3.662 \(\int \frac{e^{x^2} x}{1+e^{2 x^2}} \, dx\)

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

[Out]

ArcTan[E^x^2]/2

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Rubi [A]  time = 0.126492, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6715, 2249, 203} \[ \frac{1}{2} \tan ^{-1}\left (e^{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^x^2*x)/(1 + E^(2*x^2)),x]

[Out]

ArcTan[E^x^2]/2

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 203

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

Rubi steps

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

Mathematica [A]  time = 0.0158131, size = 10, normalized size = 1. \[ \frac{1}{2} \tan ^{-1}\left (e^{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^x^2*x)/(1 + E^(2*x^2)),x]

[Out]

ArcTan[E^x^2]/2

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Maple [A]  time = 0.021, size = 8, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ({{\rm e}^{{x}^{2}}} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(exp(x^2))

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Maxima [A]  time = 1.49493, size = 9, normalized size = 0.9 \begin{align*} \frac{1}{2} \, \arctan \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)),x, algorithm="maxima")

[Out]

1/2*arctan(e^(x^2))

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Fricas [A]  time = 0.814956, size = 28, normalized size = 2.8 \begin{align*} \frac{1}{2} \, \arctan \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)),x, algorithm="fricas")

[Out]

1/2*arctan(e^(x^2))

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Sympy [B]  time = 0.129271, size = 17, normalized size = 1.7 \begin{align*} \operatorname{RootSum}{\left (16 z^{2} + 1, \left ( i \mapsto i \log{\left (4 i + e^{x^{2}} \right )} \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)),x)

[Out]

RootSum(16*_z**2 + 1, Lambda(_i, _i*log(4*_i + exp(x**2))))

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Giac [A]  time = 1.25936, size = 9, normalized size = 0.9 \begin{align*} \frac{1}{2} \, \arctan \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)),x, algorithm="giac")

[Out]

1/2*arctan(e^(x^2))