Optimal. Leaf size=201 \[ \frac{a 2^{\frac{1}{2}-\frac{m}{2}} (e \cos (c+d x))^{-m-1} \left (\frac{(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac{m+1}{2}} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (\frac{1}{2} (-m-1),\frac{m+1}{2};\frac{1-m}{2};\frac{(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e (m+1) \left (a^2-b^2\right )}-\frac{(e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^{m+1}}{d e (m+1) (a-b)} \]
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Rubi [A] time = 0.292773, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2699, 2920, 132} \[ \frac{a 2^{\frac{1}{2}-\frac{m}{2}} (e \cos (c+d x))^{-m-1} \left (\frac{(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac{m+1}{2}} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (\frac{1}{2} (-m-1),\frac{m+1}{2};\frac{1-m}{2};\frac{(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e (m+1) \left (a^2-b^2\right )}-\frac{(e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^{m+1}}{d e (m+1) (a-b)} \]
Antiderivative was successfully verified.
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Rule 2699
Rule 2920
Rule 132
Rubi steps
\begin{align*} \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx &=-\frac{(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac{a \int \frac{(e \cos (c+d x))^{-m} (a+b \sin (c+d x))^m}{1-\sin (c+d x)} \, dx}{(a-b) e^2}\\ &=-\frac{(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac{\left (a (e \cos (c+d x))^{-1-m} (1-\sin (c+d x))^{\frac{1+m}{2}} (1+\sin (c+d x))^{\frac{1+m}{2}}\right ) \operatorname{Subst}\left (\int (1-x)^{-1+\frac{1}{2} (-1-m)} (1+x)^{\frac{1}{2} (-1-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) d e}\\ &=-\frac{(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac{2^{\frac{1}{2}-\frac{m}{2}} a (e \cos (c+d x))^{-1-m} \, _2F_1\left (\frac{1}{2} (-1-m),\frac{1+m}{2};\frac{1-m}{2};\frac{(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac{(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac{1+m}{2}} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m)}\\ \end{align*}
Mathematica [A] time = 0.940747, size = 168, normalized size = 0.84 \[ -\frac{2^{\frac{1}{2} (-m-1)} (e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^{m+1} \left (2^{\frac{m+1}{2}} (a+b)-2 a \left (\frac{(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac{m+1}{2}} \, _2F_1\left (\frac{1}{2} (-m-1),\frac{m+1}{2};\frac{1-m}{2};-\frac{(a-b) (\sin (c+d x)-1)}{2 (a+b \sin (c+d x))}\right )\right )}{d e (m+1) (a-b) (a+b)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.181, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{-2-m} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{-m - 2}{\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (e \cos \left (d x + c\right )\right )^{-m - 2}{\left (b \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{-m - 2}{\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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