Optimal. Leaf size=100 \[ \frac{a^2 (A+B) (a \sin (e+f x)+a)^{m-1}}{2 f (a-a \sin (e+f x))}-\frac{a (A (2-m)-B m) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac{1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)} \]
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Rubi [A] time = 0.132801, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2836, 78, 68} \[ \frac{a^2 (A+B) (a \sin (e+f x)+a)^{m-1}}{2 f (a-a \sin (e+f x))}-\frac{a (A (2-m)-B m) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac{1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)} \]
Antiderivative was successfully verified.
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Rule 2836
Rule 78
Rule 68
Rubi steps
\begin{align*} \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=\frac{a^3 \operatorname{Subst}\left (\int \frac{(a+x)^{-2+m} \left (A+\frac{B x}{a}\right )}{(a-x)^2} \, dx,x,a \sin (e+f x)\right )}{f}\\ &=\frac{a^2 (A+B) (a+a \sin (e+f x))^{-1+m}}{2 f (a-a \sin (e+f x))}+\frac{\left (a^2 (A (2-m)-B m)\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-2+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{2 f}\\ &=-\frac{a (A (2-m)-B m) \, _2F_1\left (1,-1+m;m;\frac{1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-1+m}}{4 f (1-m)}+\frac{a^2 (A+B) (a+a \sin (e+f x))^{-1+m}}{2 f (a-a \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 0.163781, size = 82, normalized size = 0.82 \[ -\frac{a (a (\sin (e+f x)+1))^{m-1} \left ((A (m-2)+B m) (\sin (e+f x)-1) \, _2F_1\left (1,m-1;m;\frac{1}{2} (\sin (e+f x)+1)\right )+2 (m-1) (A+B)\right )}{4 f (m-1) (\sin (e+f x)-1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.283, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{3} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \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 (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}\,{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 (B \sec \left (f x + e\right )^{3} \sin \left (f x + e\right ) + A \sec \left (f x + e\right )^{3}\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(-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 (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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