Integrand size = 13, antiderivative size = 82 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx=\frac {\left (2 a^2+b^2\right ) x}{2 b^3}-\frac {2 a^3 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2}}+\frac {a \cos (x)}{b^2}-\frac {\cos (x) \sin (x)}{2 b} \]
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Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2872, 3102, 2814, 2739, 632, 210} \[ \int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx=\frac {x \left (2 a^2+b^2\right )}{2 b^3}-\frac {2 a^3 \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2}}+\frac {a \cos (x)}{b^2}-\frac {\sin (x) \cos (x)}{2 b} \]
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Rule 210
Rule 632
Rule 2739
Rule 2814
Rule 2872
Rule 3102
Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (x) \sin (x)}{2 b}+\frac {\int \frac {a+b \sin (x)-2 a \sin ^2(x)}{a+b \sin (x)} \, dx}{2 b} \\ & = \frac {a \cos (x)}{b^2}-\frac {\cos (x) \sin (x)}{2 b}+\frac {\int \frac {a b+\left (2 a^2+b^2\right ) \sin (x)}{a+b \sin (x)} \, dx}{2 b^2} \\ & = \frac {\left (2 a^2+b^2\right ) x}{2 b^3}+\frac {a \cos (x)}{b^2}-\frac {\cos (x) \sin (x)}{2 b}-\frac {a^3 \int \frac {1}{a+b \sin (x)} \, dx}{b^3} \\ & = \frac {\left (2 a^2+b^2\right ) x}{2 b^3}+\frac {a \cos (x)}{b^2}-\frac {\cos (x) \sin (x)}{2 b}-\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b^3} \\ & = \frac {\left (2 a^2+b^2\right ) x}{2 b^3}+\frac {a \cos (x)}{b^2}-\frac {\cos (x) \sin (x)}{2 b}+\frac {\left (4 a^3\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 {x}{2}\right )\right )}{b^3} \\ & = \frac {\left (2 a^2+b^2\right ) x}{2 b^3}-\frac {2 a^3 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2}}+\frac {a \cos (x)}{b^2}-\frac {\cos (x) \sin (x)}{2 b} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx=\frac {4 a^2 x+2 b^2 x-\frac {8 a^3 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+4 a b \cos (x)-b^2 \sin (2 x)}{4 b^3} \]
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Time = 0.47 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.37
method | result | size |
default | \(-\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3} \sqrt {a^{2}-b^{2}}}+\frac {\frac {2 \left (\frac {b^{2} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}+a b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{2}+a b \right )}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2}}+\left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{3}}\) | \(112\) |
risch | \(\frac {x \,a^{2}}{b^{3}}+\frac {x}{2 b}+\frac {a \,{\mathrm e}^{i x}}{2 b^{2}}+\frac {a \,{\mathrm e}^{-i x}}{2 b^{2}}+\frac {i a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, b^{3}}-\frac {i a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, b^{3}}-\frac {\sin \left (2 x \right )}{4 b}\) | \(179\) |
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Time = 0.31 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.55 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} a^{3} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} x - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )}}, \frac {2 \, \sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} x + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )}}\right ] \]
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Timed out. \[ \int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.37 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{3}}{\sqrt {a^{2} - b^{2}} b^{3}} + \frac {{\left (2 \, a^{2} + b^{2}\right )} x}{2 \, b^{3}} + \frac {b \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right ) + 2 \, a}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2} b^{2}} \]
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Time = 7.25 (sec) , antiderivative size = 1004, normalized size of antiderivative = 12.24 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)} \, dx=\frac {\frac {2\,a}{b^2}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{b}+\frac {2\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{b^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {\mathrm {atan}\left (\frac {40\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a\,b^2+40\,a^3+\frac {48\,a^5}{b^2}}+\frac {48\,a^5\,\mathrm {tan}\left (\frac {x}{2}\right )}{48\,a^5+40\,a^3\,b^2+8\,a\,b^4}+\frac {8\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a\,b+\frac {40\,a^3}{b}+\frac {48\,a^5}{b^3}}\right )\,\left (a^2\,2{}\mathrm {i}+b^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b^3}+\frac {a^3\,\mathrm {atan}\left (\frac {\frac {a^3\,\left (\frac {8\,\left (4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6\right )}{b^5}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-8\,a^7\,b^2+4\,a^5\,b^4+7\,a^3\,b^6+2\,a\,b^8\right )}{b^6}+\frac {a^3\,\left (64\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {8\,\left (2\,a^3\,b^6+2\,a\,b^8\right )}{b^5}+\frac {a^3\,\left (32\,a^2\,b^3+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a\,b^{10}-8\,a^3\,b^8\right )}{b^6}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}\right )\,1{}\mathrm {i}}{b^3\,\sqrt {b^2-a^2}}+\frac {a^3\,\left (\frac {8\,\left (4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6\right )}{b^5}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-8\,a^7\,b^2+4\,a^5\,b^4+7\,a^3\,b^6+2\,a\,b^8\right )}{b^6}-\frac {a^3\,\left (64\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {8\,\left (2\,a^3\,b^6+2\,a\,b^8\right )}{b^5}-\frac {a^3\,\left (32\,a^2\,b^3+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a\,b^{10}-8\,a^3\,b^8\right )}{b^6}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}\right )\,1{}\mathrm {i}}{b^3\,\sqrt {b^2-a^2}}}{\frac {16\,\left (2\,a^7+a^5\,b^2\right )}{b^5}+\frac {16\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^8+8\,a^6\,b^2+2\,a^4\,b^4\right )}{b^6}+\frac {a^3\,\left (\frac {8\,\left (4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6\right )}{b^5}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-8\,a^7\,b^2+4\,a^5\,b^4+7\,a^3\,b^6+2\,a\,b^8\right )}{b^6}+\frac {a^3\,\left (64\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {8\,\left (2\,a^3\,b^6+2\,a\,b^8\right )}{b^5}+\frac {a^3\,\left (32\,a^2\,b^3+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a\,b^{10}-8\,a^3\,b^8\right )}{b^6}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}-\frac {a^3\,\left (\frac {8\,\left (4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6\right )}{b^5}+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-8\,a^7\,b^2+4\,a^5\,b^4+7\,a^3\,b^6+2\,a\,b^8\right )}{b^6}-\frac {a^3\,\left (64\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {8\,\left (2\,a^3\,b^6+2\,a\,b^8\right )}{b^5}-\frac {a^3\,\left (32\,a^2\,b^3+\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a\,b^{10}-8\,a^3\,b^8\right )}{b^6}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}\right )}{b^3\,\sqrt {b^2-a^2}}}\right )\,2{}\mathrm {i}}{b^3\,\sqrt {b^2-a^2}} \]
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