Integrand size = 19, antiderivative size = 72 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=-\frac {a^4 (a \sin (e+f x))^{-4+m}}{f (4-m)}+\frac {2 a^2 (a \sin (e+f x))^{-2+m}}{f (2-m)}+\frac {(a \sin (e+f x))^m}{f m} \]
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Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2672, 276} \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=-\frac {a^4 (a \sin (e+f x))^{m-4}}{f (4-m)}+\frac {2 a^2 (a \sin (e+f x))^{m-2}}{f (2-m)}+\frac {(a \sin (e+f x))^m}{f m} \]
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Rule 276
Rule 2672
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^{-5+m} \left (a^2-x^2\right )^2 \, dx,x,a \sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (a^4 x^{-5+m}-2 a^2 x^{-3+m}+x^{-1+m}\right ) \, dx,x,a \sin (e+f x)\right )}{f} \\ & = -\frac {a^4 (a \sin (e+f x))^{-4+m}}{f (4-m)}+\frac {2 a^2 (a \sin (e+f x))^{-2+m}}{f (2-m)}+\frac {(a \sin (e+f x))^m}{f m} \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\frac {\left (8-6 m+m^2-2 (-4+m) m \csc ^2(e+f x)+(-2+m) m \csc ^4(e+f x)\right ) (a \sin (e+f x))^m}{f (-4+m) (-2+m) m} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.02 (sec) , antiderivative size = 6931, normalized size of antiderivative = 96.26
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Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\frac {{\left ({\left (m^{2} - 6 \, m + 8\right )} \cos \left (f x + e\right )^{4} + 4 \, {\left (m - 4\right )} \cos \left (f x + e\right )^{2} + 8\right )} \left (a \sin \left (f x + e\right )\right )^{m}}{{\left (f m^{3} - 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{4} + f m^{3} - 6 \, f m^{2} - 2 \, {\left (f m^{3} - 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{2} + 8 \, f m} \]
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\[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\int \left (a \sin {\left (e + f x \right )}\right )^{m} \cot ^{5}{\left (e + f x \right )}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\frac {\frac {a^{m} \sin \left (f x + e\right )^{m}}{m} - \frac {2 \, a^{m} \sin \left (f x + e\right )^{m}}{{\left (m - 2\right )} \sin \left (f x + e\right )^{2}} + \frac {a^{m} \sin \left (f x + e\right )^{m}}{{\left (m - 4\right )} \sin \left (f x + e\right )^{4}}}{f} \]
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\[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{m} \cot \left (f x + e\right )^{5} \,d x } \]
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Time = 8.69 (sec) , antiderivative size = 219, normalized size of antiderivative = 3.04 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=-\frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^m\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-1\right )\,\left (-\frac {2\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )}{f\,m}+\frac {\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,\left (6\,m^2-4\,m+48\right )}{f\,m\,\left (m^2-6\,m+8\right )}+\frac {2\,\left (2\,{\sin \left (e+f\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-4\,m^2+8\,m+32\right )}{f\,m\,\left (m^2-6\,m+8\right )}\right )}{16\,{\sin \left (e+f\,x\right )}^4} \]
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