3.33 \(\int x \sin ^2(x) \, dx\)

Optimal. Leaf size=25 \[ \frac{x^2}{4}+\frac{\sin ^2(x)}{4}-\frac{1}{2} x \sin (x) \cos (x) \]

[Out]

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

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Rubi [A]  time = 0.0139333, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3310, 30} \[ \frac{x^2}{4}+\frac{\sin ^2(x)}{4}-\frac{1}{2} x \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0111651, size = 25, normalized size = 1. \[ \frac{x^2}{4}-\frac{1}{4} x \sin (2 x)-\frac{1}{8} \cos (2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/4 - Cos[2*x]/8 - (x*Sin[2*x])/4

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Maple [A]  time = 0.005, size = 25, normalized size = 1. \begin{align*} x \left ({\frac{x}{2}}-{\frac{\cos \left ( x \right ) \sin \left ( x \right ) }{2}} \right ) -{\frac{{x}^{2}}{4}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{2}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.952459, size = 26, normalized size = 1.04 \begin{align*} \frac{1}{4} \, x^{2} - \frac{1}{4} \, x \sin \left (2 \, x\right ) - \frac{1}{8} \, \cos \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2 - 1/4*x*sin(2*x) - 1/8*cos(2*x)

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Fricas [A]  time = 0.454715, size = 63, normalized size = 2.52 \begin{align*} -\frac{1}{2} \, x \cos \left (x\right ) \sin \left (x\right ) + \frac{1}{4} \, x^{2} - \frac{1}{4} \, \cos \left (x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.32495, size = 36, normalized size = 1.44 \begin{align*} \frac{x^{2} \sin ^{2}{\left (x \right )}}{4} + \frac{x^{2} \cos ^{2}{\left (x \right )}}{4} - \frac{x \sin{\left (x \right )} \cos{\left (x \right )}}{2} + \frac{\sin ^{2}{\left (x \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1022, size = 26, normalized size = 1.04 \begin{align*} \frac{1}{4} \, x^{2} - \frac{1}{4} \, x \sin \left (2 \, x\right ) - \frac{1}{8} \, \cos \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2 - 1/4*x*sin(2*x) - 1/8*cos(2*x)