Integrand size = 31, antiderivative size = 135 \[ \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\frac {2^{-\frac {3}{2}+m} (A (3-m)-B m) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^3(e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{3 a f (3-m)} \]
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Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2939, 2768, 72, 71} \[ \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {2^{m-\frac {3}{2}} (A (3-m)-B m) \sec ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{3 a f (3-m)}+\frac {B \sec ^3(e+f x) (a \sin (e+f x)+a)^m}{f (3-m)} \]
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Rule 71
Rule 72
Rule 2768
Rule 2939
Rubi steps \begin{align*} \text {integral}& = \frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\left (A-\frac {B m}{3-m}\right ) \int \sec ^4(e+f x) (a+a \sin (e+f x))^m \, dx \\ & = \frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\frac {\left (a^2 \left (A-\frac {B m}{3-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {5}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\frac {\left (2^{-\frac {5}{2}+m} \left (A-\frac {B m}{3-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{1+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {5}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\frac {2^{-\frac {3}{2}+m} \left (A-\frac {B m}{3-m}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^3(e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{3 a f} \\ \end{align*}
Time = 0.71 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.72 \[ \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {\sec ^3(e+f x) (a (1+\sin (e+f x)))^m \left (-6 B+2^{-\frac {1}{2}+m} (A (-3+m)+B m) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {3}{2}-m}\right )}{6 f (-3+m)} \]
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\[\int \left (\sec ^{4}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]
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\[ \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4} \,d x } \]
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Timed out. \[ \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \]
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\[ \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4} \,d x } \]
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\[ \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4} \,d x } \]
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Timed out. \[ \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^4} \,d x \]
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