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

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

[Out]

2*ArcTanh[E^(x/2)/Sqrt[-1 + E^x]]

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Rubi [A]  time = 0.0258523, antiderivative size = 20, normalized size of antiderivative = 1., 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 = {2249, 217, 206} \[ 2 \tanh ^{-1}\left (\frac{e^{x/2}}{\sqrt{e^x-1}}\right ) \]

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 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 217

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 206

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

Rubi steps

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

Mathematica [A]  time = 0.0038406, size = 20, normalized size = 1. \[ 2 \tanh ^{-1}\left (\frac{e^{x/2}}{\sqrt{e^x-1}}\right ) \]

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 = 0.021, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{{\frac{x}{2}}}}{\frac{1}{\sqrt{-1+{{\rm e}^{x}}}}}}\, dx \end{align*}

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 = 0.966361, size = 24, normalized size = 1.2 \begin{align*} 2 \, \log \left (2 \, \sqrt{e^{x} - 1} + 2 \, e^{\left (\frac{1}{2} \, x\right )}\right ) \end{align*}

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 = 1.97761, size = 47, normalized size = 2.35 \begin{align*} -2 \, \log \left (\sqrt{e^{x} - 1} - e^{\left (\frac{1}{2} \, x\right )}\right ) \end{align*}

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 = 0.618287, size = 7, normalized size = 0.35 \begin{align*} 2 \operatorname{acosh}{\left (e^{\frac{x}{2}} \right )} \end{align*}

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 = 1.14971, size = 22, normalized size = 1.1 \begin{align*} -2 \, \log \left (-\sqrt{e^{x} - 1} + e^{\left (\frac{1}{2} \, x\right )}\right ) \end{align*}

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))