3.1.67 \(\int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx\) [67]

Optimal. Leaf size=20 \[ 2 \tanh ^{-1}\left (\frac {e^{x/2}}{\sqrt {-1+e^x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2281, 223, 212} \begin {gather*} 2 \tanh ^{-1}\left (\frac {e^{x/2}}{\sqrt {e^x-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(x/2)/Sqrt[-1 + E^x],x]

[Out]

2*ArcTanh[E^(x/2)/Sqrt[-1 + E^x]]

Rule 212

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

Rule 223

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

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 20, normalized size = 1.00 \begin {gather*} 2 \tanh ^{-1}\left (\frac {e^{x/2}}{\sqrt {-1+e^x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(x/2)/Sqrt[-1 + E^x],x]

[Out]

2*ArcTanh[E^(x/2)/Sqrt[-1 + E^x]]

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{\frac {x}{2}}}{\sqrt {-1+{\mathrm e}^{x}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 1.21, size = 18, normalized size = 0.90 \begin {gather*} 2 \, \log \left (2 \, \sqrt {e^{x} - 1} + 2 \, e^{\left (\frac {1}{2} \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(2*sqrt(e^x - 1) + 2*e^(1/2*x))

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Fricas [A]
time = 0.64, size = 16, normalized size = 0.80 \begin {gather*} -2 \, \log \left (\sqrt {e^{x} - 1} - e^{\left (\frac {1}{2} \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(sqrt(e^x - 1) - e^(1/2*x))

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Sympy [A]
time = 0.34, size = 7, normalized size = 0.35 \begin {gather*} 2 \operatorname {acosh}{\left (e^{\frac {x}{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*acosh(exp(x/2))

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Giac [A]
time = 0.76, size = 16, normalized size = 0.80 \begin {gather*} -2 \, \log \left (-\sqrt {e^{x} - 1} + e^{\left (\frac {1}{2} \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(-sqrt(e^x - 1) + e^(1/2*x))

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Mupad [B]
time = 0.34, size = 16, normalized size = 0.80 \begin {gather*} \ln \left ({\mathrm {e}}^x+\sqrt {{\mathrm {e}}^x}\,\sqrt {{\mathrm {e}}^x-1}-\frac {1}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x/2)/(exp(x) - 1)^(1/2),x)

[Out]

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

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