Integrand size = 13, antiderivative size = 46 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16}+\frac {1}{32} \cos (2 x) \sin (2 x)+\frac {1}{48} \cos ^3(2 x) \sin (2 x)-\frac {1}{12} \cos ^5(2 x) \sin (2 x) \] Output:
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)
Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.65 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16}+\frac {1}{128} \sin (4 x)-\frac {1}{128} \sin (8 x)-\frac {1}{384} \sin (12 x) \] Input:
Integrate[Cos[2*x]^4*Sin[2*x]^2,x]
Output:
x/16 + Sin[4*x]/128 - Sin[8*x]/128 - Sin[12*x]/384
Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3042, 3048, 3042, 3115, 3042, 3115, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^2(2 x) \cos ^4(2 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (2 x)^2 \cos (2 x)^4dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {1}{6} \int \cos ^4(2 x)dx-\frac {1}{12} \sin (2 x) \cos ^5(2 x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} \int \sin \left (2 x+\frac {\pi }{2}\right )^4dx-\frac {1}{12} \sin (2 x) \cos ^5(2 x)\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{6} \left (\frac {3}{4} \int \cos ^2(2 x)dx+\frac {1}{8} \sin (2 x) \cos ^3(2 x)\right )-\frac {1}{12} \sin (2 x) \cos ^5(2 x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} \left (\frac {3}{4} \int \sin \left (2 x+\frac {\pi }{2}\right )^2dx+\frac {1}{8} \sin (2 x) \cos ^3(2 x)\right )-\frac {1}{12} \sin (2 x) \cos ^5(2 x)\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {1}{4} \sin (2 x) \cos (2 x)\right )+\frac {1}{8} \sin (2 x) \cos ^3(2 x)\right )-\frac {1}{12} \sin (2 x) \cos ^5(2 x)\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{6} \left (\frac {1}{8} \sin (2 x) \cos ^3(2 x)+\frac {3}{4} \left (\frac {x}{2}+\frac {1}{4} \sin (2 x) \cos (2 x)\right )\right )-\frac {1}{12} \sin (2 x) \cos ^5(2 x)\) |
Input:
Int[Cos[2*x]^4*Sin[2*x]^2,x]
Output:
-1/12*(Cos[2*x]^5*Sin[2*x]) + ((Cos[2*x]^3*Sin[2*x])/8 + (3*(x/2 + (Cos[2* x]*Sin[2*x])/4))/4)/6
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] + Simp[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]
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Time = 1.35 (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 | \(-\frac {\cos \left (2 x \right )^{5} \sin \left (2 x \right )}{12}+\frac {\left (\cos \left (2 x \right )^{3}+\frac {3 \cos \left (2 x \right )}{2}\right ) \sin \left (2 x \right )}{48}+\frac {x}{16}\) | \(36\) |
default | \(-\frac {\cos \left (2 x \right )^{5} \sin \left (2 x \right )}{12}+\frac {\left (\cos \left (2 x \right )^{3}+\frac {3 \cos \left (2 x \right )}{2}\right ) \sin \left (2 x \right )}{48}+\frac {x}{16}\) | \(36\) |
norman | \(\frac {\frac {x}{16}+\frac {47 \tan \left (x \right )^{3}}{48}-\frac {13 \tan \left (x \right )^{5}}{8}+\frac {13 \tan \left (x \right )^{7}}{8}-\frac {47 \tan \left (x \right )^{9}}{48}+\frac {\tan \left (x \right )^{11}}{16}+\frac {3 x \tan \left (x \right )^{2}}{8}+\frac {15 x \tan \left (x \right )^{4}}{16}+\frac {5 x \tan \left (x \right )^{6}}{4}+\frac {15 x \tan \left (x \right )^{8}}{16}+\frac {3 x \tan \left (x \right )^{10}}{8}+\frac {x \tan \left (x \right )^{12}}{16}-\frac {\tan \left (x \right )}{16}}{\left (1+\tan \left (x \right )^{2}\right )^{6}}\) | \(90\) |
