\(\int \sqrt {1-e^x} \, dx\) [242]

Optimal result

Integrand size = 11, antiderivative size = 28 \[ \int \sqrt {1-e^x} \, dx=2 \sqrt {1-e^x}-2 \text {arctanh}\left (\sqrt {1-e^x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2320, 52, 65, 212} \[ \int \sqrt {1-e^x} \, dx=2 \sqrt {1-e^x}-2 \text {arctanh}\left (\sqrt {1-e^x}\right ) \]

[In]

Int[Sqrt[1 - E^x],x]

[Out]

2*Sqrt[1 - E^x] - 2*ArcTanh[Sqrt[1 - E^x]]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 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 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 {\sqrt {1-x}}{x} \, dx,x,e^x\right ) \\ & = 2 \sqrt {1-e^x}+\text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,e^x\right ) \\ & = 2 \sqrt {1-e^x}-2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-e^x}\right ) \\ & = 2 \sqrt {1-e^x}-2 \text {arctanh}\left (\sqrt {1-e^x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \sqrt {1-e^x} \, dx=2 \sqrt {1-e^x}-2 \text {arctanh}\left (\sqrt {1-e^x}\right ) \]

[In]

Integrate[Sqrt[1 - E^x],x]

[Out]

2*Sqrt[1 - E^x] - 2*ArcTanh[Sqrt[1 - E^x]]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {2 \left (-1+{\mathrm e}^{x}\right )}{\sqrt {1-{\mathrm e}^{x}}}-2 \,\operatorname {arctanh}\left (\sqrt {1-{\mathrm e}^{x}}\right )\)	\(27\)
derivativedivides	\(2 \sqrt {1-{\mathrm e}^{x}}+\ln \left (\sqrt {1-{\mathrm e}^{x}}-1\right )-\ln \left (\sqrt {1-{\mathrm e}^{x}}+1\right )\)	\(36\)
default	\(2 \sqrt {1-{\mathrm e}^{x}}+\ln \left (\sqrt {1-{\mathrm e}^{x}}-1\right )-\ln \left (\sqrt {1-{\mathrm e}^{x}}+1\right )\)	\(36\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \sqrt {1-e^x} \, dx=2 \, \sqrt {-e^{x} + 1} - \log \left (\sqrt {-e^{x} + 1} + 1\right ) + \log \left (\sqrt {-e^{x} + 1} - 1\right ) \]

[In]

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

[Out]

2*sqrt(-e^x + 1) - log(sqrt(-e^x + 1) + 1) + log(sqrt(-e^x + 1) - 1)

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \sqrt {1-e^x} \, dx=2 \sqrt {1 - e^{x}} + \log {\left (\sqrt {1 - e^{x}} - 1 \right )} - \log {\left (\sqrt {1 - e^{x}} + 1 \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \sqrt {1-e^x} \, dx=2 \, \sqrt {-e^{x} + 1} - \log \left (\sqrt {-e^{x} + 1} + 1\right ) + \log \left (\sqrt {-e^{x} + 1} - 1\right ) \]

[In]

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

[Out]

2*sqrt(-e^x + 1) - log(sqrt(-e^x + 1) + 1) + log(sqrt(-e^x + 1) - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \sqrt {1-e^x} \, dx=2 \, \sqrt {-e^{x} + 1} - \log \left (\sqrt {-e^{x} + 1} + 1\right ) + \log \left (-\sqrt {-e^{x} + 1} + 1\right ) \]

[In]

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

[Out]

2*sqrt(-e^x + 1) - log(sqrt(-e^x + 1) + 1) + log(-sqrt(-e^x + 1) + 1)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \sqrt {1-e^x} \, dx=2\,\sqrt {1-{\mathrm {e}}^x}+\frac {2\,{\mathrm {e}}^{-\frac {x}{2}}\,\mathrm {asin}\left ({\mathrm {e}}^{-\frac {x}{2}}\right )\,\sqrt {1-{\mathrm {e}}^x}}{\sqrt {1-{\mathrm {e}}^{-x}}} \]

[In]

int((1 - exp(x))^(1/2),x)

[Out]

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