Integrand size = 35, antiderivative size = 137 \[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {c \operatorname {AppellF1}\left (\frac {1+n}{2},\frac {1-m}{2},-p,\frac {3+n}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (c \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{1+n} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{d f (1+n)} \]
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Time = 0.23 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3281, 525, 524} \[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {c \cos ^2(e+f x)^{\frac {1-m}{2}} (c \cos (e+f x))^{m-1} (d \sin (e+f x))^{n+1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {n+1}{2},\frac {1-m}{2},-p,\frac {n+3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{d f (n+1)} \]
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Rule 524
Rule 525
Rule 3281
Rubi steps \begin{align*} \text {integral}& = \frac {\left (c (c \cos (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \cos ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int (d x)^n \left (1-x^2\right )^{\frac {1}{2} (-1+m)} \left (a+b x^2\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (c (c \cos (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \cos ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int (d x)^n \left (1-x^2\right )^{\frac {1}{2} (-1+m)} \left (1+\frac {b x^2}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {c \operatorname {AppellF1}\left (\frac {1+n}{2},\frac {1-m}{2},-p,\frac {3+n}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (c \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{1+n} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{d f (1+n)} \\ \end{align*}
Time = 2.52 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99 \[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+n}{2},\frac {1-m}{2},-p,\frac {3+n}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (c \cos (e+f x))^m \cos ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{f (1+n)} \]
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\[\int \left (c \cos \left (f x +e \right )\right )^{m} \left (d \sin \left (f x +e \right )\right )^{n} {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]
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\[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (c \cos \left (f x + e\right )\right )^{m} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
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\[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (c \cos \left (f x + e\right )\right )^{m} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (c \cos \left (f x + e\right )\right )^{m} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\left (c\,\cos \left (e+f\,x\right )\right )}^m\,{\left (d\,\sin \left (e+f\,x\right )\right )}^n\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]
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