3.242 \(\int \sqrt{1-e^x} \, dx\)

Optimal. Leaf size=28 \[ 2 \sqrt{1-e^x}-2 \tanh ^{-1}\left (\sqrt{1-e^x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0129139, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {2282, 50, 63, 206} \[ 2 \sqrt{1-e^x}-2 \tanh ^{-1}\left (\sqrt{1-e^x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - E^x],x]

[Out]

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

Rule 2282

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]]]

Rule 50

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 63

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

Mathematica [A]  time = 0.0087402, size = 28, normalized size = 1. \[ 2 \sqrt{1-e^x}-2 \tanh ^{-1}\left (\sqrt{1-e^x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - E^x],x]

[Out]

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

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Maple [A]  time = 0.002, size = 36, normalized size = 1.3 \begin{align*} 2\,\sqrt{1-{{\rm e}^{x}}}+\ln \left ( -1+\sqrt{1-{{\rm e}^{x}}} \right ) -\ln \left ( 1+\sqrt{1-{{\rm e}^{x}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.951086, size = 47, normalized size = 1.68 \begin{align*} 2 \, \sqrt{-e^{x} + 1} - \log \left (\sqrt{-e^{x} + 1} + 1\right ) + \log \left (\sqrt{-e^{x} + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.02845, size = 95, normalized size = 3.39 \begin{align*} 2 \, \sqrt{-e^{x} + 1} - \log \left (\sqrt{-e^{x} + 1} + 1\right ) + \log \left (\sqrt{-e^{x} + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 1.24318, size = 32, normalized size = 1.14 \begin{align*} 2 \sqrt{1 - e^{x}} + \log{\left (\sqrt{1 - e^{x}} - 1 \right )} - \log{\left (\sqrt{1 - e^{x}} + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.06754, size = 50, normalized size = 1.79 \begin{align*} 2 \, \sqrt{-e^{x} + 1} - \log \left (\sqrt{-e^{x} + 1} + 1\right ) + \log \left (-\sqrt{-e^{x} + 1} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)