Integrand size = 10, antiderivative size = 34 \[ \int 8 \cos ^2(x) \sin ^4(x) \, dx=\frac {x}{2}+\frac {1}{2} \cos (x) \sin (x)-\cos ^3(x) \sin (x)-\frac {4}{3} \cos ^3(x) \sin ^3(x) \] Output:
1/2*x+1/2*cos(x)*sin(x)-cos(x)^3*sin(x)-4/3*cos(x)^3*sin(x)^3
Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int 8 \cos ^2(x) \sin ^4(x) \, dx=8 \left (\frac {x}{16}-\frac {1}{64} \sin (2 x)-\frac {1}{64} \sin (4 x)+\frac {1}{192} \sin (6 x)\right ) \] Input:
Integrate[8*Cos[x]^2*Sin[x]^4,x]
Output:
8*(x/16 - Sin[2*x]/64 - Sin[4*x]/64 + Sin[6*x]/192)
Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {27, 3042, 3048, 3042, 3048, 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.
\(\displaystyle \int 8 \sin ^4(x) \cos ^2(x) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 8 \int \cos ^2(x) \sin ^4(x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 8 \int \cos (x)^2 \sin (x)^4dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle 8 \left (\frac {1}{2} \int \cos ^2(x) \sin ^2(x)dx-\frac {1}{6} \sin ^3(x) \cos ^3(x)\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 8 \left (\frac {1}{2} \int \cos (x)^2 \sin (x)^2dx-\frac {1}{6} \sin ^3(x) \cos ^3(x)\right )\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{4} \int \cos ^2(x)dx-\frac {1}{4} \sin (x) \cos ^3(x)\right )-\frac {1}{6} \sin ^3(x) \cos ^3(x)\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{4} \int \sin \left (x+\frac {\pi }{2}\right )^2dx-\frac {1}{4} \sin (x) \cos ^3(x)\right )-\frac {1}{6} \sin ^3(x) \cos ^3(x)\right )\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\int 1dx}{2}+\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{4} \sin (x) \cos ^3(x)\right )-\frac {1}{6} \sin ^3(x) \cos ^3(x)\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {x}{2}+\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{4} \sin (x) \cos ^3(x)\right )-\frac {1}{6} \sin ^3(x) \cos ^3(x)\right )\) |
Input:
Int[8*Cos[x]^2*Sin[x]^4,x]
Output:
8*(-1/6*(Cos[x]^3*Sin[x]^3) + (-1/4*(Cos[x]^3*Sin[x]) + (x/2 + (Cos[x]*Sin [x])/2)/4)/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 0.93 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68
method | result | size |
risch | \(\frac {x}{2}+\frac {\sin \left (6 x \right )}{24}-\frac {\sin \left (4 x \right )}{8}-\frac {\sin \left (2 x \right )}{8}\) | \(23\) |
parallelrisch | \(\frac {x}{2}+\frac {\sin \left (6 x \right )}{24}-\frac {\sin \left (4 x \right )}{8}-\frac {\sin \left (2 x \right )}{8}\) | \(23\) |
default | \(\frac {x}{2}+\frac {\cos \left (x \right ) \sin \left (x \right )}{2}-\cos \left (x \right )^{3} \sin \left (x \right )-\frac {4 \cos \left (x \right )^{3} \sin \left (x \right )^{3}}{3}\) | \(29\) |
norman | \(\frac {\tan \left (\frac {x}{2}\right )^{11}+\frac {x}{2}-\frac {17 \tan \left (\frac {x}{2}\right )^{3}}{3}+38 \tan \left (\frac {x}{2}\right )^{5}-38 \tan \left (\frac {x}{2}\right )^{7}+\frac {17 \tan \left (\frac {x}{2}\right )^{9}}{3}+\frac {15 x \tan \left (\frac {x}{2}\right )^{4}}{2}+10 x \tan \left (\frac {x}{2}\right )^{6}+\frac {15 x \tan \left (\frac {x}{2}\right )^{8}}{2}+3 x \tan \left (\frac {x}{2}\right )^{10}+\frac {x \tan \left (\frac {x}{2}\right )^{12}}{2}+3 \tan \left (\frac {x}{2}\right )^{2} x -\tan \left (\frac {x}{2}\right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{6}}\) | \(114\) |
