Integrand size = 31, antiderivative size = 129 \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {2^{\frac {5}{2}+m} a^2 (B m+A (5+m)) \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-\frac {3}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{-2+m}}{5 f (5+m)}-\frac {B \cos ^5(e+f x) (a+a \sin (e+f x))^m}{f (5+m)} \]
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Time = 0.13 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2939, 2768, 72, 71} \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a^2 2^{m+\frac {5}{2}} (A (m+5)+B m) \cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-m-\frac {3}{2},\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{5 f (m+5)}-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)^m}{f (m+5)} \]
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Rule 71
Rule 72
Rule 2768
Rule 2939
Rubi steps \begin{align*} \text {integral}& = -\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 ) \text {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 ) \text {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) \operatorname {Hypergeometric2F1}\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*}
Time = 0.34 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.86 \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {\cos ^5(e+f x) (1+\sin (e+f x))^{-\frac {5}{2}-m} (a (1+\sin (e+f x)))^m \left (2^{\frac {5}{2}+m} (B m+A (5+m)) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-\frac {3}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right )+5 B (1+\sin (e+f x))^{\frac {5}{2}+m}\right )}{5 f (5+m)} \]
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\[\int \left (\cos ^{4}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]
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\[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{4} \,d x } \]
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Timed out. \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \]
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\[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{4} \,d x } \]
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\[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{4} \,d x } \]
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Timed out. \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int {\cos \left (e+f\,x\right )}^4\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]
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