3.229 \(\int \cos ^2(x) \sin (3+2 x) \, dx\)

Optimal. Leaf size=28 \[ \frac{1}{4} x \sin (3)-\frac{1}{4} \cos (2 x+3)-\frac{1}{16} \cos (4 x+3) \]

[Out]

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

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Rubi [A]  time = 0.0231948, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4574, 2638} \[ \frac{1}{4} x \sin (3)-\frac{1}{4} \cos (2 x+3)-\frac{1}{16} \cos (4 x+3) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2*Sin[3 + 2*x],x]

[Out]

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

Rule 4574

Int[Cos[w_]^(q_.)*Sin[v_]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[Sin[v]^p*Cos[w]^q, x], x] /; IGtQ[p, 0] &&
IGtQ[q, 0] && ((PolynomialQ[v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w],
x]))

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0152, size = 28, normalized size = 1. \[ \frac{1}{4} x \sin (3)-\frac{1}{4} \cos (2 x+3)-\frac{1}{16} \cos (4 x+3) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^2*Sin[3 + 2*x],x]

[Out]

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

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Maple [A]  time = 0.011, size = 23, normalized size = 0.8 \begin{align*} -{\frac{\cos \left ( 3+2\,x \right ) }{4}}-{\frac{\cos \left ( 3+4\,x \right ) }{16}}+{\frac{x\sin \left ( 3 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.95125, size = 30, normalized size = 1.07 \begin{align*} \frac{1}{4} \, x \sin \left (3\right ) - \frac{1}{16} \, \cos \left (4 \, x + 3\right ) - \frac{1}{4} \, \cos \left (2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.86913, size = 116, normalized size = 4.14 \begin{align*} -\frac{1}{2} \, \cos \left (3\right ) \cos \left (x\right )^{4} + \frac{1}{4} \, x \sin \left (3\right ) + \frac{1}{4} \,{\left (2 \, \cos \left (x\right )^{3} \sin \left (3\right ) + \cos \left (x\right ) \sin \left (3\right )\right )} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.26138, size = 75, normalized size = 2.68 \begin{align*} - \frac{x \sin ^{2}{\left (x \right )} \sin{\left (2 x + 3 \right )}}{4} - \frac{x \sin{\left (x \right )} \cos{\left (x \right )} \cos{\left (2 x + 3 \right )}}{2} + \frac{x \sin{\left (2 x + 3 \right )} \cos ^{2}{\left (x \right )}}{4} - \frac{\sin{\left (x \right )} \sin{\left (2 x + 3 \right )} \cos{\left (x \right )}}{4} - \frac{\cos ^{2}{\left (x \right )} \cos{\left (2 x + 3 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.07769, size = 30, normalized size = 1.07 \begin{align*} \frac{1}{4} \, x \sin \left (3\right ) - \frac{1}{16} \, \cos \left (4 \, x + 3\right ) - \frac{1}{4} \, \cos \left (2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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