Optimal. Leaf size=135 \[ \frac{4 \sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^{m+2} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac{5}{2};-\frac{3}{2},-n;m+\frac{7}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a^2 f (2 m+5) \sqrt{1-\sin (e+f x)}} \]
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Rubi [A] time = 0.22617, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {2918, 140, 139, 138} \[ \frac{4 \sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^{m+2} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac{5}{2};-\frac{3}{2},-n;m+\frac{7}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a^2 f (2 m+5) \sqrt{1-\sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2918
Rule 140
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx &=\frac{\cos (e+f x) \operatorname{Subst}\left (\int (a-a x)^{3/2} (a+a x)^{\frac{3}{2}+m} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{a^2 f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{\left (2 \sqrt{2} \cos (e+f x)\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{3/2} (a+a x)^{\frac{3}{2}+m} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{a f \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{\left (2 \sqrt{2} \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{a (c+d \sin (e+f x))}{a c-a d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{3/2} (a+a x)^{\frac{3}{2}+m} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^n \, dx,x,\sin (e+f x)\right )}{a f \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{4 \sqrt{2} F_1\left (\frac{5}{2}+m;-\frac{3}{2},-n;\frac{7}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{2+m} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n}}{a^2 f (5+2 m) \sqrt{1-\sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.28988, size = 160, normalized size = 1.19 \[ -\frac{4 \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \cos ^3(e+f x) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{-m-\frac{3}{2}} (a (\sin (e+f x)+1))^m (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{5}{2};-m-\frac{3}{2},-n;\frac{7}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{5 f} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.495, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{4} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4}, 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(-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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