Optimal. Leaf size=142 \[ -\frac {(e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m}{d e (2-m)}+\frac {2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{1+m}}{a d e (2-m) m}-\frac {2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e m \left (4-m^2\right )} \]
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Rubi [A]
time = 0.15, 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 = {2751, 2750}
\begin {gather*} -\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} \end {gather*}
Antiderivative was successfully verified.
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Rule 2750
Rule 2751
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 = 0.14, size = 76, normalized size = 0.54 \begin {gather*} \frac {(e \cos (c+d x))^{-m} \sec ^2(c+d x) (a (1+\sin (c+d x)))^m \left (-2+m^2-2 m \sin (c+d x)+2 \sin ^2(c+d x)\right )}{d e^3 (-2+m) m (2+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.22, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 76, normalized size = 0.54 \begin {gather*} \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 (\cos \left (d x + c\right ) e\right )^{-m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} - 4 \, d m} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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 = 6.13, size = 103, normalized size = 0.73 \begin {gather*} -\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 )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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