3.71 \(\int \cos ^4(2 x) \sin ^2(2 x) \, dx\)

Optimal. Leaf size=46 \[ \frac {x}{16}-\frac {1}{12} \sin (2 x) \cos ^5(2 x)+\frac {1}{48} \sin (2 x) \cos ^3(2 x)+\frac {1}{32} \sin (2 x) \cos (2 x) \]

[Out]

1/16*x+1/32*cos(2*x)*sin(2*x)+1/48*cos(2*x)^3*sin(2*x)-1/12*cos(2*x)^5*sin(2*x)

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2568, 2635, 8} \[ \frac {x}{16}-\frac {1}{12} \sin (2 x) \cos ^5(2 x)+\frac {1}{48} \sin (2 x) \cos ^3(2 x)+\frac {1}{32} \sin (2 x) \cos (2 x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[2*x]^4*Sin[2*x]^2,x]

[Out]

x/16 + (Cos[2*x]*Sin[2*x])/32 + (Cos[2*x]^3*Sin[2*x])/48 - (Cos[2*x]^5*Sin[2*x])/12

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rubi steps

\begin {align*} \int \cos ^4(2 x) \sin ^2(2 x) \, dx &=-\frac {1}{12} \cos ^5(2 x) \sin (2 x)+\frac {1}{6} \int \cos ^4(2 x) \, dx\\ &=\frac {1}{48} \cos ^3(2 x) \sin (2 x)-\frac {1}{12} \cos ^5(2 x) \sin (2 x)+\frac {1}{8} \int \cos ^2(2 x) \, dx\\ &=\frac {1}{32} \cos (2 x) \sin (2 x)+\frac {1}{48} \cos ^3(2 x) \sin (2 x)-\frac {1}{12} \cos ^5(2 x) \sin (2 x)+\frac {\int 1 \, dx}{16}\\ &=\frac {x}{16}+\frac {1}{32} \cos (2 x) \sin (2 x)+\frac {1}{48} \cos ^3(2 x) \sin (2 x)-\frac {1}{12} \cos ^5(2 x) \sin (2 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.65 \[ \frac {x}{16}+\frac {1}{128} \sin (4 x)-\frac {1}{128} \sin (8 x)-\frac {1}{384} \sin (12 x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[2*x]^4*Sin[2*x]^2,x]

[Out]

x/16 + Sin[4*x]/128 - Sin[8*x]/128 - Sin[12*x]/384

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fricas [A]  time = 0.43, size = 33, normalized size = 0.72 \[ -\frac {1}{96} \, {\left (8 \, \cos \left (2 \, x\right )^{5} - 2 \, \cos \left (2 \, x\right )^{3} - 3 \, \cos \left (2 \, x\right )\right )} \sin \left (2 \, x\right ) + \frac {1}{16} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)^4*sin(2*x)^2,x, algorithm="fricas")

[Out]

-1/96*(8*cos(2*x)^5 - 2*cos(2*x)^3 - 3*cos(2*x))*sin(2*x) + 1/16*x

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giac [A]  time = 0.83, size = 22, normalized size = 0.48 \[ \frac {1}{16} \, x - \frac {1}{384} \, \sin \left (12 \, x\right ) - \frac {1}{128} \, \sin \left (8 \, x\right ) + \frac {1}{128} \, \sin \left (4 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)^4*sin(2*x)^2,x, algorithm="giac")

[Out]

1/16*x - 1/384*sin(12*x) - 1/128*sin(8*x) + 1/128*sin(4*x)

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maple [A]  time = 0.02, size = 36, normalized size = 0.78 \[ -\frac {\left (\cos ^{5}\left (2 x \right )\right ) \sin \left (2 x \right )}{12}+\frac {x}{16}+\frac {\left (\cos ^{3}\left (2 x \right )+\frac {3 \cos \left (2 x \right )}{2}\right ) \sin \left (2 x \right )}{48} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*x)^4*sin(2*x)^2,x)

[Out]

-1/12*cos(2*x)^5*sin(2*x)+1/48*(cos(2*x)^3+3/2*cos(2*x))*sin(2*x)+1/16*x

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maxima [A]  time = 0.43, size = 18, normalized size = 0.39 \[ \frac {1}{96} \, \sin \left (4 \, x\right )^{3} + \frac {1}{16} \, x - \frac {1}{128} \, \sin \left (8 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)^4*sin(2*x)^2,x, algorithm="maxima")

[Out]

1/96*sin(4*x)^3 + 1/16*x - 1/128*sin(8*x)

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mupad [B]  time = 0.07, size = 37, normalized size = 0.80 \[ \frac {x}{16}-\frac {\cos \left (2\,x\right )\,\sin \left (2\,x\right )}{32}+\frac {{\sin \left (2\,x\right )}^3\,\left (\frac {{\cos \left (2\,x\right )}^3}{6}+\frac {\cos \left (2\,x\right )}{8}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*x)^4*sin(2*x)^2,x)

[Out]

x/16 - (cos(2*x)*sin(2*x))/32 + (sin(2*x)^3*(cos(2*x)/8 + cos(2*x)^3/6))/2

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sympy [A]  time = 0.07, size = 41, normalized size = 0.89 \[ \frac {x}{16} - \frac {\sin {\left (2 x \right )} \cos ^{5}{\left (2 x \right )}}{12} + \frac {\sin {\left (2 x \right )} \cos ^{3}{\left (2 x \right )}}{48} + \frac {\sin {\left (2 x \right )} \cos {\left (2 x \right )}}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)**4*sin(2*x)**2,x)

[Out]

x/16 - sin(2*x)*cos(2*x)**5/12 + sin(2*x)*cos(2*x)**3/48 + sin(2*x)*cos(2*x)/32

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