Optimal. Leaf size=72 \[ -\frac {a^4 (a \sin (e+f x))^{-4+m}}{f (4-m)}+\frac {2 a^2 (a \sin (e+f x))^{-2+m}}{f (2-m)}+\frac {(a \sin (e+f x))^m}{f m} \]
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Rubi [A]
time = 0.05, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2672, 276}
\begin {gather*} -\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} \end {gather*}
Antiderivative was successfully verified.
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Rule 276
Rule 2672
Rubi steps
\begin {align*} \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx &=\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*}
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Mathematica [A]
time = 0.35, size = 62, normalized size = 0.86 \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 0.66, size = 7703, normalized size = 106.99
method | result | size |
risch | \(\text {Expression too large to display}\) | \(7703\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.30, size = 76, normalized size = 1.06 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 117, normalized size = 1.62 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \sin {\left (e + f x \right )}\right )^{m} \cot ^{5}{\left (e + f x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 7.57, size = 219, normalized size = 3.04 \begin {gather*} -\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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