orering | \(8 \cos \left (x \right )^{2} \sin \left (x \right )^{4} x +\frac {\cos \left (x \right ) \sin \left (x \right )^{5}}{2}-\frac {4 \cos \left (x \right )^{3} \sin \left (x \right )^{3}}{3}+\frac {49 x \left (16 \sin \left (x \right )^{6}-176 \cos \left (x \right )^{2} \sin \left (x \right )^{4}+96 \sin \left (x \right )^{2} \cos \left (x \right )^{4}\right )}{144}-\frac {\cos \left (x \right )^{5} \sin \left (x \right )}{2}+\frac {7 x \left (-448 \sin \left (x \right )^{6}+5504 \cos \left (x \right )^{2} \sin \left (x \right )^{4}-4224 \sin \left (x \right )^{2} \cos \left (x \right )^{4}+192 \cos \left (x \right )^{6}\right )}{288}+\frac {x \left (13696 \sin \left (x \right )^{6}-185216 \cos \left (x \right )^{2} \sin \left (x \right )^{4}+164736 \sin \left (x \right )^{2} \cos \left (x \right )^{4}-9600 \cos \left (x \right )^{6}\right )}{2304}\) | \(141\) |
Input:
int(8*cos(x)^2*sin(x)^4,x,method=_RETURNVERBOSE)
Output:
1/2*x+1/24*sin(6*x)-1/8*sin(4*x)-1/8*sin(2*x)
Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int 8 \cos ^2(x) \sin ^4(x) \, dx=\frac {1}{6} \, {\left (8 \, \cos \left (x\right )^{5} - 14 \, \cos \left (x\right )^{3} + 3 \, \cos \left (x\right )\right )} \sin \left (x\right ) + \frac {1}{2} \, x \] Input:
integrate(8*cos(x)^2*sin(x)^4,x, algorithm="fricas")
Output:
1/6*(8*cos(x)^5 - 14*cos(x)^3 + 3*cos(x))*sin(x) + 1/2*x
Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int 8 \cos ^2(x) \sin ^4(x) \, dx=\frac {x}{2} + \frac {4 \sin ^{5}{\left (x \right )} \cos {\left (x \right )}}{3} - \frac {\sin ^{3}{\left (x \right )} \cos {\left (x \right )}}{3} - \frac {\sin {\left (x \right )} \cos {\left (x \right )}}{2} \] Input:
integrate(8*cos(x)**2*sin(x)**4,x)
Output:
x/2 + 4*sin(x)**5*cos(x)/3 - sin(x)**3*cos(x)/3 - sin(x)*cos(x)/2
Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53 \[ \int 8 \cos ^2(x) \sin ^4(x) \, dx=-\frac {1}{6} \, \sin \left (2 \, x\right )^{3} + \frac {1}{2} \, x - \frac {1}{8} \, \sin \left (4 \, x\right ) \] Input:
integrate(8*cos(x)^2*sin(x)^4,x, algorithm="maxima")
Output:
-1/6*sin(2*x)^3 + 1/2*x - 1/8*sin(4*x)
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int 8 \cos ^2(x) \sin ^4(x) \, dx=\frac {1}{2} \, x + \frac {1}{24} \, \sin \left (6 \, x\right ) - \frac {1}{8} \, \sin \left (4 \, x\right ) - \frac {1}{8} \, \sin \left (2 \, x\right ) \] Input:
integrate(8*cos(x)^2*sin(x)^4,x, algorithm="giac")
Output:
1/2*x + 1/24*sin(6*x) - 1/8*sin(4*x) - 1/8*sin(2*x)
Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int 8 \cos ^2(x) \sin ^4(x) \, dx=\frac {4\,\cos \left (x\right )\,{\sin \left (x\right )}^5}{3}+\frac {x}{2}-\frac {\sin \left (2\,x\right )}{3}+\frac {\sin \left (4\,x\right )}{24} \] Input:
int(8*cos(x)^2*sin(x)^4,x)
Output:
x/2 - sin(2*x)/3 + sin(4*x)/24 + (4*cos(x)*sin(x)^5)/3
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int 8 \cos ^2(x) \sin ^4(x) \, dx=\frac {4 \cos \left (x \right ) \sin \left (x \right )^{5}}{3}-\frac {\cos \left (x \right ) \sin \left (x \right )^{3}}{3}-\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2} \] Input:
int(8*cos(x)^2*sin(x)^4,x)
Output:
(8*cos(x)*sin(x)**5 - 2*cos(x)*sin(x)**3 - 3*cos(x)*sin(x) + 3*x)/6