3.7.59 \(\int \frac {x}{\sqrt {-1+e^{2 x^2}}} \, dx\) [659]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 18, normalized size of antiderivative = 1.00, 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 = {6847, 2320, 65, 209} \begin {gather*} \frac {1}{2} \text {ArcTan}\left (\sqrt {e^{2 x^2}-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[-1 + E^(2*x^2)],x]

[Out]

ArcTan[Sqrt[-1 + E^(2*x^2)]]/2

Rule 65

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 209

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

Rule 2320

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

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Mathematica [A]
time = 0.03, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{2} \tan ^{-1}\left (\sqrt {-1+e^{2 x^2}}\right ) \end {gather*}

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 = 0.04, size = 14, normalized size = 0.78

method	result	size
derivativedivides	\(\frac {\arctan \left (\sqrt {-1+{\mathrm e}^{2 x^{2}}}\right )}{2}\)	\(14\)
default	\(\frac {\arctan \left (\sqrt {-1+{\mathrm e}^{2 x^{2}}}\right )}{2}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{2} \, \arctan \left (\sqrt {e^{\left (2 \, x^{2}\right )} - 1}\right ) \end {gather*}

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 = 0.37, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{2} \, \arctan \left (\sqrt {e^{\left (2 \, x^{2}\right )} - 1}\right ) \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {\left (e^{x^{2}} - 1\right ) \left (e^{x^{2}} + 1\right )}}\, dx \end {gather*}

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 = 4.97, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{2} \, \arctan \left (\sqrt {e^{\left (2 \, x^{2}\right )} - 1}\right ) \end {gather*}

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))

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Mupad [B]
time = 3.71, size = 13, normalized size = 0.72 \begin {gather*} \frac {\mathrm {atan}\left (\sqrt {{\mathrm {e}}^{2\,x^2}-1}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(exp(2*x^2) - 1)^(1/2),x)

[Out]

atan((exp(2*x^2) - 1)^(1/2))/2

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