3.144 \(\int x^2 \sin ^3(k x) \, dx\)

Optimal. Leaf size=85 \[ \frac{2 x \sin ^3(k x)}{9 k^2}+\frac{4 x \sin (k x)}{3 k^2}-\frac{2 \cos ^3(k x)}{27 k^3}+\frac{14 \cos (k x)}{9 k^3}-\frac{2 x^2 \cos (k x)}{3 k}-\frac{x^2 \sin ^2(k x) \cos (k x)}{3 k} \]

[Out]

(14*Cos[k*x])/(9*k^3) - (2*x^2*Cos[k*x])/(3*k) - (2*Cos[k*x]^3)/(27*k^3) + (4*x*Sin[k*x])/(3*k^2) - (x^2*Cos[k
*x]*Sin[k*x]^2)/(3*k) + (2*x*Sin[k*x]^3)/(9*k^2)

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Rubi [A]  time = 0.0663495, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3311, 3296, 2638, 2633} \[ \frac{2 x \sin ^3(k x)}{9 k^2}+\frac{4 x \sin (k x)}{3 k^2}-\frac{2 \cos ^3(k x)}{27 k^3}+\frac{14 \cos (k x)}{9 k^3}-\frac{2 x^2 \cos (k x)}{3 k}-\frac{x^2 \sin ^2(k x) \cos (k x)}{3 k} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[k*x]^3,x]

[Out]

(14*Cos[k*x])/(9*k^3) - (2*x^2*Cos[k*x])/(3*k) - (2*Cos[k*x]^3)/(27*k^3) + (4*x*Sin[k*x])/(3*k^2) - (x^2*Cos[k
*x]*Sin[k*x]^2)/(3*k) + (2*x*Sin[k*x]^3)/(9*k^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 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]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int x^2 \sin ^3(k x) \, dx &=-\frac{x^2 \cos (k x) \sin ^2(k x)}{3 k}+\frac{2 x \sin ^3(k x)}{9 k^2}+\frac{2}{3} \int x^2 \sin (k x) \, dx-\frac{2 \int \sin ^3(k x) \, dx}{9 k^2}\\ &=-\frac{2 x^2 \cos (k x)}{3 k}-\frac{x^2 \cos (k x) \sin ^2(k x)}{3 k}+\frac{2 x \sin ^3(k x)}{9 k^2}+\frac{2 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (k x)\right )}{9 k^3}+\frac{4 \int x \cos (k x) \, dx}{3 k}\\ &=\frac{2 \cos (k x)}{9 k^3}-\frac{2 x^2 \cos (k x)}{3 k}-\frac{2 \cos ^3(k x)}{27 k^3}+\frac{4 x \sin (k x)}{3 k^2}-\frac{x^2 \cos (k x) \sin ^2(k x)}{3 k}+\frac{2 x \sin ^3(k x)}{9 k^2}-\frac{4 \int \sin (k x) \, dx}{3 k^2}\\ &=\frac{14 \cos (k x)}{9 k^3}-\frac{2 x^2 \cos (k x)}{3 k}-\frac{2 \cos ^3(k x)}{27 k^3}+\frac{4 x \sin (k x)}{3 k^2}-\frac{x^2 \cos (k x) \sin ^2(k x)}{3 k}+\frac{2 x \sin ^3(k x)}{9 k^2}\\ \end{align*}

Mathematica [A]  time = 0.0790781, size = 55, normalized size = 0.65 \[ \frac{-81 \left (k^2 x^2-2\right ) \cos (k x)+\left (9 k^2 x^2-2\right ) \cos (3 k x)-6 k x (\sin (3 k x)-27 \sin (k x))}{108 k^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[k*x]^3,x]

[Out]

(-81*(-2 + k^2*x^2)*Cos[k*x] + (-2 + 9*k^2*x^2)*Cos[3*k*x] - 6*k*x*(-27*Sin[k*x] + Sin[3*k*x]))/(108*k^3)

