Integrand size = 31, antiderivative size = 123 \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {B \sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m)}+\frac {2^{-\frac {1}{2}+m} (A (1-m)-B m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1-m)} \]
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Time = 0.13 (sec) , antiderivative size = 123, 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} \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {2^{m-\frac {1}{2}} (A (1-m)-B m) \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f (1-m)}+\frac {B \sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m)} \]
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Rule 71
Rule 72
Rule 2768
Rule 2939
Rubi steps \begin{align*} \text {integral}& = \frac {B \sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m)}+\left (A-\frac {B m}{1-m}\right ) \int \sec ^2(e+f x) (a+a \sin (e+f x))^m \, dx \\ & = \frac {B \sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m)}+\frac {\left (a^2 \left (A-\frac {B m}{1-m}\right ) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {3}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {B \sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m)}+\frac {\left (2^{-\frac {3}{2}+m} a \left (A-\frac {B m}{1-m}\right ) \sec (e+f x) \sqrt {a-a \sin (e+f x)} (a+a \sin (e+f x))^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 {3}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {B \sec (e+f x) (a+a \sin (e+f x))^m}{f (1-m)}+\frac {2^{-\frac {1}{2}+m} \left (A-\frac {B m}{1-m}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f} \\ \end{align*}
Time = 0.75 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94 \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {\sec (e+f x) (1+\sin (e+f x))^{-m} (a (1+\sin (e+f x)))^m \left (2^m (A (-1+m)+B m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sqrt {1+\sin (e+f x)}-\sqrt {2} B (1+\sin (e+f x))^m\right )}{\sqrt {2} f (-1+m)} \]
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\[\int \left (\sec ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]
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\[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
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Timed out. \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \]
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\[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
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\[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
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Timed out. \[ \int \sec ^2(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^2} \,d x \]
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