Integrand size = 36, antiderivative size = 102 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=\frac {(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {2 (3 A-2 B) \tan (e+f x)}{15 a c^3 f} \]
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Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3046, 2938, 2751, 3852, 8} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=\frac {2 (3 A-2 B) \tan (e+f x)}{15 a c^3 f}+\frac {(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2} \]
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Rule 8
Rule 2751
Rule 2938
Rule 3046
Rule 3852
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sec ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx}{a c} \\ & = \frac {(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {(3 A-2 B) \int \frac {\sec ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{5 a c^2} \\ & = \frac {(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {(2 (3 A-2 B)) \int \sec ^2(e+f x) \, dx}{15 a c^3} \\ & = \frac {(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {(2 (3 A-2 B)) \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{15 a c^3 f} \\ & = \frac {(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {2 (3 A-2 B) \tan (e+f x)}{15 a c^3 f} \\ \end{align*}
Time = 1.64 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.54 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=-\frac {\cos (e+f x) (80 B+5 (-9 A+B) \cos (e+f x)+32 (3 A-2 B) \cos (2 (e+f x))+9 A \cos (3 (e+f x))-B \cos (3 (e+f x))+120 A \sin (e+f x)-80 B \sin (e+f x)+36 A \sin (2 (e+f x))-4 B \sin (2 (e+f x))-24 A \sin (3 (e+f x))+16 B \sin (3 (e+f x)))}{240 a c^3 f (-1+\sin (e+f x))^3 (1+\sin (e+f x))} \]
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Result contains complex when optimal does not.
Time = 1.00 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {4 \left (-3 i A +12 A \,{\mathrm e}^{i \left (f x +e \right )}+10 B \,{\mathrm e}^{3 i \left (f x +e \right )}+2 i B -8 B \,{\mathrm e}^{i \left (f x +e \right )}-10 i B \,{\mathrm e}^{2 i \left (f x +e \right )}+15 i A \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) f \,c^{3} a}\) | \(111\) |
parallelrisch | \(\frac {-30 A \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (60 A -30 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-60 A +40 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-40 B \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (18 A +8 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-12 A -2 B}{15 f \,c^{3} a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(128\) |
derivativedivides | \(\frac {-\frac {2 \left (2 A +2 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {\frac {5 A}{2}+\frac {3 B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {7 A}{8}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {2 \left (\frac {9 A}{2}+\frac {7 B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {2 \left (\frac {A}{8}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a \,c^{3} f}\) | \(145\) |
default | \(\frac {-\frac {2 \left (2 A +2 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {\frac {5 A}{2}+\frac {3 B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {7 A}{8}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {2 \left (\frac {9 A}{2}+\frac {7 B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {2 \left (\frac {A}{8}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a \,c^{3} f}\) | \(145\) |
norman | \(\frac {-\frac {12 A +2 B}{15 a f c}+\frac {2 \left (6 A -7 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a f c}-\frac {2 A \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f c}-\frac {2 \left (-8 B +7 A \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f c}+\frac {2 \left (2 A -B \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f c}-\frac {2 \left (9 A -4 B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a f c}-\frac {2 \left (2 A +7 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f c}+\frac {2 \left (9 A +4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{15 a f c}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(260\) |
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Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.05 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=-\frac {4 \, {\left (3 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - {\left (2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - 9 \, A + 6 \, B\right )} \sin \left (f x + e\right ) - 6 \, A + 9 \, B}{15 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1236 vs. \(2 (85) = 170\).
Time = 4.55 (sec) , antiderivative size = 1236, normalized size of antiderivative = 12.12 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (98) = 196\).
Time = 0.23 (sec) , antiderivative size = 423, normalized size of antiderivative = 4.15 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (\frac {B {\left (\frac {4 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )}}{a c^{3} - \frac {4 \, a c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {4 \, a c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}} + \frac {3 \, A {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - 2\right )}}{a c^{3} - \frac {4 \, a c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {4 \, a c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}\right )}}{15 \, f} \]
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Time = 0.32 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.64 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=-\frac {\frac {15 \, {\left (A - B\right )}}{a c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {105 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 15 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 270 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 360 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 40 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 210 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 50 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 63 \, A - 7 \, B}{a c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{60 \, f} \]
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Time = 12.24 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.75 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=\frac {2\,\left (\frac {5\,B\,\sin \left (e+f\,x\right )}{2}-\frac {15\,A\,\cos \left (e+f\,x\right )}{4}-\frac {5\,B\,\cos \left (e+f\,x\right )}{8}-\frac {15\,A\,\sin \left (e+f\,x\right )}{4}-\frac {5\,B}{2}-3\,A\,\cos \left (2\,e+2\,f\,x\right )+\frac {3\,A\,\cos \left (3\,e+3\,f\,x\right )}{4}+2\,B\,\cos \left (2\,e+2\,f\,x\right )+\frac {B\,\cos \left (3\,e+3\,f\,x\right )}{8}+3\,A\,\sin \left (2\,e+2\,f\,x\right )+\frac {3\,A\,\sin \left (3\,e+3\,f\,x\right )}{4}+\frac {B\,\sin \left (2\,e+2\,f\,x\right )}{2}-\frac {B\,\sin \left (3\,e+3\,f\,x\right )}{2}\right )}{15\,a\,c^3\,f\,\left (\frac {\cos \left (3\,e+3\,f\,x\right )}{4}-\frac {5\,\cos \left (e+f\,x\right )}{4}+\sin \left (2\,e+2\,f\,x\right )\right )} \]
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