Integrand size = 36, antiderivative size = 102 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {2 (3 A+2 B) \tan (e+f x)}{15 a^3 c f} \]
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Time = 0.17 (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} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {2 (3 A+2 B) \tan (e+f x)}{15 a^3 c f}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(A-B) \sec (e+f x)}{5 a c f (a \sin (e+f x)+a)^2} \]
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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))}{(a+a \sin (e+f x))^2} \, dx}{a c} \\ & = -\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}+\frac {(3 A+2 B) \int \frac {\sec ^2(e+f x)}{a+a \sin (e+f x)} \, dx}{5 a^2 c} \\ & = -\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {(2 (3 A+2 B)) \int \sec ^2(e+f x) \, dx}{15 a^3 c} \\ & = -\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(2 (3 A+2 B)) \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{15 a^3 c f} \\ & = -\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {2 (3 A+2 B) \tan (e+f x)}{15 a^3 c f} \\ \end{align*}
Time = 1.68 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.53 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {\cos (e+f x) (-80 B-5 (9 A+B) \cos (e+f x)+32 (3 A+2 B) \cos (2 (e+f x))+9 A \cos (3 (e+f x))+B \cos (3 (e+f x))-120 A \sin (e+f x)-80 B \sin (e+f x)-36 A \sin (2 (e+f x))-4 B \sin (2 (e+f x))+24 A \sin (3 (e+f x))+16 B \sin (3 (e+f x)))}{240 a^3 c f (-1+\sin (e+f x)) (1+\sin (e+f x))^3} \]
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Result contains complex when optimal does not.
Time = 0.93 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {4 \left (-3 i A -12 A \,{\mathrm e}^{i \left (f x +e \right )}+10 B \,{\mathrm e}^{3 i \left (f x +e \right )}-2 i B -8 B \,{\mathrm e}^{i \left (f x +e \right )}+10 i B \,{\mathrm e}^{2 i \left (f x +e \right )}+15 i A \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a^{3} c f}\) | \(111\) |
parallelrisch | \(\frac {-30 A \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-60 A -30 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-60 A -40 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-40 B \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (18 A -8 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+12 A -2 B}{15 f c \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(128\) |
derivativedivides | \(\frac {-\frac {2 \left (\frac {A}{8}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (-2 B +2 A \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-\frac {5 A}{2}+\frac {3 B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {7 A}{8}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {9 A}{2}-\frac {7 B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{a^{3} c f}\) | \(145\) |
default | \(\frac {-\frac {2 \left (\frac {A}{8}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (-2 B +2 A \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-\frac {5 A}{2}+\frac {3 B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {7 A}{8}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {9 A}{2}-\frac {7 B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{a^{3} c f}\) | \(145\) |
norman | \(\frac {\frac {12 A -2 B}{15 c f a}-\frac {2 \left (6 A +7 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f a}-\frac {2 A \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f a}+\frac {2 \left (9 A -4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{15 c f a}+\frac {2 \left (2 A -7 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f a}-\frac {2 \left (9 A +4 B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f a}-\frac {2 \left (2 A +B \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f a}-\frac {2 \left (8 B +7 A \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\) | \(258\) |
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Time = 0.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {4 \, {\left (3 \, A + 2 \, B\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, {\left (3 \, A + 2 \, B\right )} \cos \left (f x + e\right )^{2} - 9 \, A - 6 \, B\right )} \sin \left (f x + e\right ) - 6 \, A - 9 \, B}{15 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1236 vs. \(2 (85) = 170\).
Time = 4.54 (sec) , antiderivative size = 1236, normalized size of antiderivative = 12.12 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (96) = 192\).
Time = 0.21 (sec) , antiderivative size = 423, normalized size of antiderivative = 4.15 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {2 \, {\left (\frac {B {\left (\frac {4 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 1\right )}}{a^{3} c + \frac {4 \, a^{3} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a^{3} c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a^{3} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a^{3} c \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}} - \frac {3 \, A {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + 2\right )}}{a^{3} c + \frac {4 \, a^{3} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a^{3} c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a^{3} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a^{3} c \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}\right )}}{15 \, f} \]
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Time = 0.33 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {\frac {15 \, {\left (A + B\right )}}{a^{3} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {105 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 270 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 360 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 40 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 210 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 50 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 63 \, A + 7 \, B}{a^{3} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{60 \, f} \]
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Time = 12.98 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.75 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {2\,\left (\frac {15\,A\,\cos \left (e+f\,x\right )}{4}-\frac {5\,B}{2}-\frac {5\,B\,\cos \left (e+f\,x\right )}{8}-\frac {15\,A\,\sin \left (e+f\,x\right )}{4}-\frac {5\,B\,\sin \left (e+f\,x\right )}{2}+3\,A\,\cos \left (2\,e+2\,f\,x\right )-\frac {3\,A\,\cos \left (3\,e+3\,f\,x\right )}{4}+2\,B\,\cos \left (2\,e+2\,f\,x\right )+\frac {B\,\cos \left (3\,e+3\,f\,x\right )}{8}+3\,A\,\sin \left (2\,e+2\,f\,x\right )+\frac {3\,A\,\sin \left (3\,e+3\,f\,x\right )}{4}-\frac {B\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {B\,\sin \left (3\,e+3\,f\,x\right )}{2}\right )}{15\,a^3\,c\,f\,\left (\frac {5\,\cos \left (e+f\,x\right )}{4}-\frac {\cos \left (3\,e+3\,f\,x\right )}{4}+\sin \left (2\,e+2\,f\,x\right )\right )} \]
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