3.3.29 \(\int \cos ^2(x) \sin (3+2 x) \, dx\) [229]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.00, 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 = {4670, 2718} \begin {gather*} \frac {1}{4} x \sin (3)-\frac {1}{4} \cos (2 x+3)-\frac {1}{16} \cos (4 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2*Sin[3 + 2*x],x]

[Out]

-1/4*Cos[3 + 2*x] - Cos[3 + 4*x]/16 + (x*Sin[3])/4

Rule 2718

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

Rule 4670

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]))

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*}

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Mathematica [A]
time = 0.02, size = 28, normalized size = 1.00 \begin {gather*} -\frac {1}{4} \cos (3+2 x)-\frac {1}{16} \cos (3+4 x)+\frac {1}{4} x \sin (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^2*Sin[3 + 2*x],x]

[Out]

-1/4*Cos[3 + 2*x] - Cos[3 + 4*x]/16 + (x*Sin[3])/4

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Maple [A]
time = 0.05, size = 23, normalized size = 0.82

method	result	size
default	\(-\frac {\cos \left (3+2 x \right )}{4}-\frac {\cos \left (3+4 x \right )}{16}+\frac {x \sin \left (3\right )}{4}\)	\(23\)
risch	\(-\frac {\cos \left (3+2 x \right )}{4}-\frac {\cos \left (3+4 x \right )}{16}+\frac {x \sin \left (3\right )}{4}\)	\(23\)
norman	\(\frac {x \left (\tan ^{3}\left (\frac {x}{2}\right )\right )-\frac {\left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{3}+\frac {\left (\tan ^{2}\left (\frac {3}{2}+x \right )\right )}{3}+x \tan \left (\frac {x}{2}\right ) \left (\tan ^{2}\left (\frac {3}{2}+x \right )\right )+\frac {\tan \left (\frac {x}{2}\right ) \tan \left (\frac {3}{2}+x \right )}{3}-\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right ) \tan \left (\frac {3}{2}+x \right )}{3}+\frac {\left (\tan ^{4}\left (\frac {x}{2}\right )\right ) \left (\tan ^{2}\left (\frac {3}{2}+x \right )\right )}{3}-x \tan \left (\frac {x}{2}\right )+\frac {x \tan \left (\frac {3}{2}+x \right )}{2}-\frac {1}{3}-3 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) \tan \left (\frac {3}{2}+x \right )-x \left (\tan ^{3}\left (\frac {x}{2}\right )\right ) \left (\tan ^{2}\left (\frac {3}{2}+x \right )\right )+\frac {x \left (\tan ^{4}\left (\frac {x}{2}\right )\right ) \tan \left (\frac {3}{2}+x \right )}{2}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2} \left (1+\tan ^{2}\left (\frac {3}{2}+x \right )\right )}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 1.25, size = 22, normalized size = 0.79 \begin {gather*} \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 {gather*}

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 = 1.10, size = 32, normalized size = 1.14 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (22) = 44\).
time = 0.33, size = 76, normalized size = 2.71 \begin {gather*} - \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 ^{2}{\left (x \right )} \cos {\left (2 x + 3 \right )}}{2} + \frac {3 \sin {\left (x \right )} \sin {\left (2 x + 3 \right )} \cos {\left (x \right )}}{4} \end {gather*}

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

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Giac [A]
time = 0.49, size = 22, normalized size = 0.79 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.33, size = 22, normalized size = 0.79 \begin {gather*} \frac {x\,\sin \left (3\right )}{4}-\frac {\cos \left (4\,x+3\right )}{16}-\frac {\cos \left (2\,x+3\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x + 3)*cos(x)^2,x)

[Out]

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

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