Optimal. Leaf size=171 \[ -\frac{2^{m+\frac{1}{2}} \left (A \left (m^2+3 m+2\right )+C \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac{C \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (m^2+3 m+2\right )}-\frac{C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]
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Rubi [A] time = 0.19253, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3024, 2751, 2652, 2651} \[ -\frac{2^{m+\frac{1}{2}} \left (A \left (m^2+3 m+2\right )+C \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac{C \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (m^2+3 m+2\right )}-\frac{C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]
Antiderivative was successfully verified.
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Rule 3024
Rule 2751
Rule 2652
Rule 2651
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx &=-\frac{C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\int (a+a \sin (e+f x))^m (a (C (1+m)+A (2+m))-a C \sin (e+f x)) \, dx}{a (2+m)}\\ &=\frac{C \cos (e+f x) (a+a \sin (e+f x))^m}{f \left (2+3 m+m^2\right )}-\frac{C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\left (C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=\frac{C \cos (e+f x) (a+a \sin (e+f x))^m}{f \left (2+3 m+m^2\right )}-\frac{C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\left (\left (C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=\frac{C \cos (e+f x) (a+a \sin (e+f x))^m}{f \left (2+3 m+m^2\right )}-\frac{2^{\frac{1}{2}+m} \left (C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}\\ \end{align*}
Mathematica [C] time = 2.52776, size = 385, normalized size = 2.25 \[ -\frac{\sin ^{-2 m}\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m \left (\frac{4 \sqrt{2} A \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \cos ^{2 m+1}\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{1}{2},m+\frac{1}{2};m+\frac{3}{2};\sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )\right )}{(2 m+1) \sqrt{1-\sin (e+f x)}}+\frac{C 2^{-2 m-1} e^{-3 i (e+f x)} \left (1-i e^{i (e+f x)}\right ) \left (-(-1)^{3/4} e^{-\frac{1}{2} i (e+f x)} \left (e^{i (e+f x)}+i\right )\right )^{2 m} \left ((m-2) e^{4 i (e+f x)} \, _2F_1\left (1,m-1;-m-1;-i e^{-i (e+f x)}\right )+(m+2) \, _2F_1\left (1,m+3;3-m;-i e^{-i (e+f x)}\right )\right )}{m^2-4}+\frac{2 \sqrt{2} C \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \cos ^{2 m+1}\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{1}{2},m+\frac{1}{2};m+\frac{3}{2};\sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )\right )}{(2 m+1) \sqrt{1-\sin (e+f x)}}\right )}{2 f} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 1.514, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, 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 (C \sin \left (f x + e\right )^{2} + A\right )}{\left (a \sin \left (f x + e\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 (C \cos \left (f x + e\right )^{2} - A - C\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \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 (C \sin \left (f x + e\right )^{2} + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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