3.1.98 \(\int x^2 \cos (x) \, dx\) [98]

Optimal. Leaf size=16 \[ 2 x \cos (x)-2 \sin (x)+x^2 \sin (x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3377, 2717} \begin {gather*} x^2 \sin (x)-2 \sin (x)+2 x \cos (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*x*Cos[x] - 2*Sin[x] + x^2*Sin[x]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 14, normalized size = 0.88 \begin {gather*} 2 x \cos (x)+\left (-2+x^2\right ) \sin (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*x*Cos[x] + (-2 + x^2)*Sin[x]

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Maple [A]
time = 0.02, size = 17, normalized size = 1.06

method	result	size
risch	\(2 x \cos \left (x \right )+\left (x^{2}-2\right ) \sin \left (x \right )\)	\(15\)
default	\(2 x \cos \left (x \right )-2 \sin \left (x \right )+x^{2} \sin \left (x \right )\)	\(17\)
meijerg	\(4 \sqrt {\pi }\, \left (\frac {x \cos \left (x \right )}{2 \sqrt {\pi }}-\frac {\left (-\frac {3 x^{2}}{2}+3\right ) \sin \left (x \right )}{6 \sqrt {\pi }}\right )\)	\(29\)
norman	\(\frac {2 x -2 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 x^{2} \tan \left (\frac {x}{2}\right )-4 \tan \left (\frac {x}{2}\right )}{1+\tan ^{2}\left (\frac {x}{2}\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.58, size = 14, normalized size = 0.88 \begin {gather*} 2 \, x \cos \left (x\right ) + {\left (x^{2} - 2\right )} \sin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*cos(x) + (x^2 - 2)*sin(x)

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Fricas [A]
time = 0.55, size = 14, normalized size = 0.88 \begin {gather*} 2 \, x \cos \left (x\right ) + {\left (x^{2} - 2\right )} \sin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*cos(x) + (x^2 - 2)*sin(x)

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Sympy [A]
time = 0.09, size = 17, normalized size = 1.06 \begin {gather*} x^{2} \sin {\left (x \right )} + 2 x \cos {\left (x \right )} - 2 \sin {\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*sin(x) + 2*x*cos(x) - 2*sin(x)

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Giac [A]
time = 0.98, size = 14, normalized size = 0.88 \begin {gather*} 2 \, x \cos \left (x\right ) + {\left (x^{2} - 2\right )} \sin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*cos(x) + (x^2 - 2)*sin(x)

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Mupad [B]
time = 0.03, size = 14, normalized size = 0.88 \begin {gather*} \sin \left (x\right )\,\left (x^2-2\right )+2\,x\,\cos \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(x)*(x^2 - 2) + 2*x*cos(x)

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