3.61 \(\int e^{2 x^2} x \cos (2 x^2) \, dx\)

Optimal. Leaf size=35 \[ \frac{1}{8} e^{2 x^2} \sin \left (2 x^2\right )+\frac{1}{8} e^{2 x^2} \cos \left (2 x^2\right ) \]

[Out]

(E^(2*x^2)*Cos[2*x^2])/8 + (E^(2*x^2)*Sin[2*x^2])/8

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Rubi [A]  time = 0.0774638, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {6715, 4433} \[ \frac{1}{8} e^{2 x^2} \sin \left (2 x^2\right )+\frac{1}{8} e^{2 x^2} \cos \left (2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x^2)*x*Cos[2*x^2],x]

[Out]

(E^(2*x^2)*Cos[2*x^2])/8 + (E^(2*x^2)*Sin[2*x^2])/8

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 4433

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*C
os[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin{align*} \int e^{2 x^2} x \cos \left (2 x^2\right ) \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int e^{2 x} \cos (2 x) \, dx,x,x^2\right )\\ &=\frac{1}{8} e^{2 x^2} \cos \left (2 x^2\right )+\frac{1}{8} e^{2 x^2} \sin \left (2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0370865, size = 24, normalized size = 0.69 \[ \frac{1}{8} e^{2 x^2} \left (\sin \left (2 x^2\right )+\cos \left (2 x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x^2)*x*Cos[2*x^2],x]

[Out]

(E^(2*x^2)*(Cos[2*x^2] + Sin[2*x^2]))/8

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Maple [A]  time = 0.011, size = 30, normalized size = 0.9 \begin{align*}{\frac{{{\rm e}^{2\,{x}^{2}}}\cos \left ( 2\,{x}^{2} \right ) }{8}}+{\frac{{{\rm e}^{2\,{x}^{2}}}\sin \left ( 2\,{x}^{2} \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x^2)*x*cos(2*x^2),x)

[Out]

1/8*exp(2*x^2)*cos(2*x^2)+1/8*exp(2*x^2)*sin(2*x^2)

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Maxima [A]  time = 1.00692, size = 39, normalized size = 1.11 \begin{align*} \frac{1}{8} \, \cos \left (2 \, x^{2}\right ) e^{\left (2 \, x^{2}\right )} + \frac{1}{8} \, e^{\left (2 \, x^{2}\right )} \sin \left (2 \, x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*cos(2*x^2)*e^(2*x^2) + 1/8*e^(2*x^2)*sin(2*x^2)

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Fricas [A]  time = 0.460905, size = 72, normalized size = 2.06 \begin{align*} \frac{1}{8} \, \cos \left (2 \, x^{2}\right ) e^{\left (2 \, x^{2}\right )} + \frac{1}{8} \, e^{\left (2 \, x^{2}\right )} \sin \left (2 \, x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*cos(2*x^2)*e^(2*x^2) + 1/8*e^(2*x^2)*sin(2*x^2)

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Sympy [A]  time = 13.7648, size = 29, normalized size = 0.83 \begin{align*} \frac{e^{2 x^{2}} \sin{\left (2 x^{2} \right )}}{8} + \frac{e^{2 x^{2}} \cos{\left (2 x^{2} \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x**2)*x*cos(2*x**2),x)

[Out]

exp(2*x**2)*sin(2*x**2)/8 + exp(2*x**2)*cos(2*x**2)/8

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Giac [A]  time = 1.14067, size = 28, normalized size = 0.8 \begin{align*} \frac{1}{8} \,{\left (\cos \left (2 \, x^{2}\right ) + \sin \left (2 \, x^{2}\right )\right )} e^{\left (2 \, x^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(cos(2*x^2) + sin(2*x^2))*e^(2*x^2)