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

Optimal. Leaf size=10 \[ -\frac {1}{4} \cos \left (2 x^2\right ) \]

[Out]

-1/4*cos(2*x^2)

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3379, 2638} \[ -\frac {1}{4} \cos \left (2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cos[2*x^2]/4

Rule 2638

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

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 \left (2 x^2\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \sin (2 x) \, dx,x,x^2\right )\\ &=-\frac {1}{4} \cos \left (2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \[ -\frac {1}{4} \cos \left (2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Cos[2*x^2]

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fricas [A]  time = 0.70, size = 8, normalized size = 0.80 \[ -\frac {1}{4} \, \cos \left (2 \, x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*cos(2*x^2)

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giac [A]  time = 0.14, size = 8, normalized size = 0.80 \[ -\frac {1}{4} \, \cos \left (2 \, x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*cos(2*x^2)

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maple [A]  time = 0.00, size = 9, normalized size = 0.90 \[ -\frac {\cos \left (2 x^{2}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*cos(2*x^2)

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maxima [A]  time = 0.32, size = 8, normalized size = 0.80 \[ -\frac {1}{4} \, \cos \left (2 \, x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*cos(2*x^2)

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mupad [B]  time = 0.05, size = 8, normalized size = 0.80 \[ \frac {{\sin \left (x^2\right )}^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(x^2)^2/2

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sympy [A]  time = 0.15, size = 8, normalized size = 0.80 \[ - \frac {\cos {\left (2 x^{2} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(2*x**2)/4

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