\(\int \frac {1}{\sqrt {-1+e^x}} \, dx\) [68]

Optimal result

Integrand size = 9, antiderivative size = 12 \[ \int \frac {1}{\sqrt {-1+e^x}} \, dx=2 \arctan \left (\sqrt {-1+e^x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2320, 65, 209} \[ \int \frac {1}{\sqrt {-1+e^x}} \, dx=2 \arctan \left (\sqrt {e^x-1}\right ) \]

[In]

Int[1/Sqrt[-1 + E^x],x]

[Out]

2*ArcTan[Sqrt[-1 + E^x]]

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]]]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {-1+e^x}} \, dx=2 \arctan \left (\sqrt {-1+e^x}\right ) \]

[In]

Integrate[1/Sqrt[-1 + E^x],x]

[Out]

2*ArcTan[Sqrt[-1 + E^x]]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(2 \arctan \left (\sqrt {{\mathrm e}^{x}-1}\right )\)	\(10\)
default	\(2 \arctan \left (\sqrt {{\mathrm e}^{x}-1}\right )\)	\(10\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {-1+e^x}} \, dx=2 \, \arctan \left (\sqrt {e^{x} - 1}\right ) \]

[In]

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

[Out]

2*arctan(sqrt(e^x - 1))

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {-1+e^x}} \, dx=2 \operatorname {atan}{\left (\sqrt {e^{x} - 1} \right )} \]

[In]

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

[Out]

2*atan(sqrt(exp(x) - 1))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {-1+e^x}} \, dx=2 \, \arctan \left (\sqrt {e^{x} - 1}\right ) \]

[In]

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

[Out]

2*arctan(sqrt(e^x - 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {-1+e^x}} \, dx=2 \, \arctan \left (\sqrt {e^{x} - 1}\right ) \]

[In]

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

[Out]

2*arctan(sqrt(e^x - 1))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {-1+e^x}} \, dx=2\,\mathrm {atan}\left (\sqrt {{\mathrm {e}}^x-1}\right ) \]

[In]

int(1/(exp(x) - 1)^(1/2),x)

[Out]

2*atan((exp(x) - 1)^(1/2))