Integrand size = 19, antiderivative size = 78 \[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(e+f x)\right ) \sin ^3(e+f x) (b \tan (e+f x))^{1+n}}{b f (4+n)} \]
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Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2682, 2657} \[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\frac {\sin ^3(e+f x) \cos ^2(e+f x)^{\frac {n+1}{2}} (b \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {n+4}{2},\frac {n+6}{2},\sin ^2(e+f x)\right )}{b f (n+4)} \]
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Rule 2657
Rule 2682
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\cos ^{1+n}(e+f x) \sin ^{-1-n}(e+f x) (b \tan (e+f x))^{1+n}\right ) \int \cos ^{-n}(e+f x) \sin ^{3+n}(e+f x) \, dx}{b} \\ & = \frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(e+f x)\right ) \sin ^3(e+f x) (b \tan (e+f x))^{1+n}}{b f (4+n)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 2.30 (sec) , antiderivative size = 456, normalized size of antiderivative = 5.85 \[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\frac {4 (4+n) \left (\operatorname {AppellF1}\left (1+\frac {n}{2},n,3,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-\operatorname {AppellF1}\left (1+\frac {n}{2},n,4,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \sin ^3(e+f x) (b \tan (e+f x))^n}{f (2+n) \left (-2 (4+n) \operatorname {AppellF1}\left (1+\frac {n}{2},n,4,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (3 \operatorname {AppellF1}\left (2+\frac {n}{2},n,4,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 \operatorname {AppellF1}\left (2+\frac {n}{2},n,5,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n \left (-\operatorname {AppellF1}\left (2+\frac {n}{2},1+n,3,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\operatorname {AppellF1}\left (2+\frac {n}{2},1+n,4,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) (-1+\cos (e+f x))+(4+n) \operatorname {AppellF1}\left (1+\frac {n}{2},n,3,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))\right )} \]
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\[\int \left (\sin ^{3}\left (f x +e \right )\right ) \left (b \tan \left (f x +e \right )\right )^{n}d x\]
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\[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{3} \,d x } \]
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Timed out. \[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\text {Timed out} \]
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\[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{3} \,d x } \]
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\[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{3} \,d x } \]
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Timed out. \[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \]
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