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

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

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.06, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos ^{2 m}(x) \sin ^{2 m}(x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cos \relax (x)^{2 \, m} \sin \relax (x)^{2 \, m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos \relax (x)^{2 \, m} \sin \relax (x)^{2 \, m}\,{d x} \]

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)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \left (\cos ^{2 m}\relax (x )\right ) \left (\sin ^{2 m}\relax (x )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \cos \relax (x)^{2 \, m} \sin \relax (x)^{2 \, m}\,{d x} \]

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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mupad [B]  time = 0.75, size = 52, normalized size = 0.76 \[ -\frac {{\cos \relax (x)}^{2\,m+1}\,{\sin \relax (x)}^{2\,m+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2}-m,m+\frac {1}{2};\ m+\frac {3}{2};\ {\cos \relax (x)}^2\right )}{\left (2\,m+1\right )\,{\left ({\sin \relax (x)}^2\right )}^{m+\frac {1}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(x)^(2*m + 1)*sin(x)^(2*m + 1)*hypergeom([1/2 - m, m + 1/2], m + 3/2, cos(x)^2))/((2*m + 1)*(sin(x)^2)^(m
+ 1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin ^{2 m}{\relax (x )} \cos ^{2 m}{\relax (x )}\, dx \]

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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