3.823 \(\int \cos (\frac{1}{2} (1+3 x)) \sin ^3(\frac{1}{2} (1+3 x)) \, dx\)

Optimal. Leaf size=16 \[ \frac{1}{6} \sin ^4\left (\frac{3 x}{2}+\frac{1}{2}\right ) \]

[Out]

Sin[1/2 + (3*x)/2]^4/6

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Rubi [A]  time = 0.0161971, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2564, 30} \[ \frac{1}{6} \sin ^4\left (\frac{3 x}{2}+\frac{1}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cos[(1 + 3*x)/2]*Sin[(1 + 3*x)/2]^3,x]

[Out]

Sin[1/2 + (3*x)/2]^4/6

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \cos \left (\frac{1}{2} (1+3 x)\right ) \sin ^3\left (\frac{1}{2} (1+3 x)\right ) \, dx &=\frac{2}{3} \operatorname{Subst}\left (\int x^3 \, dx,x,\sin \left (\frac{1}{2}+\frac{3 x}{2}\right )\right )\\ &=\frac{1}{6} \sin ^4\left (\frac{1}{2}+\frac{3 x}{2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0210782, size = 25, normalized size = 1.56 \[ \frac{1}{2} \left (\frac{1}{24} \cos (6 x+2)-\frac{1}{6} \cos (3 x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[(1 + 3*x)/2]*Sin[(1 + 3*x)/2]^3,x]

[Out]

(-Cos[1 + 3*x]/6 + Cos[2 + 6*x]/24)/2

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Maple [A]  time = 0.006, size = 11, normalized size = 0.7 \begin{align*}{\frac{1}{6} \left ( \sin \left ({\frac{1}{2}}+{\frac{3\,x}{2}} \right ) \right ) ^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(1/2+3/2*x)*sin(1/2+3/2*x)^3,x)

[Out]

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

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Maxima [A]  time = 0.9477, size = 14, normalized size = 0.88 \begin{align*} \frac{1}{6} \, \sin \left (\frac{3}{2} \, x + \frac{1}{2}\right )^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2+3/2*x)*sin(1/2+3/2*x)^3,x, algorithm="maxima")

[Out]

1/6*sin(3/2*x + 1/2)^4

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Fricas [A]  time = 2.326, size = 66, normalized size = 4.12 \begin{align*} \frac{1}{6} \, \cos \left (\frac{3}{2} \, x + \frac{1}{2}\right )^{4} - \frac{1}{3} \, \cos \left (\frac{3}{2} \, x + \frac{1}{2}\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2+3/2*x)*sin(1/2+3/2*x)^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.572874, size = 39, normalized size = 2.44 \begin{align*} - \frac{\sin ^{2}{\left (\frac{3 x}{2} + \frac{1}{2} \right )} \cos ^{2}{\left (\frac{3 x}{2} + \frac{1}{2} \right )}}{3} - \frac{\cos ^{4}{\left (\frac{3 x}{2} + \frac{1}{2} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2+3/2*x)*sin(1/2+3/2*x)**3,x)

[Out]

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

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Giac [A]  time = 1.10547, size = 14, normalized size = 0.88 \begin{align*} \frac{1}{6} \, \sin \left (\frac{3}{2} \, x + \frac{1}{2}\right )^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2+3/2*x)*sin(1/2+3/2*x)^3,x, algorithm="giac")

[Out]

1/6*sin(3/2*x + 1/2)^4