Optimal. Leaf size=129 \[ -\frac {a^2 2^{m+\frac {5}{2}} (A (m+5)+B m) \cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac {5}{2},-m-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{5 f (m+5)}-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)^m}{f (m+5)} \]
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Rubi [A] time = 0.20, antiderivative size = 129, normalized size of antiderivative = 1.00, 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} \[ -\frac {a^2 2^{m+\frac {5}{2}} (A (m+5)+B m) \cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac {5}{2},-m-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{5 f (m+5)}-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)^m}{f (m+5)} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 2689
Rule 2860
Rubi steps
\begin {align*} \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=-\frac {B \cos ^5(e+f x) (a+a \sin (e+f x))^m}{f (5+m)}+\left (A+\frac {B m}{5+m}\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^m \, dx\\ &=-\frac {B \cos ^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 ) \cos ^5(e+f x)\right ) \operatorname {Subst}\left (\int (a-a x)^{3/2} (a+a x)^{\frac {3}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}\\ &=-\frac {B \cos ^5(e+f x) (a+a \sin (e+f x))^m}{f (5+m)}+\frac {\left (2^{\frac {3}{2}+m} a^3 \left (A+\frac {B m}{5+m}\right ) \cos ^5(e+f x) (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 \left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {3}{2}+m} (a-a x)^{3/2} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2}}\\ &=-\frac {2^{\frac {5}{2}+m} a^2 \left (A+\frac {B m}{5+m}\right ) \cos ^5(e+f x) \, _2F_1\left (\frac {5}{2},-\frac {3}{2}-m;\frac {7}{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))^{-2+m}}{5 f}-\frac {B \cos ^5(e+f x) (a+a \sin (e+f x))^m}{f (5+m)}\\ \end {align*}
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Mathematica [A] time = 0.49, size = 111, normalized size = 0.86 \[ -\frac {\cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac {5}{2}} (a (\sin (e+f x)+1))^m \left (2^{m+\frac {5}{2}} (A (m+5)+B m) \, _2F_1\left (\frac {5}{2},-m-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-\sin (e+f x))\right )+5 B (\sin (e+f x)+1)^{m+\frac {5}{2}}\right )}{5 f (m+5)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cos \left (f x + e\right )^{4} \sin \left (f x + e\right ) + A \cos \left (f x + e\right )^{4}\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 9.51, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{4}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (e+f\,x\right )}^4\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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