Integrand size = 25, antiderivative size = 115 \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (d \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{f} \]
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Time = 0.11 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3272, 441, 440} \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {d \sin (e+f x) \cos ^2(e+f x)^{\frac {1-m}{2}} (d \cos (e+f x))^{m-1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{f} \]
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Rule 440
Rule 441
Rule 3272
Rubi steps \begin{align*} \text {integral}& = \frac {\left (d (d \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 \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 (d (d \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 \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 {d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (d \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{f} \\ \end{align*}
Time = 1.43 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.98 \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \tan (e+f x)}{f \left (3 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )+\left (2 b p \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-m}{2},1-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )-a (-1+m) \operatorname {AppellF1}\left (\frac {3}{2},\frac {3-m}{2},-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right ) \sin ^2(e+f x)\right )} \]
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\[\int \left (\cos \left (f x +e \right ) d \right )^{m} {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]
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\[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \cos \left (f x + e\right )\right )^{m} \,d x } \]
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Timed out. \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
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\[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \cos \left (f x + e\right )\right )^{m} \,d x } \]
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\[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \cos \left (f x + e\right )\right )^{m} \,d x } \]
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Timed out. \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\left (d\,\cos \left (e+f\,x\right )\right )}^m\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]
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