Optimal. Leaf size=142 \[ -\frac {2 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-2}}{a^2 d e m \left (4-m^2\right )}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-2}}{d e (2-m)}+\frac {2 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-2}}{a d e (2-m) m} \]
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Rubi [A] time = 0.22, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ -\frac {2 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-2}}{a^2 d e m \left (4-m^2\right )}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-2}}{d e (2-m)}+\frac {2 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-2}}{a d e (2-m) m} \]
Antiderivative was successfully verified.
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Rule 2671
Rule 2672
Rubi steps
\begin {align*} \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m \, dx &=-\frac {(e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m}{d e (2-m)}+\frac {2 \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m} \, dx}{a (2-m)}\\ &=-\frac {(e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m}{d e (2-m)}+\frac {2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{1+m}}{a d e (2-m) m}-\frac {2 \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{2+m} \, dx}{a^2 (2-m) m}\\ &=-\frac {(e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m}{d e (2-m)}+\frac {2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{1+m}}{a d e (2-m) m}-\frac {2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e m \left (4-m^2\right )}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 76, normalized size = 0.54 \[ \frac {\sec ^2(c+d x) \left (-2 m \sin (c+d x)+2 \sin ^2(c+d x)+m^2-2\right ) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m}}{d e^3 (m-2) m (m+2)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 75, normalized size = 0.53 \[ \frac {{\left (m^{2} \cos \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{3} - 2 \, m \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} - 4 \, d m} \]
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 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.31, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{-3-m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.13, size = 103, normalized size = 0.73 \[ -\frac {2\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (-2\,\cos \left (c+d\,x\right )\,m^2+2\,\sin \left (2\,c+2\,d\,x\right )\,m+3\,\cos \left (c+d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\right )}{d\,e^3\,m\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\left (m^2-4\right )\,\left (3\,\cos \left (c+d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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