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

Optimal. Leaf size=20 \[ \frac {\sin \left (x^2\right )}{2}-\frac {1}{2} x^2 \cos \left (x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3379, 3296, 2637} \[ \frac {\sin \left (x^2\right )}{2}-\frac {1}{2} x^2 \cos \left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2*Cos[x^2])/2 + Sin[x^2]/2

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.00 \[ \frac {\sin \left (x^2\right )}{2}-\frac {1}{2} x^2 \cos \left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(x^2*Cos[x^2]) + Sin[x^2]/2

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fricas [A]  time = 0.44, size = 16, normalized size = 0.80 \[ -\frac {1}{2} \, x^{2} \cos \left (x^{2}\right ) + \frac {1}{2} \, \sin \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2*cos(x^2) + 1/2*sin(x^2)

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giac [A]  time = 0.89, size = 16, normalized size = 0.80 \[ -\frac {1}{2} \, x^{2} \cos \left (x^{2}\right ) + \frac {1}{2} \, \sin \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2*cos(x^2) + 1/2*sin(x^2)

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maple [A]  time = 0.00, size = 17, normalized size = 0.85 \[ -\frac {x^{2} \cos \left (x^{2}\right )}{2}+\frac {\sin \left (x^{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.64, size = 16, normalized size = 0.80 \[ -\frac {1}{2} \, x^{2} \cos \left (x^{2}\right ) + \frac {1}{2} \, \sin \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2*cos(x^2) + 1/2*sin(x^2)

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mupad [B]  time = 0.21, size = 16, normalized size = 0.80 \[ \frac {\sin \left (x^2\right )}{2}-\frac {x^2\,\cos \left (x^2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(x^2)/2 - (x^2*cos(x^2))/2

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sympy [A]  time = 0.60, size = 15, normalized size = 0.75 \[ - \frac {x^{2} \cos {\left (x^{2} \right )}}{2} + \frac {\sin {\left (x^{2} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2*cos(x**2)/2 + sin(x**2)/2

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