Integrand size = 21, antiderivative size = 148 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b \left (2 a^2-b^2\right ) \csc (c+d x)}{a^4 d}+\frac {\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^3 d}+\frac {b \csc ^3(c+d x)}{3 a^2 d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^5 d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^5 d} \]
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Time = 0.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2800, 908} \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \csc ^3(c+d x)}{3 a^2 d}+\frac {\left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^5 d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^5 d}-\frac {b \left (2 a^2-b^2\right ) \csc (c+d x)}{a^4 d}+\frac {\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^3 d}-\frac {\csc ^4(c+d x)}{4 a d} \]
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Rule 908
Rule 2800
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{x^5 (a+x)} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {b^4}{a x^5}-\frac {b^4}{a^2 x^4}+\frac {-2 a^2 b^2+b^4}{a^3 x^3}+\frac {2 a^2 b^2-b^4}{a^4 x^2}+\frac {\left (a^2-b^2\right )^2}{a^5 x}-\frac {\left (a^2-b^2\right )^2}{a^5 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {b \left (2 a^2-b^2\right ) \csc (c+d x)}{a^4 d}+\frac {\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^3 d}+\frac {b \csc ^3(c+d x)}{3 a^2 d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^5 d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^5 d} \\ \end{align*}
Time = 0.63 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.78 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {12 a b \left (-2 a^2+b^2\right ) \csc (c+d x)+6 a^2 \left (2 a^2-b^2\right ) \csc ^2(c+d x)+4 a^3 b \csc ^3(c+d x)-3 a^4 \csc ^4(c+d x)+12 \left (a^2-b^2\right )^2 (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))}{12 a^5 d} \]
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Time = 0.61 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {1}{4 a \sin \left (d x +c \right )^{4}}-\frac {-2 a^{2}+b^{2}}{2 a^{3} \sin \left (d x +c \right )^{2}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{5}}-\frac {\left (2 a^{2}-b^{2}\right ) b}{a^{4} \sin \left (d x +c \right )}+\frac {b}{3 a^{2} \sin \left (d x +c \right )^{3}}-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{5}}}{d}\) | \(137\) |
default | \(\frac {-\frac {1}{4 a \sin \left (d x +c \right )^{4}}-\frac {-2 a^{2}+b^{2}}{2 a^{3} \sin \left (d x +c \right )^{2}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{5}}-\frac {\left (2 a^{2}-b^{2}\right ) b}{a^{4} \sin \left (d x +c \right )}+\frac {b}{3 a^{2} \sin \left (d x +c \right )^{3}}-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{5}}}{d}\) | \(137\) |
parallelrisch | \(\frac {-192 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+192 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \left (a^{3} \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 a^{2} b \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\left (-12 a^{3}+8 a \,b^{2}\right ) \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (56 a^{2} b -32 b^{3}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b}{3}+\left (-12 a^{3}+8 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+56 a^{2} b -32 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{192 a^{5} d}\) | \(233\) |
norman | \(\frac {-\frac {1}{64 a d}-\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 a^{2} d}+\frac {b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a^{2} d}+\frac {\left (3 a^{2}-2 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a^{3} d}+\frac {\left (3 a^{2}-2 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a^{3} d}-\frac {b \left (7 a^{2}-4 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a^{4} d}-\frac {b \left (7 a^{2}-4 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a^{4} d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{5} d}-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{5} d}\) | \(285\) |
risch | \(\frac {2 i \left (6 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-6 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+6 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+14 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+6 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-14 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-3 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d}+\frac {2 b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) b^{4}}{a^{5} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{4}}{a^{5} d}\) | \(412\) |
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Time = 0.42 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.83 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {9 \, a^{4} - 6 \, a^{2} b^{2} - 6 \, {\left (2 \, a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (5 \, a^{3} b - 3 \, a b^{3} - 3 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{5} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d\right )}} \]
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Timed out. \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} - \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}} - \frac {4 \, a^{2} b \sin \left (d x + c\right ) - 12 \, {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 3 \, a^{3} + 6 \, {\left (2 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{4} \sin \left (d x + c\right )^{4}}}{12 \, d} \]
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Time = 0.39 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.36 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5}} - \frac {12 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b} - \frac {25 \, a^{4} \sin \left (d x + c\right )^{4} - 50 \, a^{2} b^{2} \sin \left (d x + c\right )^{4} + 25 \, b^{4} \sin \left (d x + c\right )^{4} + 24 \, a^{3} b \sin \left (d x + c\right )^{3} - 12 \, a b^{3} \sin \left (d x + c\right )^{3} - 12 \, a^{4} \sin \left (d x + c\right )^{2} + 6 \, a^{2} b^{2} \sin \left (d x + c\right )^{2} - 4 \, a^{3} b \sin \left (d x + c\right ) + 3 \, a^{4}}{a^{5} \sin \left (d x + c\right )^{4}}}{12 \, d} \]
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Time = 11.62 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.90 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3}{16\,a}-\frac {b^2}{8\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b}{8\,a^2}+\frac {2\,b\,\left (\frac {3}{8\,a}-\frac {b^2}{4\,a^3}\right )}{a}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a\,b^2-3\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (14\,a^2\,b-8\,b^3\right )+\frac {a^3}{4}-\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}}{16\,a^4\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{a^5\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{a^5\,d} \]
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