3.8.13 \(\int e^{x^2} \sqrt {1-e^{2 x^2}} x \, dx\) [713]

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]

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

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Rubi [A]
time = 0.11, antiderivative size = 35, normalized size of antiderivative = 1.00, 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 = {6847, 2281, 201, 222} \begin {gather*} \frac {1}{4} \text {ArcSin}\left (e^{x^2}\right )+\frac {1}{4} e^{x^2} \sqrt {1-e^{2 x^2}} \end {gather*}

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 201

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 222

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]

Rule 2281

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 6847

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^{x^2} \sqrt {1-e^{2 x^2}} x \, dx &=\frac {1}{2} \text {Subst}\left (\int e^x \sqrt {1-e^{2 x}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {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} \text {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*}

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Mathematica [A]
time = 0.07, size = 55, normalized size = 1.57 \begin {gather*} \frac {1}{4} e^{x^2} \sqrt {1-e^{2 x^2}}-\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {1-e^{2 x^2}}}{1+e^{x^2}}\right ) \end {gather*}

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

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Maple [A]
time = 0.04, size = 27, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {\arcsin \left ({\mathrm e}^{x^{2}}\right )}{4}+\frac {{\mathrm e}^{x^{2}} \sqrt {1-{\mathrm e}^{2 x^{2}}}}{4}\)	\(27\)
default	\(\frac {\arcsin \left ({\mathrm e}^{x^{2}}\right )}{4}+\frac {{\mathrm e}^{x^{2}} \sqrt {1-{\mathrm e}^{2 x^{2}}}}{4}\)	\(27\)
risch	\(-\frac {{\mathrm e}^{x^{2}} \left (-1+{\mathrm e}^{2 x^{2}}\right )}{4 \sqrt {1-{\mathrm e}^{2 x^{2}}}}+\frac {\arcsin \left ({\mathrm e}^{x^{2}}\right )}{4}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.54, size = 26, normalized size = 0.74 \begin {gather*} \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 {gather*}

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 = 0.36, size = 43, normalized size = 1.23 \begin {gather*} \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 {gather*}

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 [A]
time = 17.33, size = 39, normalized size = 1.11 \begin {gather*} \frac {\begin {cases} \frac {\sqrt {1 - e^{2 x^{2}}} e^{x^{2}}}{2} + \frac {\operatorname {asin}{\left (e^{x^{2}} \right )}}{2} & \text {for}\: e^{x^{2}} > -1 \wedge e^{x^{2}} < 1 \end {cases}}{2} \end {gather*}

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]

Piecewise((sqrt(1 - exp(2*x**2))*exp(x**2)/2 + asin(exp(x**2))/2, (exp(x**2) > -1) & (exp(x**2) < 1)))/2

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Giac [A]
time = 4.54, size = 26, normalized size = 0.74 \begin {gather*} \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 {gather*}

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))

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Mupad [B]
time = 3.65, size = 26, normalized size = 0.74 \begin {gather*} \frac {\mathrm {asin}\left ({\mathrm {e}}^{x^2}\right )}{4}+\frac {{\mathrm {e}}^{x^2}\,\sqrt {1-{\mathrm {e}}^{2\,x^2}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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