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

Optimal. Leaf size=25 \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{e^{-x}+1}}{\sqrt{2}}\right ) \]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[1 + E^(-x)]/Sqrt[2]])

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Rubi [A]  time = 0.0749026, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2282, 1446, 1469, 627, 63, 206} \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{e^{-x}+1}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + E^(-x)]/(-E^(-x) + E^x),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[1 + E^(-x)]/Sqrt[2]])

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 1446

Int[((a_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Int[((d + e*x^n)^q*(c + a
*x^(2*n))^p)/x^(2*n*p), x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[mn2, -2*n] && IntegerQ[p]

Rule 1469

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[Sim
plify[m - n + 1], 0]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

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

Mathematica [B]  time = 0.104817, size = 112, normalized size = 4.48 \[ \frac{e^{x/2} \sqrt{e^{-x}+1} \left (\log \left (1-e^{x/2}\right )-\log \left (e^{x/2}+1\right )+\log \left (\sqrt{2} \sqrt{e^x+1}-e^{x/2}+1\right )-\log \left (\sqrt{2} \sqrt{e^x+1}+e^{x/2}+1\right )\right )}{\sqrt{2} \sqrt{e^x+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + E^(-x)]/(-E^(-x) + E^x),x]

[Out]

(E^(x/2)*Sqrt[1 + E^(-x)]*(Log[1 - E^(x/2)] - Log[1 + E^(x/2)] + Log[1 - E^(x/2) + Sqrt[2]*Sqrt[1 + E^x]] - Lo
g[1 + E^(x/2) + Sqrt[2]*Sqrt[1 + E^x]]))/(Sqrt[2]*Sqrt[1 + E^x])

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Maple [B]  time = 0.019, size = 49, normalized size = 2. \begin{align*} -{\frac{{{\rm e}^{x}}\sqrt{2}}{2}\sqrt{{\frac{1+{{\rm e}^{x}}}{{{\rm e}^{x}}}}}{\it Artanh} \left ({\frac{ \left ( 1+3\,{{\rm e}^{x}} \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ({{\rm e}^{x}} \right ) ^{2}+{{\rm e}^{x}}}}}} \right ){\frac{1}{\sqrt{ \left ( 1+{{\rm e}^{x}} \right ){{\rm e}^{x}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((1+exp(x))/exp(x))^(1/2)*exp(x)/((1+exp(x))*exp(x))^(1/2)*2^(1/2)*arctanh(1/4*(1+3*exp(x))*2^(1/2)/(exp(
x)^2+exp(x))^(1/2))

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Maxima [A]  time = 1.44735, size = 49, normalized size = 1.96 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - \sqrt{e^{\left (-x\right )} + 1}}{\sqrt{2} + \sqrt{e^{\left (-x\right )} + 1}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.1053, size = 103, normalized size = 4.12 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (\frac{2 \, \sqrt{2} \sqrt{e^{x} + 1} e^{\left (\frac{1}{2} \, x\right )} - 3 \, e^{x} - 1}{e^{x} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log((2*sqrt(2)*sqrt(e^x + 1)*e^(1/2*x) - 3*e^x - 1)/(e^x - 1))

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Sympy [A]  time = 3.72265, size = 65, normalized size = 2.6 \begin{align*} 2 \left (\begin{cases} - \frac{\sqrt{2} \operatorname{acoth}{\left (\frac{\sqrt{2} \sqrt{1 + e^{- x}}}{2} \right )}}{2} & \text{for}\: 1 + e^{- x} > 2 \\- \frac{\sqrt{2} \operatorname{atanh}{\left (\frac{\sqrt{2} \sqrt{1 + e^{- x}}}{2} \right )}}{2} & \text{for}\: 1 + e^{- x} < 2 \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*Piecewise((-sqrt(2)*acoth(sqrt(2)*sqrt(1 + exp(-x))/2)/2, 1 + exp(-x) > 2), (-sqrt(2)*atanh(sqrt(2)*sqrt(1 +
exp(-x))/2)/2, 1 + exp(-x) < 2))

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Giac [B]  time = 1.55128, size = 101, normalized size = 4.04 \begin{align*} -\frac{1}{2} \, \sqrt{2} \log \left (\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right ) + \frac{1}{2} \, \sqrt{2} \log \left (\frac{{\left | -2 \, \sqrt{2} + 2 \, \sqrt{e^{\left (2 \, x\right )} + e^{x}} - 2 \, e^{x} + 2 \right |}}{{\left | 2 \, \sqrt{2} + 2 \, \sqrt{e^{\left (2 \, x\right )} + e^{x}} - 2 \, e^{x} + 2 \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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