Integrand size = 12, antiderivative size = 118 \[ \int \frac {1}{(3+b \sin (e+f x))^3} \, dx=\frac {\left (18+b^2\right ) \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{\left (9-b^2\right )^{5/2} f}+\frac {b \cos (e+f x)}{2 \left (9-b^2\right ) f (3+b \sin (e+f x))^2}+\frac {9 b \cos (e+f x)}{2 \left (9-b^2\right )^2 f (3+b \sin (e+f x))} \]
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Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2743, 2833, 12, 2739, 632, 210} \[ \int \frac {1}{(3+b \sin (e+f x))^3} \, dx=\frac {\left (2 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac {3 a b \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac {b \cos (e+f x)}{2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2} \]
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Rule 12
Rule 210
Rule 632
Rule 2739
Rule 2743
Rule 2833
Rubi steps \begin{align*} \text {integral}& = \frac {b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}-\frac {\int \frac {-2 a+b \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx}{2 \left (a^2-b^2\right )} \\ & = \frac {b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac {\int \frac {2 a^2+b^2}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2} \\ & = \frac {b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac {\left (2 a^2+b^2\right ) \int \frac {1}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2} \\ & = \frac {b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac {\left (2 a^2+b^2\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right )^2 f} \\ & = \frac {b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}-\frac {\left (2 \left (2 a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right )^2 f} \\ & = \frac {\left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} f}+\frac {b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(3+b \sin (e+f x))^3} \, dx=\frac {\frac {2 \left (18+b^2\right ) \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{\left (9-b^2\right )^{5/2}}+\frac {b \cos (e+f x) \left (36-b^2+9 b \sin (e+f x)\right )}{\left (-9+b^2\right )^2 (3+b \sin (e+f x))^2}}{2 f} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(281\) vs. \(2(122)=244\).
Time = 1.17 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.39
method | result | size |
derivativedivides | \(\frac {\frac {\frac {b^{2} \left (5 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (4 a^{4}+7 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}-2 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}+\frac {2 b \left (4 a^{2}-b^{2}\right )}{2 a^{4}-4 a^{2} b^{2}+2 b^{4}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a \right )}^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}}{f}\) | \(282\) |
default | \(\frac {\frac {\frac {b^{2} \left (5 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (4 a^{4}+7 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}-2 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}+\frac {2 b \left (4 a^{2}-b^{2}\right )}{2 a^{4}-4 a^{2} b^{2}+2 b^{4}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a \right )}^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}}{f}\) | \(282\) |
risch | \(-\frac {i \left (-2 i b \,a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-i {\mathrm e}^{3 i \left (f x +e \right )} b^{3}+10 i a^{2} b \,{\mathrm e}^{i \left (f x +e \right )}-i b^{3} {\mathrm e}^{i \left (f x +e \right )}+6 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+3 b^{2} a \,{\mathrm e}^{2 i \left (f x +e \right )}-3 a \,b^{2}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (f x +e \right )}+2 a \,{\mathrm e}^{i \left (f x +e \right )}+i b \right )^{2} \left (a^{2}-b^{2}\right )^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} f}\) | \(472\) |
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Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (122) = 244\).
Time = 0.31 (sec) , antiderivative size = 618, normalized size of antiderivative = 5.24 \[ \int \frac {1}{(3+b \sin (e+f x))^3} \, dx=\left [-\frac {6 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (f x + e\right )}{4 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} f \sin \left (f x + e\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} f\right )}}, -\frac {3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) + {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} f \sin \left (f x + e\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} f\right )}}\right ] \]
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Timed out. \[ \int \frac {1}{(3+b \sin (e+f x))^3} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {1}{(3+b \sin (e+f x))^3} \, dx=\text {Exception raised: ValueError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (122) = 244\).
Time = 0.27 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.31 \[ \int \frac {1}{(3+b \sin (e+f x))^3} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (2 \, a^{2} + b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a b^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, b^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a b^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a^{4} b - a^{2} b^{3}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}^{2}}}{f} \]
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Time = 10.60 (sec) , antiderivative size = 395, normalized size of antiderivative = 3.35 \[ \int \frac {1}{(3+b \sin (e+f x))^3} \, dx=\frac {\frac {4\,a^2\,b-b^3}{a^4-2\,a^2\,b^2+b^4}+\frac {b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (11\,a^2\,b-2\,b^3\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (4\,a^2\,b-b^3\right )\,\left (a^2+2\,b^2\right )}{a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (5\,a^2\,b-2\,b^3\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}+\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,a^2+b^2\right )\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,a^2+b^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{2\,a^2+b^2}\right )\,\left (2\,a^2+b^2\right )}{f\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
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