\(\int 4 \cos ^4(x) \sin ^4(x) \, dx\) [835]

Optimal result

Integrand size = 10, antiderivative size = 46 \[ \int 4 \cos ^4(x) \sin ^4(x) \, dx=\frac {3 x}{32}+\frac {3}{32} \cos (x) \sin (x)+\frac {1}{16} \cos ^3(x) \sin (x)-\frac {1}{4} \cos ^5(x) \sin (x)-\frac {1}{2} \cos ^5(x) \sin ^3(x) \]

[Out]

3/32*x+3/32*cos(x)*sin(x)+1/16*cos(x)^3*sin(x)-1/4*cos(x)^5*sin(x)-1/2*cos(x)^5*sin(x)^3

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {12, 2648, 2715, 8} \[ \int 4 \cos ^4(x) \sin ^4(x) \, dx=\frac {3 x}{32}-\frac {1}{2} \sin ^3(x) \cos ^5(x)-\frac {1}{4} \sin (x) \cos ^5(x)+\frac {1}{16} \sin (x) \cos ^3(x)+\frac {3}{32} \sin (x) \cos (x) \]

[In]

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

[Out]

(3*x)/32 + (3*Cos[x]*Sin[x])/32 + (Cos[x]^3*Sin[x])/16 - (Cos[x]^5*Sin[x])/4 - (Cos[x]^5*Sin[x]^3)/2

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.52 \[ \int 4 \cos ^4(x) \sin ^4(x) \, dx=4 \left (\frac {3 x}{128}-\frac {1}{128} \sin (4 x)+\frac {\sin (8 x)}{1024}\right ) \]

[In]

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

[Out]

4*((3*x)/128 - Sin[4*x]/128 + Sin[8*x]/1024)

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.37

method	result	size
risch	\(\frac {3 x}{32}+\frac {\sin \left (8 x \right )}{256}-\frac {\sin \left (4 x \right )}{32}\)	\(17\)
parallelrisch	\(\frac {3 x}{32}+\frac {\sin \left (8 x \right )}{256}-\frac {\sin \left (4 x \right )}{32}\)	\(17\)
default	\(-\frac {\cos \left (x \right )^{5} \sin \left (x \right )^{3}}{2}-\frac {\cos \left (x \right )^{5} \sin \left (x \right )}{4}+\frac {\left (\cos \left (x \right )^{3}+\frac {3 \cos \left (x \right )}{2}\right ) \sin \left (x \right )}{16}+\frac {3 x}{32}\)	\(36\)
norman	\(\frac {\frac {3 x}{32}-\frac {23 \tan \left (\frac {x}{2}\right )^{3}}{16}+\frac {333 \tan \left (\frac {x}{2}\right )^{5}}{16}-\frac {671 \tan \left (\frac {x}{2}\right )^{7}}{16}+\frac {671 \tan \left (\frac {x}{2}\right )^{9}}{16}-\frac {333 \tan \left (\frac {x}{2}\right )^{11}}{16}+\frac {23 \tan \left (\frac {x}{2}\right )^{13}}{16}+\frac {3 \tan \left (\frac {x}{2}\right )^{15}}{16}+\frac {3 x \tan \left (\frac {x}{2}\right )^{2}}{4}+\frac {21 x \tan \left (\frac {x}{2}\right )^{4}}{8}+\frac {21 x \tan \left (\frac {x}{2}\right )^{6}}{4}+\frac {105 x \tan \left (\frac {x}{2}\right )^{8}}{16}+\frac {21 x \tan \left (\frac {x}{2}\right )^{10}}{4}+\frac {21 x \tan \left (\frac {x}{2}\right )^{12}}{8}+\frac {3 x \tan \left (\frac {x}{2}\right )^{14}}{4}+\frac {3 x \tan \left (\frac {x}{2}\right )^{16}}{32}-\frac {3 \tan \left (\frac {x}{2}\right )}{16}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{8}}\)	\(150\)

[In]

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

[Out]

3/32*x+1/256*sin(8*x)-1/32*sin(4*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67 \[ \int 4 \cos ^4(x) \sin ^4(x) \, dx=\frac {1}{32} \, {\left (16 \, \cos \left (x\right )^{7} - 24 \, \cos \left (x\right )^{5} + 2 \, \cos \left (x\right )^{3} + 3 \, \cos \left (x\right )\right )} \sin \left (x\right ) + \frac {3}{32} \, x \]

[In]

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

[Out]

1/32*(16*cos(x)^7 - 24*cos(x)^5 + 2*cos(x)^3 + 3*cos(x))*sin(x) + 3/32*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67 \[ \int 4 \cos ^4(x) \sin ^4(x) \, dx=\frac {3 x}{32} - \frac {\sin ^{3}{\left (2 x \right )} \cos {\left (2 x \right )}}{32} - \frac {3 \sin {\left (2 x \right )} \cos {\left (2 x \right )}}{64} \]

[In]

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

[Out]

3*x/32 - sin(2*x)**3*cos(2*x)/32 - 3*sin(2*x)*cos(2*x)/64

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.35 \[ \int 4 \cos ^4(x) \sin ^4(x) \, dx=\frac {3}{32} \, x + \frac {1}{256} \, \sin \left (8 \, x\right ) - \frac {1}{32} \, \sin \left (4 \, x\right ) \]

[In]

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

[Out]

3/32*x + 1/256*sin(8*x) - 1/32*sin(4*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.35 \[ \int 4 \cos ^4(x) \sin ^4(x) \, dx=\frac {3}{32} \, x + \frac {1}{256} \, \sin \left (8 \, x\right ) - \frac {1}{32} \, \sin \left (4 \, x\right ) \]

[In]

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

[Out]

3/32*x + 1/256*sin(8*x) - 1/32*sin(4*x)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int 4 \cos ^4(x) \sin ^4(x) \, dx=\frac {3\,x}{32}-\frac {\sin \left (2\,x\right )}{16}+\frac {\sin \left (4\,x\right )}{128}+4\,{\sin \left (x\right )}^5\,\left (\frac {{\cos \left (x\right )}^3}{8}+\frac {\cos \left (x\right )}{16}\right ) \]

[In]

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

[Out]

(3*x)/32 - sin(2*x)/16 + sin(4*x)/128 + 4*sin(x)^5*(cos(x)/16 + cos(x)^3/8)