Integrand size = 25, antiderivative size = 178 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=-\frac {2 d^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{27 (c-d)^3 \sqrt {c^2-d^2} f}-\frac {\cos (e+f x)}{5 (c-d) f (3+3 \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{45 (c-d)^2 f (3+3 \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f (27+27 \sin (e+f x))} \]
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Time = 0.34 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2845, 3057, 12, 2739, 632, 210} \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=-\frac {2 d^3 \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^3 f (c-d)^3 \sqrt {c^2-d^2}}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(2 c-7 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3} \]
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Rule 12
Rule 210
Rule 632
Rule 2739
Rule 2845
Rule 3057
Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {\int \frac {-a (2 c-5 d)-2 a d \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx}{5 a^2 (c-d)} \\ & = -\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}+\frac {\int \frac {a^2 \left (2 c^2-7 c d+15 d^2\right )+a^2 (2 c-7 d) d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx}{15 a^4 (c-d)^2} \\ & = -\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\int \frac {15 a^3 d^3}{c+d \sin (e+f x)} \, dx}{15 a^6 (c-d)^3} \\ & = -\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {d^3 \int \frac {1}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^3} \\ & = -\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\left (2 d^3\right ) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{a^3 (c-d)^3 f} \\ & = -\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\left (4 d^3\right ) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{a^3 (c-d)^3 f} \\ & = -\frac {2 d^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^3 (c-d)^3 \sqrt {c^2-d^2} f}-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )} \\ \end{align*}
Time = 2.53 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (2 c-7 d) (c-d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(c-d) (-2 c+7 d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+2 \left (2 c^2-9 c d+22 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-\frac {30 d^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{\sqrt {c^2-d^2}}\right )}{405 (c-d)^3 f (1+\sin (e+f x))^3} \]
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Time = 1.35 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.12
method | result | size |
derivativedivides | \(\frac {-\frac {-4 c +6 d}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 c -10 d \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (c^{2}-3 c d +3 d^{2}\right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{5 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 d^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}}{a^{3} f}\) | \(200\) |
default | \(\frac {-\frac {-4 c +6 d}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 c -10 d \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (c^{2}-3 c d +3 d^{2}\right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{5 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 d^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}}{a^{3} f}\) | \(200\) |
risch | \(-\frac {2 \left (2 c^{2}+22 d^{2}-9 c d -15 i c d \,{\mathrm e}^{3 i \left (f x +e \right )}+15 d^{2} {\mathrm e}^{4 i \left (f x +e \right )}-20 c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-145 d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+45 i d \,{\mathrm e}^{i \left (f x +e \right )} c +75 c d \,{\mathrm e}^{2 i \left (f x +e \right )}+75 i d^{2} {\mathrm e}^{3 i \left (f x +e \right )}-10 i c^{2} {\mathrm e}^{i \left (f x +e \right )}-95 i d^{2} {\mathrm e}^{i \left (f x +e \right )}\right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left (c -d \right )^{3} f \,a^{3}}-\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{3} f \,a^{3}}+\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{3} f \,a^{3}}\) | \(331\) |
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Leaf count of result is larger than twice the leaf count of optimal. 830 vs. \(2 (175) = 350\).
Time = 0.33 (sec) , antiderivative size = 1744, normalized size of antiderivative = 9.80 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\text {Exception raised: ValueError} \]
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none
Time = 0.38 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.96 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=-\frac {2 \, {\left (\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} d^{3}}{{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} \sqrt {c^{2} - d^{2}}} + \frac {15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 45 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 45 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 105 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 135 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 135 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 185 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 75 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 115 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{2} - 24 \, c d + 32 \, d^{2}}{{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}\right )}}{15 \, f} \]
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Time = 10.21 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.62 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\frac {2\,d^3\,\mathrm {atan}\left (\frac {\frac {d^3\,\left (-2\,a^3\,c^3\,d+6\,a^3\,c^2\,d^2-6\,a^3\,c\,d^3+2\,a^3\,d^4\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {2\,c\,d^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^3\,c^3-3\,a^3\,c^2\,d+3\,a^3\,c\,d^2-a^3\,d^3\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}}{2\,d^3}\right )}{a^3\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {\frac {2\,\left (7\,c^2-24\,c\,d+32\,d^2\right )}{15\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,c^2-15\,c\,d+23\,d^2\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (c^2-3\,c\,d+3\,d^2\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,c^2-7\,c\,d+9\,d^2\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (8\,c^2-27\,c\,d+37\,d^2\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \]
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