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Maple [A]  time = 0.026, size = 64, normalized size = 0.8 \begin{align*}{\frac{1}{{k}^{3}} \left ( -{\frac{{k}^{2}{x}^{2} \left ( 2+ \left ( \sin \left ( kx \right ) \right ) ^{2} \right ) \cos \left ( kx \right ) }{3}}+{\frac{4\,\cos \left ( kx \right ) }{3}}+{\frac{4\,kx\sin \left ( kx \right ) }{3}}+{\frac{2\,kx \left ( \sin \left ( kx \right ) \right ) ^{3}}{9}}+{\frac{ \left ( 4+2\, \left ( \sin \left ( kx \right ) \right ) ^{2} \right ) \cos \left ( kx \right ) }{27}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(k*x)^3,x)

[Out]

1/k^3*(-1/3*k^2*x^2*(2+sin(k*x)^2)*cos(k*x)+4/3*cos(k*x)+4/3*k*x*sin(k*x)+2/9*k*x*sin(k*x)^3+2/27*(2+sin(k*x)^
2)*cos(k*x))

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Maxima [A]  time = 0.972621, size = 74, normalized size = 0.87 \begin{align*} -\frac{6 \, k x \sin \left (3 \, k x\right ) - 162 \, k x \sin \left (k x\right ) -{\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (3 \, k x\right ) + 81 \,{\left (k^{2} x^{2} - 2\right )} \cos \left (k x\right )}{108 \, k^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(k*x)^3,x, algorithm="maxima")

[Out]

-1/108*(6*k*x*sin(3*k*x) - 162*k*x*sin(k*x) - (9*k^2*x^2 - 2)*cos(3*k*x) + 81*(k^2*x^2 - 2)*cos(k*x))/k^3

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Fricas [A]  time = 1.80226, size = 144, normalized size = 1.69 \begin{align*} \frac{{\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (k x\right )^{3} - 3 \,{\left (9 \, k^{2} x^{2} - 14\right )} \cos \left (k x\right ) - 6 \,{\left (k x \cos \left (k x\right )^{2} - 7 \, k x\right )} \sin \left (k x\right )}{27 \, k^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(k*x)^3,x, algorithm="fricas")

[Out]

1/27*((9*k^2*x^2 - 2)*cos(k*x)^3 - 3*(9*k^2*x^2 - 14)*cos(k*x) - 6*(k*x*cos(k*x)^2 - 7*k*x)*sin(k*x))/k^3

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Sympy [A]  time = 2.19431, size = 100, normalized size = 1.18 \begin{align*} \begin{cases} - \frac{x^{2} \sin ^{2}{\left (k x \right )} \cos{\left (k x \right )}}{k} - \frac{2 x^{2} \cos ^{3}{\left (k x \right )}}{3 k} + \frac{14 x \sin ^{3}{\left (k x \right )}}{9 k^{2}} + \frac{4 x \sin{\left (k x \right )} \cos ^{2}{\left (k x \right )}}{3 k^{2}} + \frac{14 \sin ^{2}{\left (k x \right )} \cos{\left (k x \right )}}{9 k^{3}} + \frac{40 \cos ^{3}{\left (k x \right )}}{27 k^{3}} & \text{for}\: k \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*sin(k*x)**3,x)

[Out]

Piecewise((-x**2*sin(k*x)**2*cos(k*x)/k - 2*x**2*cos(k*x)**3/(3*k) + 14*x*sin(k*x)**3/(9*k**2) + 4*x*sin(k*x)*
cos(k*x)**2/(3*k**2) + 14*sin(k*x)**2*cos(k*x)/(9*k**3) + 40*cos(k*x)**3/(27*k**3), Ne(k, 0)), (0, True))

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Giac [A]  time = 1.10263, size = 81, normalized size = 0.95 \begin{align*} -\frac{x \sin \left (3 \, k x\right )}{18 \, k^{2}} + \frac{3 \, x \sin \left (k x\right )}{2 \, k^{2}} + \frac{{\left (9 \, k^{2} x^{2} - 2\right )} \cos \left (3 \, k x\right )}{108 \, k^{3}} - \frac{3 \,{\left (k^{2} x^{2} - 2\right )} \cos \left (k x\right )}{4 \, k^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(k*x)^3,x, algorithm="giac")

[Out]

-1/18*x*sin(3*k*x)/k^2 + 3/2*x*sin(k*x)/k^2 + 1/108*(9*k^2*x^2 - 2)*cos(3*k*x)/k^3 - 3/4*(k^2*x^2 - 2)*cos(k*x
)/k^3