3.832 \(\int 16 \cos ^2(x) \sin ^2(x) \, dx\)

Optimal. Leaf size=18 \[ 2 x-4 \sin (x) \cos ^3(x)+2 \sin (x) \cos (x) \]

[Out]

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

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Rubi [A]  time = 0.0282526, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {12, 2568, 2635, 8} \[ 2 x-4 \sin (x) \cos ^3(x)+2 \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[16*Cos[x]^2*Sin[x]^2,x]

[Out]

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

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0079291, size = 16, normalized size = 0.89 \[ 4 \left (\frac{x}{2}-\frac{1}{8} \sin (4 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[16*Cos[x]^2*Sin[x]^2,x]

[Out]

4*(x/2 - Sin[4*x]/8)

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Maple [A]  time = 0.005, size = 19, normalized size = 1.1 \begin{align*} 2\,x+2\,\cos \left ( x \right ) \sin \left ( x \right ) -4\, \left ( \cos \left ( x \right ) \right ) ^{3}\sin \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16*cos(x)^2*sin(x)^2,x)

[Out]

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

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Maxima [A]  time = 0.959415, size = 14, normalized size = 0.78 \begin{align*} 2 \, x - \frac{1}{2} \, \sin \left (4 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*cos(x)^2*sin(x)^2,x, algorithm="maxima")

[Out]

2*x - 1/2*sin(4*x)

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Fricas [A]  time = 2.23492, size = 53, normalized size = 2.94 \begin{align*} -2 \,{\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) + 2 \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*cos(x)^2*sin(x)^2,x, algorithm="fricas")

[Out]

-2*(2*cos(x)^3 - cos(x))*sin(x) + 2*x

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Sympy [A]  time = 0.066772, size = 12, normalized size = 0.67 \begin{align*} 2 x - \sin{\left (2 x \right )} \cos{\left (2 x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*cos(x)**2*sin(x)**2,x)

[Out]

2*x - sin(2*x)*cos(2*x)

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Giac [A]  time = 1.07518, size = 14, normalized size = 0.78 \begin{align*} 2 \, x - \frac{1}{2} \, \sin \left (4 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*cos(x)^2*sin(x)^2,x, algorithm="giac")

[Out]

2*x - 1/2*sin(4*x)