3.1.75 \(\int \cos ^4(x) \sin ^2(x) \, dx\) [75]

Optimal. Leaf size=34 \[ \frac {x}{16}+\frac {1}{16} \cos (x) \sin (x)+\frac {1}{24} \cos ^3(x) \sin (x)-\frac {1}{6} \cos ^5(x) \sin (x) \]

[Out]

1/16*x+1/16*cos(x)*sin(x)+1/24*cos(x)^3*sin(x)-1/6*cos(x)^5*sin(x)

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2648, 2715, 8} \begin {gather*} \frac {x}{16}-\frac {1}{6} \sin (x) \cos ^5(x)+\frac {1}{24} \sin (x) \cos ^3(x)+\frac {1}{16} \sin (x) \cos (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/16 + (Cos[x]*Sin[x])/16 + (Cos[x]^3*Sin[x])/24 - (Cos[x]^5*Sin[x])/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*} \int \cos ^4(x) \sin ^2(x) \, dx &=-\frac {1}{6} \cos ^5(x) \sin (x)+\frac {1}{6} \int \cos ^4(x) \, dx\\ &=\frac {1}{24} \cos ^3(x) \sin (x)-\frac {1}{6} \cos ^5(x) \sin (x)+\frac {1}{8} \int \cos ^2(x) \, dx\\ &=\frac {1}{16} \cos (x) \sin (x)+\frac {1}{24} \cos ^3(x) \sin (x)-\frac {1}{6} \cos ^5(x) \sin (x)+\frac {\int 1 \, dx}{16}\\ &=\frac {x}{16}+\frac {1}{16} \cos (x) \sin (x)+\frac {1}{24} \cos ^3(x) \sin (x)-\frac {1}{6} \cos ^5(x) \sin (x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 30, normalized size = 0.88 \begin {gather*} \frac {x}{16}+\frac {1}{64} \sin (2 x)-\frac {1}{64} \sin (4 x)-\frac {1}{192} \sin (6 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 26, normalized size = 0.76

method	result	size
risch	\(\frac {x}{16}-\frac {\sin \left (6 x \right )}{192}-\frac {\sin \left (4 x \right )}{64}+\frac {\sin \left (2 x \right )}{64}\)	\(23\)
default	\(-\frac {\left (\cos ^{5}\left (x \right )\right ) \sin \left (x \right )}{6}+\frac {\left (\cos ^{3}\left (x \right )+\frac {3 \cos \left (x \right )}{2}\right ) \sin \left (x \right )}{24}+\frac {x}{16}\)	\(26\)
norman	\(\frac {\frac {x}{16}+\frac {47 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{24}-\frac {13 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4}+\frac {13 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{4}-\frac {47 \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{24}+\frac {\left (\tan ^{11}\left (\frac {x}{2}\right )\right )}{8}+\frac {3 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8}+\frac {15 x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{16}+\frac {5 x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{4}+\frac {15 x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{16}+\frac {3 x \left (\tan ^{10}\left (\frac {x}{2}\right )\right )}{8}+\frac {x \left (\tan ^{12}\left (\frac {x}{2}\right )\right )}{16}-\frac {\tan \left (\frac {x}{2}\right )}{8}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{6}}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*cos(x)^5*sin(x)+1/24*(cos(x)^3+3/2*cos(x))*sin(x)+1/16*x

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Maxima [A]
time = 0.78, size = 18, normalized size = 0.53 \begin {gather*} \frac {1}{48} \, \sin \left (2 \, x\right )^{3} + \frac {1}{16} \, x - \frac {1}{64} \, \sin \left (4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.81, size = 25, normalized size = 0.74 \begin {gather*} -\frac {1}{48} \, {\left (8 \, \cos \left (x\right )^{5} - 2 \, \cos \left (x\right )^{3} - 3 \, \cos \left (x\right )\right )} \sin \left (x\right ) + \frac {1}{16} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(8*cos(x)^5 - 2*cos(x)^3 - 3*cos(x))*sin(x) + 1/16*x

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Sympy [A]
time = 0.01, size = 31, normalized size = 0.91 \begin {gather*} \frac {x}{16} - \frac {\sin {\left (x \right )} \cos ^{5}{\left (x \right )}}{6} + \frac {\sin {\left (x \right )} \cos ^{3}{\left (x \right )}}{24} + \frac {\sin {\left (x \right )} \cos {\left (x \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.83, size = 22, normalized size = 0.65 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.02, size = 26, normalized size = 0.76 \begin {gather*} \left (\frac {{\cos \left (x\right )}^3}{6}+\frac {\cos \left (x\right )}{8}\right )\,{\sin \left (x\right )}^3-\frac {\cos \left (x\right )\,\sin \left (x\right )}{16}+\frac {x}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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