Integrand size = 26, antiderivative size = 75 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {\sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {4 \tan (e+f x)}{5 a^3 c^2 f}+\frac {4 \tan ^3(e+f x)}{15 a^3 c^2 f} \]
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Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2815, 2751, 3852} \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {4 \tan ^3(e+f x)}{15 a^3 c^2 f}+\frac {4 \tan (e+f x)}{5 a^3 c^2 f}-\frac {\sec ^3(e+f x)}{5 c^2 f \left (a^3 \sin (e+f x)+a^3\right )} \]
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Rule 2751
Rule 2815
Rule 3852
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sec ^4(e+f x)}{a+a \sin (e+f x)} \, dx}{a^2 c^2} \\ & = -\frac {\sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {4 \int \sec ^4(e+f x) \, dx}{5 a^3 c^2} \\ & = -\frac {\sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {4 \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{5 a^3 c^2 f} \\ & = -\frac {\sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {4 \tan (e+f x)}{5 a^3 c^2 f}+\frac {4 \tan ^3(e+f x)}{15 a^3 c^2 f} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(154\) vs. \(2(75)=150\).
Time = 1.16 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.05 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1050 \cos (e+f x)+256 \cos (2 (e+f x))+350 \cos (3 (e+f x))+128 \cos (4 (e+f x))-768 \sin (e+f x)+350 \sin (2 (e+f x))-256 \sin (3 (e+f x))+175 \sin (4 (e+f x)))}{1920 a^3 c^2 f (-1+\sin (e+f x))^2 (1+\sin (e+f x))^3} \]
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Result contains complex when optimal does not.
Time = 1.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {16 \left (6 \,{\mathrm e}^{3 i \left (f x +e \right )}+2 i {\mathrm e}^{2 i \left (f x +e \right )}+2 \,{\mathrm e}^{i \left (f x +e \right )}+i\right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2} a^{3}}\) | \(77\) |
parallelrisch | \(\frac {\frac {2}{5}+\frac {2 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\frac {10 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-2 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-2 \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {26 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15}-\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5}-\frac {14 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}}{f \,c^{2} a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(129\) |
derivativedivides | \(\frac {-\frac {1}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {5}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {5}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {3}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {11}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} c^{2} f}\) | \(133\) |
default | \(\frac {-\frac {1}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {5}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {5}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {3}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {11}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} c^{2} f}\) | \(133\) |
norman | \(\frac {\frac {10 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}+\frac {2}{5 a c f}-\frac {2 \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {2 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {14 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a c f}-\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 a c f}+\frac {2 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}-\frac {26 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 a c f}}{a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\) | \(198\) |
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Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {8 \, \cos \left (f x + e\right )^{4} - 4 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right ) - 1}{15 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1418 vs. \(2 (66) = 132\).
Time = 4.83 (sec) , antiderivative size = 1418, normalized size of antiderivative = 18.91 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (69) = 138\).
Time = 0.20 (sec) , antiderivative size = 335, normalized size of antiderivative = 4.47 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {13 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {25 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {15 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - 3\right )}}{15 \, {\left (a^{3} c^{2} + \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {6 \, a^{3} c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {6 \, a^{3} c^{2} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a^{3} c^{2} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )} f} \]
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Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {\frac {5 \, {\left (15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13\right )}}{a^{3} c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}} + \frac {165 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 480 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 650 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 400 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 113}{a^{3} c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{120 \, f} \]
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Time = 7.61 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.71 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {2\,\left (15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-25\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+13\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+9\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-3\right )}{15\,a^3\,c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^5} \]
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