3.1.25 \(\int x^2 \sin (2 x) \, dx\) [25]

Optimal. Leaf size=29 \[ \frac {1}{4} \cos (2 x)-\frac {1}{2} x^2 \cos (2 x)+\frac {1}{2} x \sin (2 x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3377, 2718} \begin {gather*} -\frac {1}{2} x^2 \cos (2 x)+\frac {1}{2} x \sin (2 x)+\frac {1}{4} \cos (2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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 \sin (2 x) \, dx &=-\frac {1}{2} x^2 \cos (2 x)+\int x \cos (2 x) \, dx\\ &=-\frac {1}{2} x^2 \cos (2 x)+\frac {1}{2} x \sin (2 x)-\frac {1}{2} \int \sin (2 x) \, dx\\ &=\frac {1}{4} \cos (2 x)-\frac {1}{2} x^2 \cos (2 x)+\frac {1}{2} x \sin (2 x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 25, normalized size = 0.86 \begin {gather*} -\frac {1}{4} \left (-1+2 x^2\right ) \cos (2 x)+\frac {1}{2} x \sin (2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathics [A]
time = 1.89, size = 23, normalized size = 0.79 \begin {gather*} \frac {x \text {Sin}\left [2 x\right ]}{2}-\frac {x^2 \text {Cos}\left [2 x\right ]}{2}+\frac {\text {Cos}\left [2 x\right ]}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^2*Sin[2*x],x]')

[Out]

x Sin[2 x] / 2 - x ^ 2 Cos[2 x] / 2 + Cos[2 x] / 4

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Maple [A]
time = 0.02, size = 24, normalized size = 0.83

method	result	size
risch	\(\left (-\frac {x^{2}}{2}+\frac {1}{4}\right ) \cos \left (2 x \right )+\frac {x \sin \left (2 x \right )}{2}\)	\(21\)
derivativedivides	\(\frac {\cos \left (2 x \right )}{4}-\frac {x^{2} \cos \left (2 x \right )}{2}+\frac {x \sin \left (2 x \right )}{2}\)	\(24\)
default	\(\frac {\cos \left (2 x \right )}{4}-\frac {x^{2} \cos \left (2 x \right )}{2}+\frac {x \sin \left (2 x \right )}{2}\)	\(24\)
norman	\(\frac {x \tan \left (x \right )-\frac {x^{2}}{2}+\frac {x^{2} \left (\tan ^{2}\left (x \right )\right )}{2}+\frac {1}{2}}{1+\tan ^{2}\left (x \right )}\)	\(30\)
meijerg	\(\frac {\sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-2 x^{2}+1\right ) \cos \left (2 x \right )}{2 \sqrt {\pi }}+\frac {x \sin \left (2 x \right )}{\sqrt {\pi }}\right )}{2}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 21, normalized size = 0.72 \begin {gather*} -\frac {1}{4} \, {\left (2 \, x^{2} - 1\right )} \cos \left (2 \, x\right ) + \frac {1}{2} \, x \sin \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 21, normalized size = 0.72 \begin {gather*} -\frac {1}{4} \, {\left (2 \, x^{2} - 1\right )} \cos \left (2 \, x\right ) + \frac {1}{2} \, x \sin \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.10, size = 24, normalized size = 0.83 \begin {gather*} - \frac {x^{2} \cos {\left (2 x \right )}}{2} + \frac {x \sin {\left (2 x \right )}}{2} + \frac {\cos {\left (2 x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 25, normalized size = 0.86 \begin {gather*} \frac {1}{8} \left (-4 x^{2}+2\right ) \cos \left (2 x\right )+\frac {4}{8} x \sin \left (2 x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(2*x),x)

[Out]

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

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Mupad [B]
time = 0.03, size = 24, normalized size = 0.83 \begin {gather*} \frac {x\,\sin \left (2\,x\right )}{2}+\left (2\,{\sin \left (x\right )}^2-1\right )\,\left (\frac {x^2}{2}-\frac {1}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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