3.2.44 \(\int x^2 \sin ^3(k x) \, dx\) [144]

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

[Out]

14/9*cos(k*x)/k^3-2/3*x^2*cos(k*x)/k-2/27*cos(k*x)^3/k^3+4/3*x*sin(k*x)/k^2-1/3*x^2*cos(k*x)*sin(k*x)^2/k+2/9*
x*sin(k*x)^3/k^2

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Rubi [A]
time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3392, 3377, 2718, 2713} \begin {gather*} -\frac {2 \cos ^3(k x)}{27 k^3}+\frac {14 \cos (k x)}{9 k^3}+\frac {2 x \sin ^3(k x)}{9 k^2}+\frac {4 x \sin (k x)}{3 k^2}-\frac {2 x^2 \cos (k x)}{3 k}-\frac {x^2 \sin ^2(k x) \cos (k x)}{3 k} \end {gather*}

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 2713

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]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

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 3392

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]

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 \text {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*}

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Mathematica [A]
time = 0.05, size = 55, normalized size = 0.65 \begin {gather*} \frac {-81 \left (-2+k^2 x^2\right ) \cos (k x)+\left (-2+9 k^2 x^2\right ) \cos (3 k x)-6 k x (-27 \sin (k x)+\sin (3 k x))}{108 k^3} \end {gather*}

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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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 2.98, size = 63, normalized size = 0.74 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {6 k x \left (6+\text {Sin}\left [k x\right ]^2\right ) \text {Sin}\left [k x\right ]+9 k^2 x^2 \text {Cos}\left [k x\right ] \left (-3+\text {Cos}\left [k x\right ]^2\right )+2 \text {Cos}\left [k x\right ] \left (20+\text {Sin}\left [k x\right ]^2\right )}{27 k^3},k\text {!=}0\right \}\right \},0\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Sin[k*x]^3*x^2,x]')

[Out]

Piecewise[{{(6 k x (6 + Sin[k x] ^ 2) Sin[k x] + 9 k ^ 2 x ^ 2 Cos[k x] (-3 + Cos[k x] ^ 2) + 2 Cos[k x] (20 +
Sin[k x] ^ 2)) / (27 k ^ 3), k != 0}}, 0]

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Maple [A]
time = 0.07, size = 64, normalized size = 0.75

method	result	size
risch	\(-\frac {3 \left (x^{2} k^{2}-2\right ) \cos \left (k x \right )}{4 k^{3}}+\frac {3 x \sin \left (k x \right )}{2 k^{2}}+\frac {\left (9 x^{2} k^{2}-2\right ) \cos \left (3 k x \right )}{108 k^{3}}-\frac {x \sin \left (3 k x \right )}{18 k^{2}}\)	\(61\)
derivativedivides	\(\frac {-\frac {k^{2} x^{2} \left (2+\sin ^{2}\left (k x \right )\right ) \cos \left (k x \right )}{3}+\frac {4 \cos \left (k x \right )}{3}+\frac {4 k x \sin \left (k x \right )}{3}+\frac {2 k x \left (\sin ^{3}\left (k x \right )\right )}{9}+\frac {2 \left (2+\sin ^{2}\left (k x \right )\right ) \cos \left (k x \right )}{27}}{k^{3}}\)	\(64\)
default	\(\frac {-\frac {k^{2} x^{2} \left (2+\sin ^{2}\left (k x \right )\right ) \cos \left (k x \right )}{3}+\frac {4 \cos \left (k x \right )}{3}+\frac {4 k x \sin \left (k x \right )}{3}+\frac {2 k x \left (\sin ^{3}\left (k x \right )\right )}{9}+\frac {2 \left (2+\sin ^{2}\left (k x \right )\right ) \cos \left (k x \right )}{27}}{k^{3}}\)	\(64\)
norman	\(\frac {-\frac {2 x^{2}}{3 k}+\frac {80}{27 k^{3}}+\frac {8 x \tan \left (\frac {k x}{2}\right )}{3 k^{2}}+\frac {64 x \left (\tan ^{3}\left (\frac {k x}{2}\right )\right )}{9 k^{2}}+\frac {8 x \left (\tan ^{5}\left (\frac {k x}{2}\right )\right )}{3 k^{2}}-\frac {2 x^{2} \left (\tan ^{2}\left (\frac {k x}{2}\right )\right )}{k}+\frac {2 x^{2} \left (\tan ^{4}\left (\frac {k x}{2}\right )\right )}{k}+\frac {2 x^{2} \left (\tan ^{6}\left (\frac {k x}{2}\right )\right )}{3 k}+\frac {8 \left (\tan ^{4}\left (\frac {k x}{2}\right )\right )}{3 k^{3}}+\frac {56 \left (\tan ^{2}\left (\frac {k x}{2}\right )\right )}{9 k^{3}}}{\left (1+\tan ^{2}\left (\frac {k x}{2}\right )\right )^{3}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(k*x)^3,x,method=_RETURNVERBOSE)

[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 = 0.26, size = 55, normalized size = 0.65 \begin {gather*} -\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 {gather*}

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 = 0.32, size = 59, normalized size = 0.69 \begin {gather*} \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 {gather*}

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 = 0.28, size = 100, normalized size = 1.18 \begin {gather*} \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 {gather*}

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 = 0.00, size = 103, normalized size = 1.21 \begin {gather*} \frac {4 \left (-3 x^{2} k^{2}+6\right ) k^{3} \cos \left (k x\right )}{\left (-4 k^{3}\right )^{2}}+\frac {4\cdot 6 x k k^{3} \sin \left (k x\right )}{\left (-4 k^{3}\right )^{2}}+\frac {108 \left (9 x^{2} k^{2}-2\right ) k^{3} \cos \left (3 k x\right )}{\left (-108 k^{3}\right )^{2}}-\frac {108\cdot 6 x k k^{3} \sin \left (3 k x\right )}{\left (-108 k^{3}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Mupad [B]
time = 0.23, size = 67, normalized size = 0.79 \begin {gather*} \frac {\frac {14\,\cos \left (k\,x\right )}{9}-\frac {2\,{\cos \left (k\,x\right )}^3}{27}+k\,\left (\frac {14\,x\,\sin \left (k\,x\right )}{9}-\frac {2\,x\,{\cos \left (k\,x\right )}^2\,\sin \left (k\,x\right )}{9}\right )+k^2\,\left (\frac {x^2\,{\cos \left (k\,x\right )}^3}{3}-x^2\,\cos \left (k\,x\right )\right )}{k^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((14*cos(k*x))/9 - (2*cos(k*x)^3)/27 + k*((14*x*sin(k*x))/9 - (2*x*cos(k*x)^2*sin(k*x))/9) + k^2*((x^2*cos(k*x
)^3)/3 - x^2*cos(k*x)))/k^3

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