\(\int \cos ^2(x) \sin ^4(x) \, dx\) [64]

Optimal result

Integrand size = 9, antiderivative size = 36 \[ \int \cos ^2(x) \sin ^4(x) \, dx=\frac {x}{16}+\frac {1}{16} \cos (x) \sin (x)-\frac {1}{8} \cos ^3(x) \sin (x)-\frac {1}{6} \cos ^3(x) \sin ^3(x) \]

[Out]

1/16*x+1/16*cos(x)*sin(x)-1/8*cos(x)^3*sin(x)-1/6*cos(x)^3*sin(x)^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2648, 2715, 8} \[ \int \cos ^2(x) \sin ^4(x) \, dx=\frac {x}{16}-\frac {1}{6} \sin ^3(x) \cos ^3(x)-\frac {1}{8} \sin (x) \cos ^3(x)+\frac {1}{16} \sin (x) \cos (x) \]

[In]

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

[Out]

x/16 + (Cos[x]*Sin[x])/16 - (Cos[x]^3*Sin[x])/8 - (Cos[x]^3*Sin[x]^3)/6

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}{6} \cos ^3(x) \sin ^3(x)+\frac {1}{2} \int \cos ^2(x) \sin ^2(x) \, dx \\ & = -\frac {1}{8} \cos ^3(x) \sin (x)-\frac {1}{6} \cos ^3(x) \sin ^3(x)+\frac {1}{8} \int \cos ^2(x) \, dx \\ & = \frac {1}{16} \cos (x) \sin (x)-\frac {1}{8} \cos ^3(x) \sin (x)-\frac {1}{6} \cos ^3(x) \sin ^3(x)+\frac {\int 1 \, dx}{16} \\ & = \frac {x}{16}+\frac {1}{16} \cos (x) \sin (x)-\frac {1}{8} \cos ^3(x) \sin (x)-\frac {1}{6} \cos ^3(x) \sin ^3(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.83 \[ \int \cos ^2(x) \sin ^4(x) \, dx=\frac {x}{16}-\frac {1}{64} \sin (2 x)-\frac {1}{64} \sin (4 x)+\frac {1}{192} \sin (6 x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.64

method	result	size
risch	\(\frac {x}{16}+\frac {\sin \left (6 x \right )}{192}-\frac {\sin \left (4 x \right )}{64}-\frac {\sin \left (2 x \right )}{64}\)	\(23\)
parallelrisch	\(\frac {x}{16}+\frac {\sin \left (6 x \right )}{192}-\frac {\sin \left (4 x \right )}{64}-\frac {\sin \left (2 x \right )}{64}\)	\(23\)
default	\(\frac {x}{16}+\frac {\cos \left (x \right ) \sin \left (x \right )}{16}-\frac {\left (\cos ^{3}\left (x \right )\right ) \sin \left (x \right )}{8}-\frac {\left (\sin ^{3}\left (x \right )\right ) \left (\cos ^{3}\left (x \right )\right )}{6}\)	\(29\)
norman	\(\frac {\frac {x}{16}-\frac {17 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{24}+\frac {19 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4}-\frac {19 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{4}+\frac {17 \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{24}+\frac {\left (\tan ^{11}\left (\frac {x}{2}\right )\right )}{8}+\frac {3 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8}+\frac {15 x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{16}+\frac {5 x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{4}+\frac {15 x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{16}+\frac {3 x \left (\tan ^{10}\left (\frac {x}{2}\right )\right )}{8}+\frac {x \left (\tan ^{12}\left (\frac {x}{2}\right )\right )}{16}-\frac {\tan \left (\frac {x}{2}\right )}{8}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{6}}\)	\(116\)

[In]

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

[Out]

1/16*x+1/192*sin(6*x)-1/64*sin(4*x)-1/64*sin(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \cos ^2(x) \sin ^4(x) \, dx=\frac {1}{48} \, {\left (8 \, \cos \left (x\right )^{5} - 14 \, \cos \left (x\right )^{3} + 3 \, \cos \left (x\right )\right )} \sin \left (x\right ) + \frac {1}{16} \, x \]

[In]

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

[Out]

1/48*(8*cos(x)^5 - 14*cos(x)^3 + 3*cos(x))*sin(x) + 1/16*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \cos ^2(x) \sin ^4(x) \, dx=\frac {x}{16} + \frac {\sin ^{5}{\left (x \right )} \cos {\left (x \right )}}{6} - \frac {\sin ^{3}{\left (x \right )} \cos {\left (x \right )}}{24} - \frac {\sin {\left (x \right )} \cos {\left (x \right )}}{16} \]

[In]

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

[Out]

x/16 + sin(x)**5*cos(x)/6 - sin(x)**3*cos(x)/24 - sin(x)*cos(x)/16

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.50 \[ \int \cos ^2(x) \sin ^4(x) \, dx=-\frac {1}{48} \, \sin \left (2 \, x\right )^{3} + \frac {1}{16} \, x - \frac {1}{64} \, \sin \left (4 \, x\right ) \]

[In]

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

[Out]

-1/48*sin(2*x)^3 + 1/16*x - 1/64*sin(4*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.61 \[ \int \cos ^2(x) \sin ^4(x) \, dx=\frac {1}{16} \, x + \frac {1}{192} \, \sin \left (6 \, x\right ) - \frac {1}{64} \, \sin \left (4 \, x\right ) - \frac {1}{64} \, \sin \left (2 \, x\right ) \]

[In]

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

[Out]

1/16*x + 1/192*sin(6*x) - 1/64*sin(4*x) - 1/64*sin(2*x)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \cos ^2(x) \sin ^4(x) \, dx=\frac {\cos \left (x\right )\,{\sin \left (x\right )}^5}{6}+\frac {x}{16}-\frac {\sin \left (2\,x\right )}{24}+\frac {\sin \left (4\,x\right )}{192} \]

[In]

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

[Out]

x/16 - sin(2*x)/24 + sin(4*x)/192 + (cos(x)*sin(x)^5)/6