3.659 \(\int \frac{x}{\sqrt{-1+e^{2 x^2}}} \, dx\)

Optimal. Leaf size=18 \[ \frac{1}{2} \tan ^{-1}\left (\sqrt{e^{2 x^2}-1}\right ) \]

[Out]

ArcTan[Sqrt[-1 + E^(2*x^2)]]/2

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Rubi [A]  time = 0.0605359, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {6715, 2282, 63, 203} \[ \frac{1}{2} \tan ^{-1}\left (\sqrt{e^{2 x^2}-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[-1 + E^(2*x^2)],x]

[Out]

ArcTan[Sqrt[-1 + E^(2*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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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{x}{\sqrt{-1+e^{2 x^2}}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+e^{2 x}}} \, dx,x,x^2\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} x} \, dx,x,e^{2 x^2}\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-1+e^{2 x^2}}\right )\\ &=\frac{1}{2} \tan ^{-1}\left (\sqrt{-1+e^{2 x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0162912, size = 18, normalized size = 1. \[ \frac{1}{2} \tan ^{-1}\left (\sqrt{e^{2 x^2}-1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[-1 + E^(2*x^2)],x]

[Out]

ArcTan[Sqrt[-1 + E^(2*x^2)]]/2

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Maple [A]  time = 0.06, size = 14, normalized size = 0.8 \begin{align*}{\frac{1}{2}\arctan \left ( \sqrt{-1+{{\rm e}^{2\,{x}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45893, size = 18, normalized size = 1. \begin{align*} \frac{1}{2} \, \arctan \left (\sqrt{e^{\left (2 \, x^{2}\right )} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(sqrt(e^(2*x^2) - 1))

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Fricas [A]  time = 0.789069, size = 45, normalized size = 2.5 \begin{align*} \frac{1}{2} \, \arctan \left (\sqrt{e^{\left (2 \, x^{2}\right )} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(sqrt(e^(2*x^2) - 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\left (e^{x^{2}} - 1\right ) \left (e^{x^{2}} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/sqrt((exp(x**2) - 1)*(exp(x**2) + 1)), x)

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Giac [A]  time = 1.23556, size = 18, normalized size = 1. \begin{align*} \frac{1}{2} \, \arctan \left (\sqrt{e^{\left (2 \, x^{2}\right )} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(sqrt(e^(2*x^2) - 1))