Optimal. Leaf size=149 \[ -\frac{2^{\frac{p}{2}-\frac{3}{2}} (\sin (e+f x)+1)^{\frac{3-p}{2}-2} (g \cos (e+f x))^{p+1} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{p+1}{2};\frac{5-p}{2},-n;\frac{p+3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{a^2 f g (p+1)} \]
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Rubi [A] time = 0.229998, antiderivative size = 153, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {2920, 139, 138} \[ -\frac{g 2^{\frac{p-3}{2}} (1-\sin (e+f x)) (\sin (e+f x)+1)^{\frac{1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{p+1}{2};\frac{5-p}{2},-n;\frac{p+3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{a^2 f (p+1)} \]
Antiderivative was successfully verified.
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Rule 2920
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \frac{(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx &=\frac{\left (g (g \cos (e+f x))^{-1+p} (1-\sin (e+f x))^{\frac{1-p}{2}} (1+\sin (e+f x))^{\frac{1-p}{2}}\right ) \operatorname{Subst}\left (\int (1-x)^{\frac{1}{2} (-1+p)} (1+x)^{-2+\frac{1}{2} (-1+p)} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{a^2 f}\\ &=\frac{\left (g (g \cos (e+f x))^{-1+p} (1-\sin (e+f x))^{\frac{1-p}{2}} (1+\sin (e+f x))^{\frac{1-p}{2}} (c+d \sin (e+f x))^n \left (-\frac{c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \operatorname{Subst}\left (\int (1-x)^{\frac{1}{2} (-1+p)} (1+x)^{-2+\frac{1}{2} (-1+p)} \left (-\frac{c}{-c-d}-\frac{d x}{-c-d}\right )^n \, dx,x,\sin (e+f x)\right )}{a^2 f}\\ &=-\frac{2^{\frac{1}{2} (-3+p)} g F_1\left (\frac{1+p}{2};\frac{5-p}{2},-n;\frac{3+p}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right ) (g \cos (e+f x))^{-1+p} (1-\sin (e+f x)) (1+\sin (e+f x))^{\frac{1-p}{2}} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n}}{a^2 f (1+p)}\\ \end{align*}
Mathematica [F] time = 10.3247, size = 0, normalized size = 0. \[ \int \frac{(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.53, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( g\cos \left ( fx+e \right ) \right ) ^{p} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}}{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{2}}}\, 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 \frac{\left (g \cos \left (f x + e\right )\right )^{p}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{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 (-\frac{\left (g \cos \left (f x + e\right )\right )^{p}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, 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 \frac{\left (g \cos \left (f x + e\right )\right )^{p}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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