3.833 \(\int 8 \cos ^2(x) \sin ^4(x) \, dx\)

Optimal. Leaf size=34 \[ \frac {x}{2}-\frac {4}{3} \sin ^3(x) \cos ^3(x)-\sin (x) \cos ^3(x)+\frac {1}{2} \sin (x) \cos (x) \]

[Out]

1/2*x+1/2*cos(x)*sin(x)-cos(x)^3*sin(x)-4/3*cos(x)^3*sin(x)^3

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {12, 2568, 2635, 8} \[ \frac {x}{2}-\frac {4}{3} \sin ^3(x) \cos ^3(x)-\sin (x) \cos ^3(x)+\frac {1}{2} \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[8*Cos[x]^2*Sin[x]^4,x]

[Out]

x/2 + (Cos[x]*Sin[x])/2 - Cos[x]^3*Sin[x] - (4*Cos[x]^3*Sin[x]^3)/3

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 8 \cos ^2(x) \sin ^4(x) \, dx &=8 \int \cos ^2(x) \sin ^4(x) \, dx\\ &=-\frac {4}{3} \cos ^3(x) \sin ^3(x)+4 \int \cos ^2(x) \sin ^2(x) \, dx\\ &=-\cos ^3(x) \sin (x)-\frac {4}{3} \cos ^3(x) \sin ^3(x)+\int \cos ^2(x) \, dx\\ &=\frac {1}{2} \cos (x) \sin (x)-\cos ^3(x) \sin (x)-\frac {4}{3} \cos ^3(x) \sin ^3(x)+\frac {\int 1 \, dx}{2}\\ &=\frac {x}{2}+\frac {1}{2} \cos (x) \sin (x)-\cos ^3(x) \sin (x)-\frac {4}{3} \cos ^3(x) \sin ^3(x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 32, normalized size = 0.94 \[ 8 \left (\frac {x}{16}-\frac {1}{64} \sin (2 x)-\frac {1}{64} \sin (4 x)+\frac {1}{192} \sin (6 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[8*Cos[x]^2*Sin[x]^4,x]

[Out]

8*(x/16 - Sin[2*x]/64 - Sin[4*x]/64 + Sin[6*x]/192)

________________________________________________________________________________________

fricas [A]  time = 2.39, size = 25, normalized size = 0.74 \[ \frac {1}{6} \, {\left (8 \, \cos \relax (x)^{5} - 14 \, \cos \relax (x)^{3} + 3 \, \cos \relax (x)\right )} \sin \relax (x) + \frac {1}{2} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*cos(x)^2*sin(x)^4,x, algorithm="fricas")

[Out]

1/6*(8*cos(x)^5 - 14*cos(x)^3 + 3*cos(x))*sin(x) + 1/2*x

________________________________________________________________________________________

giac [A]  time = 0.12, size = 22, normalized size = 0.65 \[ \frac {1}{2} \, x + \frac {1}{24} \, \sin \left (6 \, x\right ) - \frac {1}{8} \, \sin \left (4 \, x\right ) - \frac {1}{8} \, \sin \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*cos(x)^2*sin(x)^4,x, algorithm="giac")

[Out]

1/2*x + 1/24*sin(6*x) - 1/8*sin(4*x) - 1/8*sin(2*x)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 29, normalized size = 0.85 \[ \frac {x}{2}+\frac {\cos \relax (x ) \sin \relax (x )}{2}-\left (\cos ^{3}\relax (x )\right ) \sin \relax (x )-\frac {4 \left (\cos ^{3}\relax (x )\right ) \left (\sin ^{3}\relax (x )\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*cos(x)^2*sin(x)^4,x)

[Out]

1/2*x+1/2*cos(x)*sin(x)-cos(x)^3*sin(x)-4/3*cos(x)^3*sin(x)^3

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 18, normalized size = 0.53 \[ -\frac {1}{6} \, \sin \left (2 \, x\right )^{3} + \frac {1}{2} \, x - \frac {1}{8} \, \sin \left (4 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*cos(x)^2*sin(x)^4,x, algorithm="maxima")

[Out]

-1/6*sin(2*x)^3 + 1/2*x - 1/8*sin(4*x)

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 24, normalized size = 0.71 \[ \frac {4\,\cos \relax (x)\,{\sin \relax (x)}^5}{3}+\frac {x}{2}-\frac {\sin \left (2\,x\right )}{3}+\frac {\sin \left (4\,x\right )}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*cos(x)^2*sin(x)^4,x)

[Out]

x/2 - sin(2*x)/3 + sin(4*x)/24 + (4*cos(x)*sin(x)^5)/3

________________________________________________________________________________________

sympy [A]  time = 0.05, size = 32, normalized size = 0.94 \[ \frac {x}{2} + \frac {4 \sin ^{5}{\relax (x )} \cos {\relax (x )}}{3} - \frac {\sin ^{3}{\relax (x )} \cos {\relax (x )}}{3} - \frac {\sin {\relax (x )} \cos {\relax (x )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*cos(x)**2*sin(x)**4,x)

[Out]

x/2 + 4*sin(x)**5*cos(x)/3 - sin(x)**3*cos(x)/3 - sin(x)*cos(x)/2

________________________________________________________________________________________