Integrand size = 29, antiderivative size = 148 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {2 a^2 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 d}+\frac {\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d} \]
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Time = 0.32 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2968, 3129, 3128, 3102, 2814, 2739, 632, 210} \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 a^2 \sqrt {a^2-b^2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d}+\frac {a x \left (2 a^2-b^2\right )}{2 b^4}+\frac {\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac {a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac {\sin ^2(c+d x) \cos (c+d x)}{3 b d} \]
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Rule 210
Rule 632
Rule 2739
Rule 2814
Rule 2968
Rule 3102
Rule 3128
Rule 3129
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx \\ & = \frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d}+\frac {\int \frac {\sin (c+d x) \left (-2 a+b \sin (c+d x)+3 a \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 b} \\ & = -\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d}+\frac {\int \frac {3 a^2-a b \sin (c+d x)-2 \left (3 a^2-b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{6 b^2} \\ & = \frac {\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d}+\frac {\int \frac {3 a^2 b+3 a \left (2 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{6 b^3} \\ & = \frac {a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d}-\frac {\left (a^2 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^4} \\ & = \frac {a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d}-\frac {\left (2 a^2 \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d} \\ & = \frac {a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d}+\frac {\left (4 a^2 \left (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} (c+d x)\right )\right )}{b^4 d} \\ & = \frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {2 a^2 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 d}+\frac {\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{3 b d} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {-12 a^3 c+6 a b^2 c-12 a^3 d x+6 a b^2 d x+24 a^2 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+3 b \left (-4 a^2+b^2\right ) \cos (c+d x)+b^3 \cos (3 (c+d x))+3 a b^2 \sin (2 (c+d x))}{12 b^4 d} \]
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Time = 0.46 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.24
method | result | size |
derivativedivides | \(\frac {-\frac {2 a^{2} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{4}}+\frac {\frac {2 \left (\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}}{2}+\left (a^{2} b -b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a \,b^{2}}{2}+a^{2} b -\frac {b^{3}}{3}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+a \left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\) | \(184\) |
default | \(\frac {-\frac {2 a^{2} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{4}}+\frac {\frac {2 \left (\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}}{2}+\left (a^{2} b -b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a \,b^{2}}{2}+a^{2} b -\frac {b^{3}}{3}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+a \left (2 a^{2}-b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\) | \(184\) |
risch | \(\frac {a^{3} x}{b^{4}}-\frac {a x}{2 b^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{8 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{8 b d}-\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}+\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}-\frac {\cos \left (3 d x +3 c \right )}{12 b d}-\frac {a \sin \left (2 d x +2 c \right )}{4 b^{2} d}\) | \(232\) |
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Time = 0.40 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.13 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} b \cos \left (d x + c\right ) - 3 \, \sqrt {-a^{2} + b^{2}} a^{2} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x}{6 \, b^{4} d}, -\frac {2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} b \cos \left (d x + c\right ) - 6 \, \sqrt {a^{2} - b^{2}} a^{2} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x}{6 \, b^{4} d}\right ] \]
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Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.43 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {3 \, {\left (2 \, a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {12 \, {\left (a^{4} - a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} - 2 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \]
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Time = 12.09 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.52 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2\,\cos \left (c+d\,x\right )}{b^3\,d}-\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^2\,d}-\frac {\frac {\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{12}}{b\,d}+\frac {2\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}+\frac {2\,a^2\,\mathrm {atanh}\left (\frac {-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{\sqrt {b^2-a^2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )\,\sqrt {b^2-a^2}}{b^4\,d} \]
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