{2} x B \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {46 a^{2} x B \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {46 a^{2} x B \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {38 a^{2} x B \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {25 a^{2} x B \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {13 a^{2} x B \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {5 a^{2} x B \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {a^{2} x B \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(566\) |
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Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (111) = 222\).
Time = 0.26 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.47 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {60 \, B a^{2} f x - {\left (15 \, B a^{2} f x - {\left (3 \, A + 43 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (A + B\right )} a^{2} - {\left (45 \, B a^{2} f x - {\left (9 \, A - 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B a^{2} f x - {\left (A + 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) - {\left (60 \, B a^{2} f x + 12 \, {\left (A + B\right )} a^{2} - {\left (15 \, B a^{2} f x + {\left (3 \, A + 43 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B a^{2} f x + {\left (A - 9 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1647 vs. \(2 (102) = 204\).
Time = 7.78 (sec) , antiderivative size = 1647, normalized size of antiderivative = 14.71 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1139 vs. \(2 (111) = 222\).
Time = 0.33 (sec) , antiderivative size = 1139, normalized size of antiderivative = 10.17 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \]
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Time = 0.32 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.35 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {\frac {15 \, {\left (f x + e\right )} B a^{2}}{c^{3}} + \frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 60 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 170 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 100 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A a^{2} + 23 \, B a^{2}\right )}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{15 \, f} \]
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Time = 15.85 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.08 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {B\,a^2\,x}{c^3}-\frac {\frac {a^2\,\left (6\,A+46\,B-15\,B\,\left (e+f\,x\right )\right )}{15}-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {a^2\,\left (120\,B-150\,B\,\left (e+f\,x\right )\right )}{15}+10\,B\,a^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (30\,A+30\,B-75\,B\,\left (e+f\,x\right )\right )}{15}+5\,B\,a^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (60\,A+340\,B-150\,B\,\left (e+f\,x\right )\right )}{15}+10\,B\,a^2\,\left (e+f\,x\right )\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {a^2\,\left (200\,B-75\,B\,\left (e+f\,x\right )\right )}{15}+5\,B\,a^2\,\left (e+f\,x\right )\right )+B\,a^2\,\left (e+f\,x\right )}{c^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^5} \]
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