Integrand size = 21, antiderivative size = 83 \[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sin ^2(e+f x)\right ) (a \sin (e+f x))^m (b \tan (e+f x))^{1+n}}{b f (1+m+n)} \]
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Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2682, 2657} \[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\frac {\cos ^2(e+f x)^{\frac {n+1}{2}} (a \sin (e+f x))^m (b \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {1}{2} (m+n+1),\frac {1}{2} (m+n+3),\sin ^2(e+f x)\right )}{b f (m+n+1)} \]
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Rule 2657
Rule 2682
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a \cos ^{1+n}(e+f x) (a \sin (e+f x))^{-1-n} (b \tan (e+f x))^{1+n}\right ) \int \cos ^{-n}(e+f x) (a \sin (e+f x))^{m+n} \, dx}{b} \\ & = \frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sin ^2(e+f x)\right ) (a \sin (e+f x))^m (b \tan (e+f x))^{1+n}}{b f (1+m+n)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.82 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.13 \[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\frac {(3+m+n) \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),n,1+m,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x) (a \sin (e+f x))^m (b \tan (e+f x))^n}{f (1+m+n) \left ((3+m+n) \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),n,1+m,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((1+m) \operatorname {AppellF1}\left (\frac {1}{2} (3+m+n),n,2+m,\frac {1}{2} (5+m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n \operatorname {AppellF1}\left (\frac {1}{2} (3+m+n),1+n,1+m,\frac {1}{2} (5+m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]
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\[\int \left (\sin \left (f x +e \right ) a \right )^{m} \left (b \tan \left (f x +e \right )\right )^{n}d x\]
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\[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\int \left (a \sin {\left (e + f x \right )}\right )^{m} \left (b \tan {\left (e + f x \right )}\right )^{n}\, dx \]
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\[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\int {\left (a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \]
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