### 3.25 $$\int x^2 \sin (2 x) \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0226745, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {3296, 2638} $-\frac{1}{2} x^2 \cos (2 x)+\frac{1}{2} x \sin (2 x)+\frac{1}{4} \cos (2 x)$

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 3296

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]

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 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*}

Mathematica [A]  time = 0.0240711, size = 25, normalized size = 0.86 $\frac{1}{2} x \sin (2 x)-\frac{1}{4} \left (2 x^2-1\right ) \cos (2 x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 24, normalized size = 0.8 \begin{align*}{\frac{\cos \left ( 2\,x \right ) }{4}}-{\frac{{x}^{2}\cos \left ( 2\,x \right ) }{2}}+{\frac{x\sin \left ( 2\,x \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.947915, size = 28, normalized size = 0.97 \begin{align*} -\frac{1}{4} \,{\left (2 \, x^{2} - 1\right )} \cos \left (2 \, x\right ) + \frac{1}{2} \, x \sin \left (2 \, x\right ) \end{align*}

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 = 1.93983, size = 59, normalized size = 2.03 \begin{align*} -\frac{1}{4} \,{\left (2 \, x^{2} - 1\right )} \cos \left (2 \, x\right ) + \frac{1}{2} \, x \sin \left (2 \, x\right ) \end{align*}

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 = 0.292341, size = 24, normalized size = 0.83 \begin{align*} - \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{align*}

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 = 1.04342, size = 28, normalized size = 0.97 \begin{align*} -\frac{1}{4} \,{\left (2 \, x^{2} - 1\right )} \cos \left (2 \, x\right ) + \frac{1}{2} \, x \sin \left (2 \, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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