Optimal. Leaf size=148 \[ \frac{B \sec ^3(e+f x) (a \sin (e+f x)+a)^{m+2}}{a^2 c^2 f (1-m)}+\frac{2^{m+\frac{1}{2}} (A (1-m)-B (m+2)) \sec ^3(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{1}{2}-m;-\frac{1}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (1-m)} \]
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Rubi [A] time = 0.330138, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2967, 2860, 2689, 70, 69} \[ \frac{B \sec ^3(e+f x) (a \sin (e+f x)+a)^{m+2}}{a^2 c^2 f (1-m)}+\frac{2^{m+\frac{1}{2}} (A (1-m)-B (m+2)) \sec ^3(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{1}{2}-m;-\frac{1}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (1-m)} \]
Antiderivative was successfully verified.
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Rule 2967
Rule 2860
Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx &=\frac{\int \sec ^4(e+f x) (a+a \sin (e+f x))^{2+m} (A+B \sin (e+f x)) \, dx}{a^2 c^2}\\ &=\frac{B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}+\frac{\left (A-\frac{B (2+m)}{1-m}\right ) \int \sec ^4(e+f x) (a+a \sin (e+f x))^{2+m} \, dx}{a^2 c^2}\\ &=\frac{B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}+\frac{\left (\left (A-\frac{B (2+m)}{1-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{c^2 f}\\ &=\frac{B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}+\frac{\left (2^{-\frac{1}{2}+m} \left (A-\frac{B (2+m)}{1-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{1+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{\frac{1}{2}-m}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{x}{2}\right )^{-\frac{1}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{c^2 f}\\ &=\frac{2^{\frac{1}{2}+m} \left (A-\frac{B (2+m)}{1-m}\right ) \, _2F_1\left (-\frac{3}{2},\frac{1}{2}-m;-\frac{1}{2};\frac{1}{2} (1-\sin (e+f x))\right ) \sec ^3(e+f x) (1+\sin (e+f x))^{\frac{1}{2}-m} (a+a \sin (e+f x))^{1+m}}{3 a c^2 f}+\frac{B \sec ^3(e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f (1-m)}\\ \end{align*}
Mathematica [C] time = 23.3283, size = 8371, normalized size = 56.56 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.754, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) }{ \left ( c-c\sin \left ( fx+e \right ) \right ) ^{2}}}\, 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 \frac{{\left (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 )}^{2}}\,{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 (-\frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, 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 \frac{{\left (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 )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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