Integrand size = 27, antiderivative size = 60 \[ \int \frac {\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {b \log (\sin (c+d x))}{a^2 d}-\frac {\left (1-\frac {b^2}{a^2}\right ) \log (a+b \sin (c+d x))}{b d} \]
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Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2916, 12, 908} \[ \int \frac {\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (1-\frac {b^2}{a^2}\right ) \log (a+b \sin (c+d x))}{b d}-\frac {b \log (\sin (c+d x))}{a^2 d}-\frac {\csc (c+d x)}{a d} \]
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Rule 12
Rule 908
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {b^2 \left (b^2-x^2\right )}{x^2 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = \frac {\text {Subst}\left (\int \frac {b^2-x^2}{x^2 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {b^2}{a x^2}-\frac {b^2}{a^2 x}+\frac {-a^2+b^2}{a^2 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{b d} \\ & = -\frac {\csc (c+d x)}{a d}-\frac {b \log (\sin (c+d x))}{a^2 d}-\frac {\left (1-\frac {b^2}{a^2}\right ) \log (a+b \sin (c+d x))}{b d} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-a b \csc (c+d x)-b^2 \log (\sin (c+d x))+\left (-a^2+b^2\right ) \log (a+b \sin (c+d x))}{a^2 b d} \]
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Time = 0.34 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(\frac {\frac {\left (-a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{2} b}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}}{d}\) | \(59\) |
default | \(\frac {\frac {\left (-a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{2} b}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}}{d}\) | \(59\) |
parallelrisch | \(\frac {2 a^{2} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+2 \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) b^{2}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}-\csc \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a b}{2 a^{2} b d}\) | \(131\) |
risch | \(\frac {i x}{b}+\frac {2 i c}{b d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b d}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{2} d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(143\) |
norman | \(\frac {-\frac {1}{2 a d}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{2} b d}\) | \(164\) |
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Time = 0.38 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15 \[ \int \frac {\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + a b}{a^{2} b d \sin \left (d x + c\right )} \]
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Timed out. \[ \int \frac {\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int \frac {\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {b \log \left (\sin \left (d x + c\right )\right )}{a^{2}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b} + \frac {1}{a \sin \left (d x + c\right )}}{d} \]
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Time = 0.33 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b} + \frac {1}{a \sin \left (d x + c\right )}}{d} \]
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Time = 11.61 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.97 \[ \int \frac {\cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (\frac {b}{a^2}-\frac {1}{b}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b\,d}-\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d} \]
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