Integrand size = 10, antiderivative size = 14 \[ \int \frac {3}{5+4 \sin (x)} \, dx=x+2 \arctan \left (\frac {\cos (x)}{2+\sin (x)}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2736} \[ \int \frac {3}{5+4 \sin (x)} \, dx=2 \arctan \left (\frac {\cos (x)}{\sin (x)+2}\right )+x \]
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Rule 12
Rule 2736
Rubi steps \begin{align*} \text {integral}& = 3 \int \frac {1}{5+4 \sin (x)} \, dx \\ & = x+2 \arctan \left (\frac {\cos (x)}{2+\sin (x)}\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(14)=28\).
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 5.64 \[ \int \frac {3}{5+4 \sin (x)} \, dx=3 \left (-\frac {1}{3} \arctan \left (\frac {2 \cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )+2 \sin \left (\frac {x}{2}\right )}\right )+\frac {1}{3} \arctan \left (\frac {\cos \left (\frac {x}{2}\right )+2 \sin \left (\frac {x}{2}\right )}{2 \cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )}\right )\right ) \]
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Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86
method | result | size |
default | \(2 \arctan \left (\frac {5 \tan \left (\frac {x}{2}\right )}{3}+\frac {4}{3}\right )\) | \(12\) |
risch | \(i \ln \left ({\mathrm e}^{i x}+2 i\right )-i \ln \left ({\mathrm e}^{i x}+\frac {i}{2}\right )\) | \(26\) |
parallelrisch | \(-i \left (\ln \left (5 \tan \left (\frac {x}{2}\right )+4-3 i\right )-\ln \left (5 \tan \left (\frac {x}{2}\right )+4+3 i\right )\right )\) | \(29\) |
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Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {3}{5+4 \sin (x)} \, dx=\arctan \left (\frac {5 \, \sin \left (x\right ) + 4}{3 \, \cos \left (x\right )}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\).
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.93 \[ \int \frac {3}{5+4 \sin (x)} \, dx=2 \operatorname {atan}{\left (\frac {5 \tan {\left (\frac {x}{2} \right )}}{3} + \frac {4}{3} \right )} + 2 \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \]
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Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {3}{5+4 \sin (x)} \, dx=2 \, \arctan \left (\frac {5 \, \sin \left (x\right )}{3 \, {\left (\cos \left (x\right ) + 1\right )}} + \frac {4}{3}\right ) \]
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Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {3}{5+4 \sin (x)} \, dx=x + 2 \, \arctan \left (-\frac {2 \, \cos \left (x\right ) + \sin \left (x\right ) + 2}{\cos \left (x\right ) - 2 \, \sin \left (x\right ) - 4}\right ) \]
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Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {3}{5+4 \sin (x)} \, dx=x+2\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}+\frac {4}{3}\right )-2\,\mathrm {atan}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right ) \]
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