3.713 \(\int e^{x^2} \sqrt{1-e^{2 x^2}} x \, dx\)

Optimal. Leaf size=35 \[ \frac{1}{4} e^{x^2} \sqrt{1-e^{2 x^2}}+\frac{1}{4} \sin ^{-1}\left (e^{x^2}\right ) \]

[Out]

(E^x^2*Sqrt[1 - E^(2*x^2)])/4 + ArcSin[E^x^2]/4

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Rubi [A]  time = 0.160398, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6715, 2249, 195, 216} \[ \frac{1}{4} e^{x^2} \sqrt{1-e^{2 x^2}}+\frac{1}{4} \sin ^{-1}\left (e^{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x^2*Sqrt[1 - E^(2*x^2)]*x,x]

[Out]

(E^x^2*Sqrt[1 - E^(2*x^2)])/4 + ArcSin[E^x^2]/4

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 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0275902, size = 32, normalized size = 0.91 \[ \frac{1}{4} \left (e^{x^2} \sqrt{1-e^{2 x^2}}+\sin ^{-1}\left (e^{x^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x^2*Sqrt[1 - E^(2*x^2)]*x,x]

[Out]

(E^x^2*Sqrt[1 - E^(2*x^2)] + ArcSin[E^x^2])/4

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Maple [A]  time = 0.058, size = 27, normalized size = 0.8 \begin{align*}{\frac{{{\rm e}^{{x}^{2}}}}{4}\sqrt{1- \left ({{\rm e}^{{x}^{2}}} \right ) ^{2}}}+{\frac{\arcsin \left ({{\rm e}^{{x}^{2}}} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2)*x*(1-exp(2*x^2))^(1/2),x)

[Out]

1/4*exp(x^2)*(1-exp(x^2)^2)^(1/2)+1/4*arcsin(exp(x^2))

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Maxima [A]  time = 1.46789, size = 35, normalized size = 1. \begin{align*} \frac{1}{4} \, \sqrt{-e^{\left (2 \, x^{2}\right )} + 1} e^{\left (x^{2}\right )} + \frac{1}{4} \, \arcsin \left (e^{\left (x^{2}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x*(1-exp(2*x^2))^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(-e^(2*x^2) + 1)*e^(x^2) + 1/4*arcsin(e^(x^2))

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Fricas [A]  time = 0.932436, size = 113, normalized size = 3.23 \begin{align*} \frac{1}{4} \, \sqrt{-e^{\left (2 \, x^{2}\right )} + 1} e^{\left (x^{2}\right )} - \frac{1}{2} \, \arctan \left ({\left (\sqrt{-e^{\left (2 \, x^{2}\right )} + 1} - 1\right )} e^{\left (-x^{2}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x*(1-exp(2*x^2))^(1/2),x, algorithm="fricas")

[Out]

1/4*sqrt(-e^(2*x^2) + 1)*e^(x^2) - 1/2*arctan((sqrt(-e^(2*x^2) + 1) - 1)*e^(-x^2))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x**2)*x*(1-exp(2*x**2))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.30599, size = 35, normalized size = 1. \begin{align*} \frac{1}{4} \, \sqrt{-e^{\left (2 \, x^{2}\right )} + 1} e^{\left (x^{2}\right )} + \frac{1}{4} \, \arcsin \left (e^{\left (x^{2}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x*(1-exp(2*x^2))^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(-e^(2*x^2) + 1)*e^(x^2) + 1/4*arcsin(e^(x^2))