Optimal. Leaf size=84 \[ \frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \, _2F_1\left (\frac {1-m}{2},\frac {4+m}{2};\frac {6+m}{2};\sin ^2(a+b x)\right ) \sin ^3(a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (4+m)} \]
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Rubi [A]
time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4395, 2657}
\begin {gather*} \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}} \, _2F_1\left (\frac {1-m}{2},\frac {m+4}{2};\frac {m+6}{2};\sin ^2(a+b x)\right )}{b (m+4)} \end {gather*}
Antiderivative was successfully verified.
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Rule 2657
Rule 4395
Rubi steps
\begin {align*} \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx &=\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}} \, _2F_1\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*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 5.72, size = 602, normalized size = 7.17 \begin {gather*} \frac {32 (4+m) \left (F_1\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 )-F_1\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) F_1\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 F_1\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 F_1\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 F_1\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 F_1\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 F_1\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 F_1\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) F_1\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 )} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.57, size = 0, normalized size = 0.00 \[\int \left (\sin ^{3}\left (x b +a \right )\right ) \left (\sin ^{m}\left (2 x b +2 a \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{3}{\left (a + b x \right )} \sin ^{m}{\left (2 a + 2 b x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (a+b\,x\right )}^3\,{\sin \left (2\,a+2\,b\,x\right )}^m \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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