Integrand size = 14, antiderivative size = 35 \[ \int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx=\frac {b x}{a^2+b^2}-\frac {a \log (a \cos (x)+b \sin (x))}{a^2+b^2} \]
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Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3176, 3212} \[ \int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx=\frac {b x}{a^2+b^2}-\frac {a \log (a \cos (x)+b \sin (x))}{a^2+b^2} \]
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Rule 3176
Rule 3212
Rubi steps \begin{align*} \text {integral}& = \frac {b x}{a^2+b^2}-\frac {a \int \frac {b \cos (x)-a \sin (x)}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2} \\ & = \frac {b x}{a^2+b^2}-\frac {a \log (a \cos (x)+b \sin (x))}{a^2+b^2} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34 \[ \int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx=\frac {2 (-i a+b) x+2 i a \arctan (\tan (x))-a \log \left ((a \cos (x)+b \sin (x))^2\right )}{2 \left (a^2+b^2\right )} \]
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Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.31
method | result | size |
parallelrisch | \(\frac {a \ln \left (\frac {1}{\cos \left (x \right )+1}\right )-a \ln \left (\frac {-a \cos \left (x \right )-b \sin \left (x \right )}{\cos \left (x \right )+1}\right )+x b}{a^{2}+b^{2}}\) | \(46\) |
default | \(-\frac {a \ln \left (a +b \tan \left (x \right )\right )}{a^{2}+b^{2}}+\frac {\frac {a \ln \left (1+\tan \left (x \right )^{2}\right )}{2}+b \arctan \left (\tan \left (x \right )\right )}{a^{2}+b^{2}}\) | \(47\) |
risch | \(\frac {i x}{i b -a}+\frac {2 i a x}{a^{2}+b^{2}}-\frac {a \ln \left ({\mathrm e}^{2 i x}-\frac {i b +a}{i b -a}\right )}{a^{2}+b^{2}}\) | \(67\) |
norman | \(\frac {\frac {b x}{a^{2}+b^{2}}+\frac {b x \tan \left (\frac {x}{2}\right )^{2}}{a^{2}+b^{2}}}{1+\tan \left (\frac {x}{2}\right )^{2}}+\frac {a \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )}{a^{2}+b^{2}}-\frac {a \ln \left (\tan \left (\frac {x}{2}\right )^{2} a -2 b \tan \left (\frac {x}{2}\right )-a \right )}{a^{2}+b^{2}}\) | \(96\) |
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Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.31 \[ \int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx=\frac {2 \, b x - a \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \]
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Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 165, normalized size of antiderivative = 4.71 \[ \int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx=\begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \\- \frac {\log {\left (\cos {\left (x \right )} \right )}}{a} & \text {for}\: b = 0 \\\frac {i x \sin {\left (x \right )}}{2 i b \sin {\left (x \right )} + 2 b \cos {\left (x \right )}} + \frac {x \cos {\left (x \right )}}{2 i b \sin {\left (x \right )} + 2 b \cos {\left (x \right )}} - \frac {\sin {\left (x \right )}}{2 i b \sin {\left (x \right )} + 2 b \cos {\left (x \right )}} & \text {for}\: a = - i b \\- \frac {i x \sin {\left (x \right )}}{- 2 i b \sin {\left (x \right )} + 2 b \cos {\left (x \right )}} + \frac {x \cos {\left (x \right )}}{- 2 i b \sin {\left (x \right )} + 2 b \cos {\left (x \right )}} - \frac {\sin {\left (x \right )}}{- 2 i b \sin {\left (x \right )} + 2 b \cos {\left (x \right )}} & \text {for}\: a = i b \\- \frac {a \log {\left (\frac {a \cos {\left (x \right )}}{b} + \sin {\left (x \right )} \right )}}{a^{2} + b^{2}} + \frac {b x}{a^{2} + b^{2}} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (35) = 70\).
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.51 \[ \int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx=\frac {2 \, b \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{2} + b^{2}} - \frac {a \log \left (-a - \frac {2 \, b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}{a^{2} + b^{2}} + \frac {a \log \left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}{a^{2} + b^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {a b \log \left ({\left | b \tan \left (x\right ) + a \right |}\right )}{a^{2} b + b^{3}} + \frac {b x}{a^{2} + b^{2}} + \frac {a \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \]
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Time = 23.05 (sec) , antiderivative size = 970, normalized size of antiderivative = 27.71 \[ \int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {a\,\ln \left (a\,\cos \left (x\right )+b\,\sin \left (x\right )\right )}{a^2+b^2}-\frac {2\,b\,\mathrm {atan}\left (\frac {\left (a^4+2\,a^2\,b^2+b^4\right )\,\left (\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {\left (4\,a^4-13\,a^2\,b^2+b^4\right )\,\left (\frac {b\,\left (64\,a\,b^2+\frac {a\,\left (32\,a^2\,b^2-64\,a^4+\frac {a\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{a^2+b^2}\right )}{a^2+b^2}\right )}{a^2+b^2}-\frac {b^3\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^3}+\frac {a\,\left (\frac {b\,\left (32\,a^2\,b^2-64\,a^4+\frac {a\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {a\,b\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^2}\right )}{a^2+b^2}\right )}{{\left (4\,a^4+5\,a^2\,b^2+b^4\right )}^2}-\frac {6\,a\,b\,\left (2\,a^2-b^2\right )\,\left (64\,a^2+\frac {a\,\left (64\,a\,b^2+\frac {a\,\left (32\,a^2\,b^2-64\,a^4+\frac {a\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{a^2+b^2}\right )}{a^2+b^2}\right )}{a^2+b^2}-\frac {b\,\left (\frac {b\,\left (32\,a^2\,b^2-64\,a^4+\frac {a\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {a\,b\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^2}\right )}{a^2+b^2}-\frac {a\,b^2\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^3}\right )}{{\left (4\,a^4+5\,a^2\,b^2+b^4\right )}^2}\right )-\frac {\left (4\,a^4-13\,a^2\,b^2+b^4\right )\,\left (\frac {b\,\left (32\,a^2\,b-\frac {a\,\left (64\,a^3\,b-32\,a\,b^3+\frac {a\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{a^2+b^2}\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {b^3\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^3}-\frac {a\,\left (\frac {b\,\left (64\,a^3\,b-32\,a\,b^3+\frac {a\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {a\,b\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^2}\right )}{a^2+b^2}\right )}{{\left (4\,a^4+5\,a^2\,b^2+b^4\right )}^2}+\frac {6\,a\,b\,\left (2\,a^2-b^2\right )\,\left (\frac {a\,\left (32\,a^2\,b-\frac {a\,\left (64\,a^3\,b-32\,a\,b^3+\frac {a\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{a^2+b^2}\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {b\,\left (\frac {b\,\left (64\,a^3\,b-32\,a\,b^3+\frac {a\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {a\,b\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^2}\right )}{a^2+b^2}+\frac {a\,b^2\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^3}\right )}{{\left (4\,a^4+5\,a^2\,b^2+b^4\right )}^2}\right )}{32\,a\,b}\right )}{a^2+b^2} \]
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