Integrand size = 8, antiderivative size = 40 \[ \int \frac {1}{a+b \sin (x)} \, dx=\frac {2 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]
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Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2739, 632, 210} \[ \int \frac {1}{a+b \sin (x)} \, dx=\frac {2 \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]
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Rule 210
Rule 632
Rule 2739
Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right ) \\ & = -\left (4 \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 )\right ) \\ & = \frac {2 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a+b \sin (x)} \, dx=\frac {2 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]
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Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\) | \(39\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}}\) | \(119\) |
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Time = 0.31 (sec) , antiderivative size = 148, normalized size of antiderivative = 3.70 \[ \int \frac {1}{a+b \sin (x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right )}{2 \, {\left (a^{2} - b^{2}\right )}}, -\frac {\arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right )}{\sqrt {a^{2} - b^{2}}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (31) = 62\).
Time = 1.72 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.50 \[ \int \frac {1}{a+b \sin (x)} \, dx=\begin {cases} \tilde {\infty } \log {\left (\tan {\left (\frac {x}{2} \right )} \right )} & \text {for}\: a = 0 \wedge b = 0 \\\frac {\log {\left (\tan {\left (\frac {x}{2} \right )} \right )}}{b} & \text {for}\: a = 0 \\\frac {2}{b \tan {\left (\frac {x}{2} \right )} - b} & \text {for}\: a = - b \\- \frac {2}{b \tan {\left (\frac {x}{2} \right )} + b} & \text {for}\: a = b \\\frac {\log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{\sqrt {- a^{2} + b^{2}}} - \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{\sqrt {- a^{2} + b^{2}}} & \text {otherwise} \end {cases} \]
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Exception generated. \[ \int \frac {1}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \frac {1}{a+b \sin (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}}} \]
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Time = 6.45 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.12 \[ \int \frac {1}{a+b \sin (x)} \, dx=\frac {2\,\mathrm {atan}\left (\frac {b}{\sqrt {a^2-b^2}}+\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]
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