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]

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

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Rubi [A]  time = 0.08, antiderivative size = 35, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.73, size = 29, normalized size = 0.83 \[ \frac {1}{8} \, \cos \left (2 \, x^{2}\right ) e^{\left (2 \, x^{2}\right )} + \frac {1}{8} \, e^{\left (2 \, x^{2}\right )} \sin \left (2 \, x^{2}\right ) \]

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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giac [A]  time = 0.14, size = 21, normalized size = 0.60 \[ \frac {1}{8} \, {\left (\cos \left (2 \, x^{2}\right ) + \sin \left (2 \, x^{2}\right )\right )} e^{\left (2 \, x^{2}\right )} \]

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)

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maple [A]  time = 0.06, size = 30, normalized size = 0.86 \[ \frac {{\mathrm e}^{2 x^{2}} \cos \left (2 x^{2}\right )}{8}+\frac {{\mathrm e}^{2 x^{2}} \sin \left (2 x^{2}\right )}{8} \]

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 = 0.36, size = 29, normalized size = 0.83 \[ \frac {1}{8} \, \cos \left (2 \, x^{2}\right ) e^{\left (2 \, x^{2}\right )} + \frac {1}{8} \, e^{\left (2 \, x^{2}\right )} \sin \left (2 \, x^{2}\right ) \]

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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mupad [B]  time = 0.08, size = 21, normalized size = 0.60 \[ \frac {{\mathrm {e}}^{2\,x^2}\,\left (\cos \left (2\,x^2\right )+\sin \left (2\,x^2\right )\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.23, size = 29, normalized size = 0.83 \[ \frac {e^{2 x^{2}} \sin {\left (2 x^{2} \right )}}{8} + \frac {e^{2 x^{2}} \cos {\left (2 x^{2} \right )}}{8} \]

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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