Integrand size = 20, antiderivative size = 84 \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {4+m}{2},\frac {6+m}{2},\sin ^2(a+b x)\right ) \sin ^3(a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (4+m)} \]
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Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4395, 2657} \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {\sin ^3(a+b x) \tan (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m+4}{2},\frac {m+6}{2},\sin ^2(a+b x)\right )}{b (m+4)} \]
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Rule 2657
Rule 4395
Rubi steps \begin{align*} \text {integral}& = \left (\cos ^{-m}(a+b x) \sin ^{-m}(a+b x) \sin ^m(2 a+2 b x)\right ) \int \cos ^m(a+b x) \sin ^{3+m}(a+b x) \, dx \\ & = \frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {4+m}{2},\frac {6+m}{2},\sin ^2(a+b x)\right ) \sin ^3(a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (4+m)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 6.70 (sec) , antiderivative size = 602, normalized size of antiderivative = 7.17 \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {32 (4+m) \left (\operatorname {AppellF1}\left (1+\frac {m}{2},-m,3+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-\operatorname {AppellF1}\left (1+\frac {m}{2},-m,4+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \cos ^6\left (\frac {1}{2} (a+b x)\right ) \sin ^4\left (\frac {1}{2} (a+b x)\right ) \sin ^m(2 (a+b x))}{b (2+m) \left (-2 (4+m) \operatorname {AppellF1}\left (1+\frac {m}{2},-m,4+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) \cos ^2\left (\frac {1}{2} (a+b x)\right )+2 \left (m \operatorname {AppellF1}\left (2+\frac {m}{2},1-m,3+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-m \operatorname {AppellF1}\left (2+\frac {m}{2},1-m,4+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+3 \operatorname {AppellF1}\left (2+\frac {m}{2},-m,4+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 m \operatorname {AppellF1}\left (2+\frac {m}{2},-m,4+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-4 \operatorname {AppellF1}\left (2+\frac {m}{2},-m,5+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-2 m \operatorname {AppellF1}\left (2+\frac {m}{2},-m,5+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) (-1+\cos (a+b x))+(4+m) \operatorname {AppellF1}\left (1+\frac {m}{2},-m,3+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) (1+\cos (a+b x))\right )} \]
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\[\int \sin \left (x b +a \right )^{3} \sin \left (2 x b +2 a \right )^{m}d x\]
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\[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right )^{3} \,d x } \]
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Timed out. \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\text {Timed out} \]
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\[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right )^{3} \,d x } \]
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\[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right )^{3} \,d x } \]
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Timed out. \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\int {\sin \left (a+b\,x\right )}^3\,{\sin \left (2\,a+2\,b\,x\right )}^m \,d x \]
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