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]

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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 30

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

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]

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*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ \frac {x^2}{4}-\frac {1}{4} x \sin (2 x)-\frac {1}{8} \cos (2 x) \]

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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fricas [A]  time = 0.43, size = 19, normalized size = 0.76 \[ -\frac {1}{2} \, x \cos \relax (x) \sin \relax (x) + \frac {1}{4} \, x^{2} - \frac {1}{4} \, \cos \relax (x)^{2} \]

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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giac [A]  time = 0.01, size = 19, normalized size = 0.76 \[ \frac {1}{4} \, x^{2} - \frac {1}{4} \, x \sin \left (2 \, x\right ) - \frac {1}{8} \, \cos \left (2 \, x\right ) \]

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)

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maple [A]  time = 0.01, size = 25, normalized size = 1.00 \[ -\frac {x^{2}}{4}+\frac {\left (\sin ^{2}\relax (x )\right )}{4}+\left (-\frac {\cos \relax (x ) \sin \relax (x )}{2}+\frac {x}{2}\right ) x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 19, normalized size = 0.76 \[ \frac {1}{4} \, x^{2} - \frac {1}{4} \, x \sin \left (2 \, x\right ) - \frac {1}{8} \, \cos \left (2 \, x\right ) \]

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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mupad [B]  time = 0.09, size = 19, normalized size = 0.76 \[ \frac {{\sin \relax (x)}^2}{4}-\frac {x\,\sin \left (2\,x\right )}{4}+\frac {x^2}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(x)^2/4 - (x*sin(2*x))/4 + x^2/4

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sympy [A]  time = 0.36, size = 36, normalized size = 1.44 \[ \frac {x^{2} \sin ^{2}{\relax (x )}}{4} + \frac {x^{2} \cos ^{2}{\relax (x )}}{4} - \frac {x \sin {\relax (x )} \cos {\relax (x )}}{2} + \frac {\sin ^{2}{\relax (x )}}{4} \]

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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