3.799 \(\int x \sin ^3(x^2) \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{6} \cos ^3\left (x^2\right )-\frac {\cos \left (x^2\right )}{2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3379, 2633} \[ \frac {1}{6} \cos ^3\left (x^2\right )-\frac {\cos \left (x^2\right )}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cos[x^2]/2 + Cos[x^2]^3/6

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]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ \frac {1}{24} \cos \left (3 x^2\right )-\frac {3 \cos \left (x^2\right )}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[x^2])/8 + Cos[3*x^2]/24

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fricas [A]  time = 0.68, size = 15, normalized size = 0.79 \[ \frac {1}{6} \, \cos \left (x^{2}\right )^{3} - \frac {1}{2} \, \cos \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*cos(x^2)^3 - 1/2*cos(x^2)

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giac [A]  time = 0.14, size = 15, normalized size = 0.79 \[ \frac {1}{6} \, \cos \left (x^{2}\right )^{3} - \frac {1}{2} \, \cos \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*cos(x^2)^3 - 1/2*cos(x^2)

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maple [A]  time = 0.02, size = 15, normalized size = 0.79 \[ -\frac {\left (2+\sin ^{2}\left (x^{2}\right )\right ) \cos \left (x^{2}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 15, normalized size = 0.79 \[ \frac {1}{24} \, \cos \left (3 \, x^{2}\right ) - \frac {3}{8} \, \cos \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*cos(3*x^2) - 3/8*cos(x^2)

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mupad [B]  time = 2.95, size = 14, normalized size = 0.74 \[ \frac {\cos \left (x^2\right )\,\left ({\cos \left (x^2\right )}^2-3\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(x^2)*(cos(x^2)^2 - 3))/6

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sympy [A]  time = 0.48, size = 22, normalized size = 1.16 \[ - \frac {\sin ^{2}{\left (x^{2} \right )} \cos {\left (x^{2} \right )}}{2} - \frac {\cos ^{3}{\left (x^{2} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sin(x**2)**2*cos(x**2)/2 - cos(x**2)**3/3

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