Integrand size = 18, antiderivative size = 48 \[ \int \frac {1}{a+b \cos (c+d x) \sin (c+d x)} \, dx=\frac {2 \arctan \left (\frac {b+2 a \tan (c+d x)}{\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2} d} \]
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Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2745, 2739, 632, 210} \[ \int \frac {1}{a+b \cos (c+d x) \sin (c+d x)} \, dx=\frac {2 \arctan \left (\frac {2 a \tan (c+d x)+b}{\sqrt {4 a^2-b^2}}\right )}{d \sqrt {4 a^2-b^2}} \]
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Rule 210
Rule 632
Rule 2739
Rule 2745
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{a+\frac {1}{2} b \sin (2 c+2 d x)} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {1}{a+b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (2 c+2 d x)\right )\right )}{d} \\ & = -\frac {2 \text {Subst}\left (\int \frac {1}{-4 a^2+b^2-x^2} \, dx,x,b+2 a \tan \left (\frac {1}{2} (2 c+2 d x)\right )\right )}{d} \\ & = \frac {2 \arctan \left (\frac {b+2 a \tan (c+d x)}{\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2} d} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a+b \cos (c+d x) \sin (c+d x)} \, dx=\frac {2 \arctan \left (\frac {b+2 a \tan (c+d x)}{\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2} d} \]
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Time = 0.83 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {2 \arctan \left (\frac {b +2 a \tan \left (d x +c \right )}{\sqrt {4 a^{2}-b^{2}}}\right )}{d \sqrt {4 a^{2}-b^{2}}}\) | \(45\) |
default | \(\frac {2 \arctan \left (\frac {b +2 a \tan \left (d x +c \right )}{\sqrt {4 a^{2}-b^{2}}}\right )}{d \sqrt {4 a^{2}-b^{2}}}\) | \(45\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \sqrt {-4 a^{2}+b^{2}}-4 a^{2}+b^{2}}{b \sqrt {-4 a^{2}+b^{2}}}\right )}{\sqrt {-4 a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \sqrt {-4 a^{2}+b^{2}}+4 a^{2}-b^{2}}{b \sqrt {-4 a^{2}+b^{2}}}\right )}{\sqrt {-4 a^{2}+b^{2}}\, d}\) | \(135\) |
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none
Time = 0.28 (sec) , antiderivative size = 290, normalized size of antiderivative = 6.04 \[ \int \frac {1}{a+b \cos (c+d x) \sin (c+d x)} \, dx=\left [-\frac {\sqrt {-4 \, a^{2} + b^{2}} \log \left (-\frac {2 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - b^{2} + {\left (2 \, b \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - b\right )} \sqrt {-4 \, a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a^{2}}\right )}{2 \, {\left (4 \, a^{2} - b^{2}\right )} d}, -\frac {\arctan \left (-\frac {{\left (4 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b\right )} \sqrt {4 \, a^{2} - b^{2}}}{2 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, a^{2} + b^{2}}\right )}{\sqrt {4 \, a^{2} - b^{2}} d}\right ] \]
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Timed out. \[ \int \frac {1}{a+b \cos (c+d x) \sin (c+d x)} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {1}{a+b \cos (c+d x) \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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none
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.27 \[ \int \frac {1}{a+b \cos (c+d x) \sin (c+d x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {2 \, a \tan \left (d x + c\right ) + b}{\sqrt {4 \, a^{2} - b^{2}}}\right )\right )}}{\sqrt {4 \, a^{2} - b^{2}} d} \]
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Time = 27.35 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92 \[ \int \frac {1}{a+b \cos (c+d x) \sin (c+d x)} \, dx=\frac {2\,\mathrm {atan}\left (\frac {b+2\,a\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {4\,a^2-b^2}}\right )}{d\,\sqrt {4\,a^2-b^2}} \]
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