Optimal. Leaf size=508 \[ -\frac{b g (g \cos (e+f x))^{p-1} \left (-\frac{b (1-\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{b (\sin (e+f x)+1)}{a+b \sin (e+f x)}\right )^{\frac{1-p}{2}} F_1\left (1-p;\frac{1-p}{2},\frac{1-p}{2};2-p;\frac{a+b}{a+b \sin (e+f x)},\frac{a-b}{a+b \sin (e+f x)}\right )}{f (1-p) (b c-a d)^2}+\frac{b g (g \cos (e+f x))^{p-1} \left (-\frac{d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{d (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} F_1\left (1-p;\frac{1-p}{2},\frac{1-p}{2};2-p;\frac{c+d}{c+d \sin (e+f x)},\frac{c-d}{c+d \sin (e+f x)}\right )}{f (1-p) (b c-a d)^2}+\frac{g (g \cos (e+f x))^{p-1} \left (-\frac{d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{d (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} F_1\left (2-p;\frac{1-p}{2},\frac{1-p}{2};3-p;\frac{c+d}{c+d \sin (e+f x)},\frac{c-d}{c+d \sin (e+f x)}\right )}{f (2-p) (b c-a d) (c+d \sin (e+f x))} \]
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Rubi [A] time = 0.524493, antiderivative size = 508, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {2924, 2703} \[ -\frac{b g (g \cos (e+f x))^{p-1} \left (-\frac{b (1-\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{b (\sin (e+f x)+1)}{a+b \sin (e+f x)}\right )^{\frac{1-p}{2}} F_1\left (1-p;\frac{1-p}{2},\frac{1-p}{2};2-p;\frac{a+b}{a+b \sin (e+f x)},\frac{a-b}{a+b \sin (e+f x)}\right )}{f (1-p) (b c-a d)^2}+\frac{b g (g \cos (e+f x))^{p-1} \left (-\frac{d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{d (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} F_1\left (1-p;\frac{1-p}{2},\frac{1-p}{2};2-p;\frac{c+d}{c+d \sin (e+f x)},\frac{c-d}{c+d \sin (e+f x)}\right )}{f (1-p) (b c-a d)^2}+\frac{g (g \cos (e+f x))^{p-1} \left (-\frac{d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{d (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} F_1\left (2-p;\frac{1-p}{2},\frac{1-p}{2};3-p;\frac{c+d}{c+d \sin (e+f x)},\frac{c-d}{c+d \sin (e+f x)}\right )}{f (2-p) (b c-a d) (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 2924
Rule 2703
Rubi steps
\begin{align*} \int \frac{(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx &=\int \left (\frac{b^2 (g \cos (e+f x))^p}{(b c-a d)^2 (a+b \sin (e+f x))}-\frac{d (g \cos (e+f x))^p}{(b c-a d) (c+d \sin (e+f x))^2}-\frac{b d (g \cos (e+f x))^p}{(b c-a d)^2 (c+d \sin (e+f x))}\right ) \, dx\\ &=\frac{b^2 \int \frac{(g \cos (e+f x))^p}{a+b \sin (e+f x)} \, dx}{(b c-a d)^2}-\frac{(b d) \int \frac{(g \cos (e+f x))^p}{c+d \sin (e+f x)} \, dx}{(b c-a d)^2}-\frac{d \int \frac{(g \cos (e+f x))^p}{(c+d \sin (e+f x))^2} \, dx}{b c-a d}\\ &=-\frac{b g F_1\left (1-p;\frac{1-p}{2},\frac{1-p}{2};2-p;\frac{a+b}{a+b \sin (e+f x)},\frac{a-b}{a+b \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac{b (1-\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{b (1+\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac{1-p}{2}}}{(b c-a d)^2 f (1-p)}+\frac{b g F_1\left (1-p;\frac{1-p}{2},\frac{1-p}{2};2-p;\frac{c+d}{c+d \sin (e+f x)},\frac{c-d}{c+d \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac{d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{d (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}}}{(b c-a d)^2 f (1-p)}+\frac{g F_1\left (2-p;\frac{1-p}{2},\frac{1-p}{2};3-p;\frac{c+d}{c+d \sin (e+f x)},\frac{c-d}{c+d \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac{d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}} \left (\frac{d (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1-p}{2}}}{(b c-a d) f (2-p) (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [B] time = 35.8543, size = 38759, normalized size = 76.3 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 1.034, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( g\cos \left ( fx+e \right ) \right ) ^{p}}{ \left ( a+b\sin \left ( fx+e \right ) \right ) \left ( c+d\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(-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 (\frac{\left (g \cos \left (f x + e\right )\right )^{p}}{a c^{2} + 2 \, b c d + a d^{2} -{\left (2 \, b c d + a d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (b d^{2} \cos \left (f x + e\right )^{2} - b c^{2} - 2 \, a c d - b d^{2}\right )} \sin \left (f x + e\right )}, 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 (b \sin \left (f x + e\right ) + a\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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