orering | \(x \cos \left (2 x \right )^{4} \sin \left (2 x \right )^{2}+\frac {\cos \left (2 x \right )^{3} \sin \left (2 x \right )^{3}}{12}-\frac {\cos \left (2 x \right )^{5} \sin \left (2 x \right )}{32}+\frac {49 x \left (48 \cos \left (2 x \right )^{2} \sin \left (2 x \right )^{4}-88 \cos \left (2 x \right )^{4} \sin \left (2 x \right )^{2}+8 \cos \left (2 x \right )^{6}\right )}{576}+\frac {\cos \left (2 x \right ) \sin \left (2 x \right )^{5}}{32}+\frac {7 x \left (384 \sin \left (2 x \right )^{6}-8448 \cos \left (2 x \right )^{2} \sin \left (2 x \right )^{4}+11008 \cos \left (2 x \right )^{4} \sin \left (2 x \right )^{2}-896 \cos \left (2 x \right )^{6}\right )}{4608}+\frac {x \left (-76800 \sin \left (2 x \right )^{6}+1317888 \cos \left (2 x \right )^{2} \sin \left (2 x \right )^{4}-1481728 \cos \left (2 x \right )^{4} \sin \left (2 x \right )^{2}+109568 \cos \left (2 x \right )^{6}\right )}{147456}\) | \(190\) |
Input:
int(cos(2*x)^4*sin(2*x)^2,x,method=_RETURNVERBOSE)
Output:
1/16*x-1/384*sin(12*x)-1/128*sin(8*x)+1/128*sin(4*x)
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=-\frac {1}{96} \, {\left (8 \, \cos \left (2 \, x\right )^{5} - 2 \, \cos \left (2 \, x\right )^{3} - 3 \, \cos \left (2 \, x\right )\right )} \sin \left (2 \, x\right ) + \frac {1}{16} \, x \] Input:
integrate(cos(2*x)^4*sin(2*x)^2,x, algorithm="fricas")
Output:
-1/96*(8*cos(2*x)^5 - 2*cos(2*x)^3 - 3*cos(2*x))*sin(2*x) + 1/16*x
Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16} - \frac {\sin {\left (2 x \right )} \cos ^{5}{\left (2 x \right )}}{12} + \frac {\sin {\left (2 x \right )} \cos ^{3}{\left (2 x \right )}}{48} + \frac {\sin {\left (2 x \right )} \cos {\left (2 x \right )}}{32} \] Input:
integrate(cos(2*x)**4*sin(2*x)**2,x)
Output:
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
Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.39 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {1}{96} \, \sin \left (4 \, x\right )^{3} + \frac {1}{16} \, x - \frac {1}{128} \, \sin \left (8 \, x\right ) \] Input:
integrate(cos(2*x)^4*sin(2*x)^2,x, algorithm="maxima")
Output:
1/96*sin(4*x)^3 + 1/16*x - 1/128*sin(8*x)
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.48 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {1}{16} \, x - \frac {1}{384} \, \sin \left (12 \, x\right ) - \frac {1}{128} \, \sin \left (8 \, x\right ) + \frac {1}{128} \, \sin \left (4 \, x\right ) \] Input:
integrate(cos(2*x)^4*sin(2*x)^2,x, algorithm="giac")
Output:
1/16*x - 1/384*sin(12*x) - 1/128*sin(8*x) + 1/128*sin(4*x)
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=\frac {x}{16}-\frac {\cos \left (2\,x\right )\,\sin \left (2\,x\right )}{32}+\frac {{\sin \left (2\,x\right )}^3\,\left (\frac {{\cos \left (2\,x\right )}^3}{6}+\frac {\cos \left (2\,x\right )}{8}\right )}{2} \] Input:
int(cos(2*x)^4*sin(2*x)^2,x)
Output:
x/16 - (cos(2*x)*sin(2*x))/32 + (sin(2*x)^3*(cos(2*x)/8 + cos(2*x)^3/6))/2
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \cos ^4(2 x) \sin ^2(2 x) \, dx=-\frac {\cos \left (2 x \right ) \sin \left (2 x \right )^{5}}{12}+\frac {7 \cos \left (2 x \right ) \sin \left (2 x \right )^{3}}{48}-\frac {\cos \left (2 x \right ) \sin \left (2 x \right )}{32}+\frac {x}{16} \] Input:
int(cos(2*x)^4*sin(2*x)^2,x)
Output:
( - 8*cos(2*x)*sin(2*x)**5 + 14*cos(2*x)*sin(2*x)**3 - 3*cos(2*x)*sin(2*x) + 6*x)/96