Integrand size = 26, antiderivative size = 34 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx=\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \]
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Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2815, 2750} \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx=\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \]
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Rule 2750
Rule 2815
Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^5} \, dx \\ & = \frac {a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(34)=68\).
Time = 2.40 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.38 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-10 \sin \left (\frac {1}{2} (e+f x)\right )-5 \sin \left (\frac {3}{2} (e+f x)\right )+\sin \left (\frac {5}{2} (e+f x)\right )\right )}{10 c^3 f (-1+\sin (e+f x))^3} \]
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Time = 0.66 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56
method | result | size |
parallelrisch | \(-\frac {2 a^{2} \left (5 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+10 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+1\right )}{5 f \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(53\) |
risch | \(\frac {2 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-4 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 a^{2}}{5}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3}}\) | \(55\) |
derivativedivides | \(\frac {2 a^{2} \left (-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {16}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\right )}{f \,c^{3}}\) | \(88\) |
default | \(\frac {2 a^{2} \left (-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {16}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\right )}{f \,c^{3}}\) | \(88\) |
norman | \(\frac {-\frac {2 a^{2}}{5 c f}-\frac {24 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}-\frac {52 a^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {2 a^{2} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(133\) |
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Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (33) = 66\).
Time = 0.25 (sec) , antiderivative size = 168, normalized size of antiderivative = 4.94 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx=\frac {a^{2} \cos \left (f x + e\right )^{3} + 3 \, a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} + {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2}\right )} \sin \left (f x + e\right )}{5 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (29) = 58\).
Time = 4.30 (sec) , antiderivative size = 354, normalized size of antiderivative = 10.41 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx=\begin {cases} - \frac {10 a^{2} \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 c^{3} f} - \frac {20 a^{2} \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 c^{3} f} - \frac {2 a^{2}}{5 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 c^{3} f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )^{2}}{\left (- c \sin {\left (e \right )} + c\right )^{3}} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (33) = 66\).
Time = 0.22 (sec) , antiderivative size = 557, normalized size of antiderivative = 16.38 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (\frac {a^{2} {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \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}} - 7\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {6 \, a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {2 \, a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \]
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Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.68 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (5 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 10 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a^{2}\right )}}{5 \, c^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}} \]
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Time = 6.44 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.71 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx=\frac {2\,a^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{5\,c^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \]
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