3.241 \(\int \frac {1}{\sqrt {1+e^x}} \, dx\)

Optimal. Leaf size=12 \[ -2 \tanh ^{-1}\left (\sqrt {e^x+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2282, 63, 207} \[ -2 \tanh ^{-1}\left (\sqrt {e^x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + E^x],x]

[Out]

-2*ArcTanh[Sqrt[1 + E^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 207

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

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]]]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 12, normalized size = 1.00 \[ -2 \tanh ^{-1}\left (\sqrt {e^x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 + E^x],x]

[Out]

-2*ArcTanh[Sqrt[1 + E^x]]

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fricas [B]  time = 0.43, size = 21, normalized size = 1.75 \[ -\log \left (\sqrt {e^{x} + 1} + 1\right ) + \log \left (\sqrt {e^{x} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.97, size = 21, normalized size = 1.75 \[ -\log \left (\sqrt {e^{x} + 1} + 1\right ) + \log \left (\sqrt {e^{x} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 10, normalized size = 0.83 \[ -2 \arctanh \left (\sqrt {{\mathrm e}^{x}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.64, size = 21, normalized size = 1.75 \[ -\log \left (\sqrt {e^{x} + 1} + 1\right ) + \log \left (\sqrt {e^{x} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 9, normalized size = 0.75 \[ -2\,\mathrm {atanh}\left (\sqrt {{\mathrm {e}}^x+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(exp(x) + 1)^(1/2),x)

[Out]

-2*atanh((exp(x) + 1)^(1/2))

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sympy [B]  time = 1.41, size = 26, normalized size = 2.17 \[ \log {\left (-1 + \frac {1}{\sqrt {e^{x} + 1}} \right )} - \log {\left (1 + \frac {1}{\sqrt {e^{x} + 1}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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