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

Optimal. Leaf size=41 \[ \frac{x^3}{6}-\frac{1}{2} x^2 \sin (x) \cos (x)-\frac{x}{4}+\frac{1}{2} x \sin ^2(x)+\frac{1}{4} \sin (x) \cos (x) \]

[Out]

-x/4 + x^3/6 + (Cos[x]*Sin[x])/4 - (x^2*Cos[x]*Sin[x])/2 + (x*Sin[x]^2)/2

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Rubi [A]  time = 0.0304728, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3311, 30, 2635, 8} \[ \frac{x^3}{6}-\frac{1}{2} x^2 \sin (x) \cos (x)-\frac{x}{4}+\frac{1}{2} x \sin ^2(x)+\frac{1}{4} \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/4 + x^3/6 + (Cos[x]*Sin[x])/4 - (x^2*Cos[x]*Sin[x])/2 + (x*Sin[x]^2)/2

Rule 3311

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

Rule 30

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

Rule 2635

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x^2 \sin ^2(x) \, dx &=-\frac{1}{2} x^2 \cos (x) \sin (x)+\frac{1}{2} x \sin ^2(x)+\frac{\int x^2 \, dx}{2}-\frac{1}{2} \int \sin ^2(x) \, dx\\ &=\frac{x^3}{6}+\frac{1}{4} \cos (x) \sin (x)-\frac{1}{2} x^2 \cos (x) \sin (x)+\frac{1}{2} x \sin ^2(x)-\frac{\int 1 \, dx}{4}\\ &=-\frac{x}{4}+\frac{x^3}{6}+\frac{1}{4} \cos (x) \sin (x)-\frac{1}{2} x^2 \cos (x) \sin (x)+\frac{1}{2} x \sin ^2(x)\\ \end{align*}

Mathematica [A]  time = 0.0369686, size = 29, normalized size = 0.71 \[ \frac{1}{24} \left (4 x^3+\left (3-6 x^2\right ) \sin (2 x)-6 x \cos (2 x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^3 - 6*x*Cos[2*x] + (3 - 6*x^2)*Sin[2*x])/24

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Maple [A]  time = 0., size = 37, normalized size = 0.9 \begin{align*}{x}^{2} \left ({\frac{x}{2}}-{\frac{\cos \left ( x \right ) \sin \left ( x \right ) }{2}} \right ) -{\frac{x \left ( \cos \left ( x \right ) \right ) ^{2}}{2}}+{\frac{\cos \left ( x \right ) \sin \left ( x \right ) }{4}}+{\frac{x}{4}}-{\frac{{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*(1/2*x-1/2*cos(x)*sin(x))-1/2*x*cos(x)^2+1/4*cos(x)*sin(x)+1/4*x-1/3*x^3

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Maxima [A]  time = 0.941102, size = 35, normalized size = 0.85 \begin{align*} \frac{1}{6} \, x^{3} - \frac{1}{4} \, x \cos \left (2 \, x\right ) - \frac{1}{8} \,{\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/4*x*cos(2*x) - 1/8*(2*x^2 - 1)*sin(2*x)

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Fricas [A]  time = 1.81543, size = 89, normalized size = 2.17 \begin{align*} \frac{1}{6} \, x^{3} - \frac{1}{2} \, x \cos \left (x\right )^{2} - \frac{1}{4} \,{\left (2 \, x^{2} - 1\right )} \cos \left (x\right ) \sin \left (x\right ) + \frac{1}{4} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/2*x*cos(x)^2 - 1/4*(2*x^2 - 1)*cos(x)*sin(x) + 1/4*x

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Sympy [A]  time = 0.624869, size = 56, normalized size = 1.37 \begin{align*} \frac{x^{3} \sin ^{2}{\left (x \right )}}{6} + \frac{x^{3} \cos ^{2}{\left (x \right )}}{6} - \frac{x^{2} \sin{\left (x \right )} \cos{\left (x \right )}}{2} + \frac{x \sin ^{2}{\left (x \right )}}{4} - \frac{x \cos ^{2}{\left (x \right )}}{4} + \frac{\sin{\left (x \right )} \cos{\left (x \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*sin(x)**2/6 + x**3*cos(x)**2/6 - x**2*sin(x)*cos(x)/2 + x*sin(x)**2/4 - x*cos(x)**2/4 + sin(x)*cos(x)/4

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Giac [A]  time = 1.08424, size = 35, normalized size = 0.85 \begin{align*} \frac{1}{6} \, x^{3} - \frac{1}{4} \, x \cos \left (2 \, x\right ) - \frac{1}{8} \,{\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/4*x*cos(2*x) - 1/8*(2*x^2 - 1)*sin(2*x)