Integrand size = 33, antiderivative size = 127 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c+3 A d-8 B d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(2 A c+3 B c+3 A d+7 B d) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )} \]
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Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3047, 3098, 2829, 2727} \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {(2 A c+3 A d+3 B c+7 B d) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(2 A c+3 A d+3 B c-8 B d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]
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Rule 2727
Rule 2829
Rule 3047
Rule 3098
Rubi steps \begin{align*} \text {integral}& = \int \frac {A c+(B c+A d) \sin (e+f x)+B d \sin ^2(e+f x)}{(a+a \sin (e+f x))^3} \, dx \\ & = -\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {\int \frac {-a (2 A c+3 B c+3 A d-3 B d)-5 a B d \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c+3 A d-8 B d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}+\frac {(2 A c+3 B c+3 A d+7 B d) \int \frac {1}{a+a \sin (e+f x)} \, dx}{15 a^2} \\ & = -\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c+3 A d-8 B d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(2 A c+3 B c+3 A d+7 B d) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )} \\ \end{align*}
Time = 1.72 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (15 (A d+B (c+2 d)) \cos \left (\frac {1}{2} (e+f x)\right )-5 (2 A c+3 B c+3 A d+4 B d) \cos \left (\frac {3}{2} (e+f x)\right )-2 (-3 (3 A c+2 B c+2 A d+8 B d)+(2 A c+3 B c+3 A d-8 B d) \cos (e+f x)+(2 A c+3 B c+3 A d+7 B d) \cos (2 (e+f x))) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{30 a^3 f (1+\sin (e+f x))^3} \]
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Time = 0.66 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\frac {-30 A \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +\left (\left (-60 c -30 d \right ) A -30 B c \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\left (-80 c -30 d \right ) A -30 B \left (c +\frac {4 d}{3}\right )\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\left (-40 c -30 d \right ) A -30 B \left (c +\frac {2 d}{3}\right )\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (-14 c -6 d \right ) A -6 B \left (c +\frac {2 d}{3}\right )}{15 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(139\) |
derivativedivides | \(\frac {-\frac {2 \left (8 A c -6 d A -6 B c +4 d B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (4 A c -4 d A -4 B c +4 d B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A c +2 d A +2 B c}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 A c}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-8 A c +8 d A +8 B c -8 d B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}}{a^{3} f}\) | \(151\) |
default | \(\frac {-\frac {2 \left (8 A c -6 d A -6 B c +4 d B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (4 A c -4 d A -4 B c +4 d B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A c +2 d A +2 B c}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 A c}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-8 A c +8 d A +8 B c -8 d B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}}{a^{3} f}\) | \(151\) |
risch | \(-\frac {2 \left (3 B c +2 A c +15 B d \,{\mathrm e}^{4 i \left (f x +e \right )}-15 i B c \,{\mathrm e}^{i \left (f x +e \right )}+30 i B d \,{\mathrm e}^{3 i \left (f x +e \right )}+15 i B c \,{\mathrm e}^{3 i \left (f x +e \right )}-10 i A c \,{\mathrm e}^{i \left (f x +e \right )}-20 i B d \,{\mathrm e}^{i \left (f x +e \right )}-15 i d \,{\mathrm e}^{i \left (f x +e \right )} A +15 i A d \,{\mathrm e}^{3 i \left (f x +e \right )}-20 A c \,{\mathrm e}^{2 i \left (f x +e \right )}-15 A d \,{\mathrm e}^{2 i \left (f x +e \right )}-15 B c \,{\mathrm e}^{2 i \left (f x +e \right )}-40 B d \,{\mathrm e}^{2 i \left (f x +e \right )}+3 d A +7 d B \right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(203\) |
norman | \(\frac {-\frac {14 A c +6 d A +6 B c +4 d B}{15 f a}-\frac {2 A c \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {2 \left (2 A c +d A +B c \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {2 \left (34 A c +11 d A +11 B c +14 d B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 f a}-\frac {2 \left (16 A c +9 d A +9 B c +2 d B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {2 \left (14 A c +3 d A +3 B c +4 d B \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {2 \left (14 A c +9 d A +9 B c +4 d B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {2 \left (18 A c +7 d A +7 B c +8 d B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 f a}-\frac {\left (8 A c +6 d A +6 B c +4 d B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(324\) |
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Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (121) = 242\).
Time = 0.25 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.13 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {{\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 7 \, B\right )} d\right )} \cos \left (f x + e\right )^{3} - {\left (2 \, {\left (2 \, A + 3 \, B\right )} c + {\left (6 \, A - B\right )} d\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} c + 3 \, {\left (A - B\right )} d - 3 \, {\left ({\left (3 \, A + 2 \, B\right )} c + {\left (2 \, A + 3 \, B\right )} d\right )} \cos \left (f x + e\right ) - {\left ({\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 7 \, B\right )} d\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} c + 3 \, {\left (A - B\right )} d + 3 \, {\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 2 \, B\right )} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1819 vs. \(2 (121) = 242\).
Time = 4.37 (sec) , antiderivative size = 1819, normalized size of antiderivative = 14.32 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 733 vs. \(2 (121) = 242\).
Time = 0.23 (sec) , antiderivative size = 733, normalized size of antiderivative = 5.77 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \]
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Time = 0.32 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, A c + 3 \, B c + 3 \, A d + 2 \, B d\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \]
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Time = 14.02 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.93 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {53\,A\,c}{4}+3\,A\,d+3\,B\,c+\frac {13\,B\,d}{4}-4\,A\,c\,\cos \left (e+f\,x\right )+\frac {3\,A\,d\,\cos \left (e+f\,x\right )}{2}+\frac {3\,B\,c\,\cos \left (e+f\,x\right )}{2}+B\,d\,\cos \left (e+f\,x\right )+\frac {25\,A\,c\,\sin \left (e+f\,x\right )}{2}+\frac {15\,A\,d\,\sin \left (e+f\,x\right )}{2}+\frac {15\,B\,c\,\sin \left (e+f\,x\right )}{2}+\frac {5\,B\,d\,\sin \left (e+f\,x\right )}{2}-\frac {9\,A\,c\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {3\,A\,d\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {3\,B\,c\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {9\,B\,d\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,A\,c\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,B\,d\,\sin \left (2\,e+2\,f\,x\right )}{4}\right )}{15\,a^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}-\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \]
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