Optimal. Leaf size=135 \[ \frac{2^{m-\frac{5}{2}} (A (5-m)-B m) \sec ^5(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m+2} \, _2F_1\left (-\frac{5}{2},\frac{7}{2}-m;-\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{5 a^2 f (5-m)}+\frac{B \sec ^5(e+f x) (a \sin (e+f x)+a)^m}{f (5-m)} \]
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Rubi [A] time = 0.193075, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2860, 2689, 70, 69} \[ \frac{2^{m-\frac{5}{2}} (A (5-m)-B m) \sec ^5(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m+2} \, _2F_1\left (-\frac{5}{2},\frac{7}{2}-m;-\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{5 a^2 f (5-m)}+\frac{B \sec ^5(e+f x) (a \sin (e+f x)+a)^m}{f (5-m)} \]
Antiderivative was successfully verified.
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Rule 2860
Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \sec ^6(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=\frac{B \sec ^5(e+f x) (a+a \sin (e+f x))^m}{f (5-m)}+\left (A-\frac{B m}{5-m}\right ) \int \sec ^6(e+f x) (a+a \sin (e+f x))^m \, dx\\ &=\frac{B \sec ^5(e+f x) (a+a \sin (e+f x))^m}{f (5-m)}+\frac{\left (a^2 \left (A-\frac{B m}{5-m}\right ) \sec ^5(e+f x) (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{7}{2}+m}}{(a-a x)^{7/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{B \sec ^5(e+f x) (a+a \sin (e+f x))^m}{f (5-m)}+\frac{\left (2^{-\frac{7}{2}+m} \left (A-\frac{B m}{5-m}\right ) \sec ^5(e+f x) (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{2+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{7}{2}+m}}{(a-a x)^{7/2}} \, dx,x,\sin (e+f x)\right )}{a f}\\ &=\frac{B \sec ^5(e+f x) (a+a \sin (e+f x))^m}{f (5-m)}+\frac{2^{-\frac{5}{2}+m} \left (A-\frac{B m}{5-m}\right ) \, _2F_1\left (-\frac{5}{2},\frac{7}{2}-m;-\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right ) \sec ^5(e+f x) (1+\sin (e+f x))^{\frac{1}{2}-m} (a+a \sin (e+f x))^{2+m}}{5 a^2 f}\\ \end{align*}
Mathematica [F] time = 3.50869, size = 0, normalized size = 0. \[ \int \sec ^6(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.404, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{6} \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 )^{6}\,{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 )^{6} \sin \left (f x + e\right ) + A \sec \left (f x + e\right )^{6}\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 )^{6}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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