Integrand size = 29, antiderivative size = 105 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {\left (2 a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^3 b^2 d}+\frac {\sin (c+d x)}{b d} \]
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Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 12, 908} \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \csc (c+d x)}{a^2 d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^3 b^2 d}-\frac {\left (2 a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {\csc ^2(c+d x)}{2 a d}+\frac {\sin (c+d x)}{b d} \]
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Rule 12
Rule 908
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {b^3 \left (b^2-x^2\right )^2}{x^3 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{x^3 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^2 d} \\ & = \frac {\text {Subst}\left (\int \left (1+\frac {b^4}{a x^3}-\frac {b^4}{a^2 x^2}+\frac {-2 a^2 b^2+b^4}{a^3 x}-\frac {\left (a^2-b^2\right )^2}{a^3 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{b^2 d} \\ & = \frac {b \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {\left (2 a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^3 b^2 d}+\frac {\sin (c+d x)}{b d} \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 b \csc (c+d x)}{a^2}-\frac {\csc ^2(c+d x)}{a}+\frac {\frac {2 b^2 \left (-2 a^2+b^2\right ) \log (\sin (c+d x))-2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^3}+2 b \sin (c+d x)}{b^2}}{2 d} \]
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Time = 0.55 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )}{b}-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (-2 a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}+\frac {\left (-a^{4}+2 a^{2} b^{2}-b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{3} b^{2}}}{d}\) | \(99\) |
default | \(\frac {\frac {\sin \left (d x +c \right )}{b}-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (-2 a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}+\frac {\left (-a^{4}+2 a^{2} b^{2}-b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{3} b^{2}}}{d}\) | \(99\) |
parallelrisch | \(\frac {-8 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+8 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}+8 \left (-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b^{2} a^{2} \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-3\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a \,b^{3} \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-3 a^{2} b^{2} \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 a^{3} b \sin \left (d x +c \right )}{8 a^{3} b^{2} d}\) | \(182\) |
risch | \(\frac {i a x}{b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {2 i a c}{b^{2} d}+\frac {2 i \left (-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{3} d}\) | \(270\) |
norman | \(\frac {-\frac {1}{8 a d}-\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2} d}+\frac {b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} d}+\frac {\left (4 a^{2}+3 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} b d}+\frac {\left (4 a^{2}+3 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} b d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{3} b^{2} d}\) | \(285\) |
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Time = 0.41 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^{2} b^{2} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, {\left (2 \, a^{2} b^{2} - b^{4} - {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{3} b^{2} d \cos \left (d x + c\right )^{2} - a^{3} b^{2} d\right )}} \]
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Timed out. \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, \sin \left (d x + c\right )}{b} - \frac {2 \, {\left (2 \, a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{3} b^{2}} + \frac {2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]
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Time = 0.37 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, \sin \left (d x + c\right )}{b} - \frac {2 \, {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{3} b^{2}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} + 2 \, a b \sin \left (d x + c\right ) - a^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \]
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Time = 11.82 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.27 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}-\frac {\frac {a}{2}-2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a^2+b^2\right )}{b}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{a^3\,d}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{a^3\,b^2\,d} \]
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