3.352 \(\int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx\)

Optimal. Leaf size=68 \[ \frac{\sin ^{2 m+1}(x) \cos ^{2 m-1}(x) \cos ^2(x)^{\frac{1}{2}-m} \, _2F_1\left (\frac{1}{2} (1-2 m),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\sin ^2(x)\right )}{2 m+1} \]

[Out]

(Cos[x]^(-1 + 2*m)*(Cos[x]^2)^(1/2 - m)*Hypergeometric2F1[(1 - 2*m)/2, (1 + 2*m)/2, (3 + 2*m)/2, Sin[x]^2]*Sin
[x]^(1 + 2*m))/(1 + 2*m)

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Rubi [A]  time = 0.0374064, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2577} \[ \frac{\sin ^{2 m+1}(x) \cos ^{2 m-1}(x) \cos ^2(x)^{\frac{1}{2}-m} \text{Hypergeometric2F1}\left (\frac{1}{2} (1-2 m),\frac{1}{2} (2 m+1),\frac{1}{2} (2 m+3),\sin ^2(x)\right )}{2 m+1} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^(2*m)*Sin[x]^(2*m),x]

[Out]

(Cos[x]^(-1 + 2*m)*(Cos[x]^2)^(1/2 - m)*Hypergeometric2F1[(1 - 2*m)/2, (1 + 2*m)/2, (3 + 2*m)/2, Sin[x]^2]*Sin
[x]^(1 + 2*m))/(1 + 2*m)

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

\begin{align*} \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx &=\frac{\cos ^{-1+2 m}(x) \cos ^2(x)^{\frac{1}{2}-m} \, _2F_1\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*}

Mathematica [A]  time = 0.0662521, size = 58, normalized size = 0.85 \[ \frac{\sin ^{2 m+1}(x) \cos ^{2 m-1}(x) \cos ^2(x)^{\frac{1}{2}-m} \, _2F_1\left (\frac{1}{2}-m,m+\frac{1}{2};m+\frac{3}{2};\sin ^2(x)\right )}{2 m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^(2*m)*Sin[x]^(2*m),x]

[Out]

(Cos[x]^(-1 + 2*m)*(Cos[x]^2)^(1/2 - m)*Hypergeometric2F1[1/2 - m, 1/2 + m, 3/2 + m, Sin[x]^2]*Sin[x]^(1 + 2*m
))/(1 + 2*m)

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Maple [F]  time = 0.298, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( x \right ) \right ) ^{2\,m} \left ( \sin \left ( x \right ) \right ) ^{2\,m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^(2*m)*sin(x)^(2*m),x)

[Out]

int(cos(x)^(2*m)*sin(x)^(2*m),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (x\right )^{2 \, m} \sin \left (x\right )^{2 \, m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^(2*m)*sin(x)^(2*m),x, algorithm="maxima")

[Out]

integrate(cos(x)^(2*m)*sin(x)^(2*m), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\cos \left (x\right )^{2 \, m} \sin \left (x\right )^{2 \, m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^(2*m)*sin(x)^(2*m),x, algorithm="fricas")

[Out]

integral(cos(x)^(2*m)*sin(x)^(2*m), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin ^{2 m}{\left (x \right )} \cos ^{2 m}{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**(2*m)*sin(x)**(2*m),x)

[Out]

Integral(sin(x)**(2*m)*cos(x)**(2*m), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (x\right )^{2 \, m} \sin \left (x\right )^{2 \, m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^(2*m)*sin(x)^(2*m),x, algorithm="giac")

[Out]

integrate(cos(x)^(2*m)*sin(x)^(2*m), x)