Integrand size = 35, antiderivative size = 330 \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=-\frac {g \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {a+b}{a+b \sin (e+f x)},\frac {a-b}{a+b \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac {b (1-\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {b (1+\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}}}{(b c-a d) f (1-p)}+\frac {g \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {c+d}{c+d \sin (e+f x)},\frac {c-d}{c+d \sin (e+f x)}\right ) (g \cos (e+f x))^{-1+p} \left (-\frac {d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {d (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}}}{(b c-a d) f (1-p)} \]
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Time = 0.27 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {3003, 2782} \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {g (g \cos (e+f x))^{p-1} \left (-\frac {d (1-\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {d (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {c+d}{c+d \sin (e+f x)},\frac {c-d}{c+d \sin (e+f x)}\right )}{f (1-p) (b c-a d)}-\frac {g (g \cos (e+f x))^{p-1} \left (-\frac {b (1-\sin (e+f x))}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}} \left (\frac {b (\sin (e+f x)+1)}{a+b \sin (e+f x)}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (1-p,\frac {1-p}{2},\frac {1-p}{2},2-p,\frac {a+b}{a+b \sin (e+f x)},\frac {a-b}{a+b \sin (e+f x)}\right )}{f (1-p) (b c-a d)} \]
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Rule 2782
Rule 3003
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {b (g \cos (e+f x))^p}{(b c-a d) (a+b \sin (e+f x))}-\frac {d (g \cos (e+f x))^p}{(b c-a d) (c+d \sin (e+f x))}\right ) \, dx \\ & = \frac {b \int \frac {(g \cos (e+f x))^p}{a+b \sin (e+f x)} \, dx}{b c-a d}-\frac {d \int \frac {(g \cos (e+f x))^p}{c+d \sin (e+f x)} \, dx}{b c-a d} \\ & = -\frac {g \operatorname {AppellF1}\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) f (1-p)}+\frac {g \operatorname {AppellF1}\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) f (1-p)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(5085\) vs. \(2(330)=660\).
Time = 33.67 (sec) , antiderivative size = 5085, normalized size of antiderivative = 15.41 \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\text {Result too large to show} \]
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\[\int \frac {\left (g \cos \left (f x +e \right )\right )^{p}}{\left (a +b \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )}d x\]
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\[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p}}{{\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\text {Timed out} \]
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\[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p}}{{\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}} \,d x } \]
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\[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p}}{{\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {(g \cos (e+f x))^p}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^p}{\left (a+b\,\sin \left (e+f\,x\right )\right )\,\left (c+d\,\sin \left (e+f\,x\right )\right )} \,d x \]
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