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

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx=2 \text {arctanh}\left (\frac {e^{x/2}}{\sqrt {-1+e^x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2281, 223, 212} \[ \int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx=2 \text {arctanh}\left (\frac {e^{x/2}}{\sqrt {e^x-1}}\right ) \]

[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*} \text {integral}& = 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 \text {arctanh}\left (\frac {e^{x/2}}{\sqrt {-1+e^x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx=-2 \log \left (-e^{x/2}+\sqrt {-1+e^x}\right ) \]

[In]

Integrate[E^(x/2)/Sqrt[-1 + E^x],x]

[Out]

-2*Log[-E^(x/2) + Sqrt[-1 + E^x]]

Maple [F]

\[\int \frac {{\mathrm e}^{\frac {x}{2}}}{\sqrt {-1+{\mathrm e}^{x}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx=-2 \, \log \left (\sqrt {e^{x} - 1} - e^{\left (\frac {1}{2} \, x\right )}\right ) \]

[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))

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx=2 \log {\left (2 \sqrt {e^{x} - 1} + 2 e^{\frac {x}{2}} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx=2 \, \log \left (2 \, \sqrt {e^{x} - 1} + 2 \, e^{\left (\frac {1}{2} \, x\right )}\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx=-2 \, \log \left (-\sqrt {e^{x} - 1} + e^{\left (\frac {1}{2} \, x\right )}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{x/2}}{\sqrt {-1+e^x}} \, dx=\ln \left ({\mathrm {e}}^x+\sqrt {{\mathrm {e}}^x}\,\sqrt {{\mathrm {e}}^x-1}-\frac {1}{2}\right ) \]

[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)