\(\int \cos ^4(2 x) \sin ^2(2 x) \, dx\) [71]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16}+\frac {1}{32} \cos (2 x) \sin (2 x)+\frac {1}{48} \cos ^3(2 x) \sin (2 x)-\frac {1}{12} \cos ^5(2 x) \sin (2 x) \]

[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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2648, 2715, 8} \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16}-\frac {1}{12} \sin (2 x) \cos ^5(2 x)+\frac {1}{48} \sin (2 x) \cos ^3(2 x)+\frac {1}{32} \sin (2 x) \cos (2 x) \]

[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 2648

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 2715

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] && Integ
erQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.65 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16}+\frac {1}{128} \sin (4 x)-\frac {1}{128} \sin (8 x)-\frac {1}{384} \sin (12 x) \]

[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

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.50

method	result	size
risch	\(\frac {x}{16}-\frac {\sin \left (12 x \right )}{384}-\frac {\sin \left (8 x \right )}{128}+\frac {\sin \left (4 x \right )}{128}\)	\(23\)
parallelrisch	\(\frac {x}{16}-\frac {\sin \left (12 x \right )}{384}-\frac {\sin \left (8 x \right )}{128}+\frac {\sin \left (4 x \right )}{128}\)	\(23\)
derivativedivides	\(-\frac {\left (\cos ^{5}\left (2 x \right )\right ) \sin \left (2 x \right )}{12}+\frac {\left (\cos ^{3}\left (2 x \right )+\frac {3 \cos \left (2 x \right )}{2}\right ) \sin \left (2 x \right )}{48}+\frac {x}{16}\)	\(36\)
default	\(-\frac {\left (\cos ^{5}\left (2 x \right )\right ) \sin \left (2 x \right )}{12}+\frac {\left (\cos ^{3}\left (2 x \right )+\frac {3 \cos \left (2 x \right )}{2}\right ) \sin \left (2 x \right )}{48}+\frac {x}{16}\)	\(36\)
norman	\(\frac {\frac {x}{16}+\frac {47 \left (\tan ^{3}\left (x \right )\right )}{48}-\frac {13 \left (\tan ^{5}\left (x \right )\right )}{8}+\frac {13 \left (\tan ^{7}\left (x \right )\right )}{8}-\frac {47 \left (\tan ^{9}\left (x \right )\right )}{48}+\frac {\left (\tan ^{11}\left (x \right )\right )}{16}+\frac {3 x \left (\tan ^{2}\left (x \right )\right )}{8}+\frac {15 x \left (\tan ^{4}\left (x \right )\right )}{16}+\frac {5 x \left (\tan ^{6}\left (x \right )\right )}{4}+\frac {15 x \left (\tan ^{8}\left (x \right )\right )}{16}+\frac {3 x \left (\tan ^{10}\left (x \right )\right )}{8}+\frac {x \left (\tan ^{12}\left (x \right )\right )}{16}-\frac {\tan \left (x \right )}{16}}{\left (1+\tan ^{2}\left (x \right )\right )^{6}}\)	\(90\)

[In]

int(cos(2*x)^4*sin(2*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=-\frac {1}{96} \, {\left (8 \, \cos \left (2 \, x\right )^{5} - 2 \, \cos \left (2 \, x\right )^{3} - 3 \, \cos \left (2 \, x\right )\right )} \sin \left (2 \, x\right ) + \frac {1}{16} \, x \]

[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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16} - \frac {\sin {\left (2 x \right )} \cos ^{5}{\left (2 x \right )}}{12} + \frac {\sin {\left (2 x \right )} \cos ^{3}{\left (2 x \right )}}{48} + \frac {\sin {\left (2 x \right )} \cos {\left (2 x \right )}}{32} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.39 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {1}{96} \, \sin \left (4 \, x\right )^{3} + \frac {1}{16} \, x - \frac {1}{128} \, \sin \left (8 \, x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.48 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {1}{16} \, x - \frac {1}{384} \, \sin \left (12 \, x\right ) - \frac {1}{128} \, \sin \left (8 \, x\right ) + \frac {1}{128} \, \sin \left (4 \, x\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16}-\frac {\cos \left (2\,x\right )\,\sin \left (2\,x\right )}{32}+\frac {{\sin \left (2\,x\right )}^3\,\left (\frac {{\cos \left (2\,x\right )}^3}{6}+\frac {\cos \left (2\,x\right )}{8}\right )}{2} \]

[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