Integrand size = 13, antiderivative size = 68 \[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx=\frac {\cos ^{-1+2 m}(x) \cos ^2(x)^{\frac {1}{2}-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\sin ^2(x)\right ) \sin ^{1+2 m}(x)}{1+2 m} \]
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Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2657} \[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx=\frac {\sin ^{2 m+1}(x) \cos ^{2 m-1}(x) \cos ^2(x)^{\frac {1}{2}-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 m),\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+3),\sin ^2(x)\right )}{2 m+1} \]
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Rule 2657
Rubi steps \begin{align*} \text {integral}& = \frac {\cos ^{-1+2 m}(x) \cos ^2(x)^{\frac {1}{2}-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 m),\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m),\sin ^2(x)\right ) \sin ^{1+2 m}(x)}{1+2 m} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx=\frac {\cos ^{-1+2 m}(x) \cos ^2(x)^{\frac {1}{2}-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-m,\frac {1}{2}+m,\frac {3}{2}+m,\sin ^2(x)\right ) \sin ^{1+2 m}(x)}{1+2 m} \]
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\[\int \left (\cos ^{2 m}\left (x \right )\right ) \left (\sin ^{2 m}\left (x \right )\right )d x\]
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\[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx=\int { \cos \left (x\right )^{2 \, m} \sin \left (x\right )^{2 \, m} \,d x } \]
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\[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx=\int \sin ^{2 m}{\left (x \right )} \cos ^{2 m}{\left (x \right )}\, dx \]
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\[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx=\int { \cos \left (x\right )^{2 \, m} \sin \left (x\right )^{2 \, m} \,d x } \]
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\[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx=\int { \cos \left (x\right )^{2 \, m} \sin \left (x\right )^{2 \, m} \,d x } \]
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Time = 0.77 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx=-\frac {{\cos \left (x\right )}^{2\,m+1}\,{\sin \left (x\right )}^{2\,m+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2}-m,m+\frac {1}{2};\ m+\frac {3}{2};\ {\cos \left (x\right )}^2\right )}{\left (2\,m+1\right )\,{\left ({\sin \left (x\right )}^2\right )}^{m+\frac {1}{2}}} \]
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