Optimal. Leaf size=269 \[ \frac{c 2^{n+\frac{1}{2}} \cos (e+f x) ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) (1-\sin (e+f x))^{\frac{1}{2}-n} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} \, _2F_1\left (\frac{1}{2} (2 m+1),\frac{1}{2} (1-2 n);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) (m+n+1) (m+n+2)}-\frac{(B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1) (m+n+2)}+\frac{C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)} \]
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Rubi [A] time = 0.745312, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3039, 2973, 2745, 2689, 70, 69} \[ \frac{c 2^{n+\frac{1}{2}} \cos (e+f x) ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) (1-\sin (e+f x))^{\frac{1}{2}-n} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} \, _2F_1\left (\frac{1}{2} (2 m+1),\frac{1}{2} (1-2 n);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) (m+n+1) (m+n+2)}-\frac{(B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1) (m+n+2)}+\frac{C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)} \]
Antiderivative was successfully verified.
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Rule 3039
Rule 2973
Rule 2745
Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}-\frac{\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n (-a c (C (1-m+n)+A (2+m+n))-a c (C+2 C m+B (2+m+n)) \sin (e+f x)) \, dx}{a c (2+m+n)}\\ &=-\frac{(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}+\frac{((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx}{(1+m+n) (2+m+n)}\\ &=-\frac{(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}+\frac{\left (((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{-m+n} \, dx}{(1+m+n) (2+m+n)}\\ &=-\frac{(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}+\frac{\left (c^2 ((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac{1}{2} (-1-2 m)+m} (c+c \sin (e+f x))^{\frac{1}{2} (-1-2 m)}\right ) \operatorname{Subst}\left (\int (c-c x)^{-m+\frac{1}{2} (-1+2 m)+n} (c+c x)^{\frac{1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f (1+m+n) (2+m+n)}\\ &=-\frac{(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}+\frac{\left (2^{-\frac{1}{2}+n} c^2 ((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac{1}{2}+\frac{1}{2} (-1-2 m)+m+n} \left (\frac{c-c \sin (e+f x)}{c}\right )^{\frac{1}{2}-n} (c+c \sin (e+f x))^{\frac{1}{2} (-1-2 m)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{-m+\frac{1}{2} (-1+2 m)+n} (c+c x)^{\frac{1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f (1+m+n) (2+m+n)}\\ &=\frac{2^{\frac{1}{2}+n} c ((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \cos (e+f x) \, _2F_1\left (\frac{1}{2} (1+2 m),\frac{1}{2} (1-2 n);\frac{1}{2} (3+2 m);\frac{1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac{1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m) (1+m+n) (2+m+n)}-\frac{(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}\\ \end{align*}
Mathematica [C] time = 16.3394, size = 6226, normalized size = 23.14 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 3.537, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{n} \left ( A+B\sin \left ( fx+e \right ) +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} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{n}\,{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} - B \sin \left (f x + e\right ) - A - C\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{n}, 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 (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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