Optimal. Leaf size=85 \[ -\frac{\cos ^2(a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac{1-m}{2}} \sin ^m(2 a+2 b x) \text{Hypergeometric2F1}\left (\frac{1-m}{2},\frac{m+3}{2},\frac{m+5}{2},\cos ^2(a+b x)\right )}{b (m+3)} \]
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Rubi [A] time = 0.0707474, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4309, 2576} \[ -\frac{\cos ^2(a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac{1-m}{2}} \sin ^m(2 a+2 b x) \, _2F_1\left (\frac{1-m}{2},\frac{m+3}{2};\frac{m+5}{2};\cos ^2(a+b x)\right )}{b (m+3)} \]
Antiderivative was successfully verified.
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Rule 4309
Rule 2576
Rubi steps
\begin{align*} \int \cos ^2(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 ^{2+m}(a+b x) \sin ^m(a+b x) \, dx\\ &=-\frac{\cos ^2(a+b x) \cot (a+b x) \, _2F_1\left (\frac{1-m}{2},\frac{3+m}{2};\frac{5+m}{2};\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{\frac{1-m}{2}} \sin ^m(2 a+2 b x)}{b (3+m)}\\ \end{align*}
Mathematica [C] time = 7.96872, size = 890, normalized size = 10.47 \[ \frac{4 (m+3) \left (4 F_1\left (\frac{m+1}{2};-m,2 (m+1);\frac{m+3}{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 (\frac{m+1}{2};-m,2 m+1;\frac{m+3}{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 (\frac{m+1}{2};-m,2 m+3;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) \cos ^3\left (\frac{1}{2} (a+b x)\right ) \cos ^2(a+b x) \sin \left (\frac{1}{2} (a+b x)\right ) \sin ^m(2 (a+b x))}{b (m+1) \left (8 (m+3) F_1\left (\frac{m+1}{2};-m,2 (m+1);\frac{m+3}{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 (m+3) F_1\left (\frac{m+1}{2};-m,2 m+1;\frac{m+3}{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 )-8 (m+3) F_1\left (\frac{m+1}{2};-m,2 m+3;\frac{m+3}{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 (4 m F_1\left (\frac{m+3}{2};1-m,2 (m+1);\frac{m+5}{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 (\frac{m+3}{2};1-m,2 m+1;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 m F_1\left (\frac{m+3}{2};1-m,2 m+3;\frac{m+5}{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 (\frac{m+3}{2};-m,2 (m+1);\frac{m+5}{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 (\frac{m+3}{2};-m,2 (m+1);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-8 m F_1\left (\frac{m+3}{2};-m,2 (m+2);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-12 F_1\left (\frac{m+3}{2};-m,2 (m+2);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+8 m F_1\left (\frac{m+3}{2};-m,2 m+3;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+8 F_1\left (\frac{m+3}{2};-m,2 m+3;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) (\cos (a+b x)-1)\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.875, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( bx+a \right ) \right ) ^{2} \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (2 \, b x + 2 \, a\right )^{m} \cos \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sin \left (2 \, b x + 2 \, a\right )^{m} \cos \left (b x + a\right )^{2}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (2 \, b x + 2 \, a\right )^{m} \cos \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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