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]

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

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Rubi [A]  time = 0.07, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, 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 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])

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 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 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 {\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*}

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Mathematica [B]  time = 0.13, 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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fricas [A]  time = 0.39, size = 34, normalized size = 1.36 \[ \frac {1}{2} \, \sqrt {2} \log \left (\frac {2 \, \sqrt {2} \sqrt {e^{x} + 1} e^{\left (\frac {1}{2} \, x\right )} - 3 \, e^{x} - 1}{e^{x} - 1}\right ) \]

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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giac [B]  time = 0.25, size = 75, normalized size = 3.00 \[ -\frac {1}{2} \, \sqrt {2} \log \left (\frac {\sqrt {2} - 1}{\sqrt {2} + 1}\right ) + \frac {1}{2} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 2 \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} - 2 \, e^{x} + 2 \right |}}{{\left | 2 \, \sqrt {2} + 2 \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} - 2 \, e^{x} + 2 \right |}}\right ) \]

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))

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maple [B]  time = 0.03, size = 49, normalized size = 1.96 \[ -\frac {\sqrt {\left ({\mathrm e}^{x}+1\right ) {\mathrm e}^{-x}}\, \sqrt {2}\, \arctanh \left (\frac {\left (3 \,{\mathrm e}^{x}+1\right ) \sqrt {2}}{4 \sqrt {{\mathrm e}^{x}+{\mathrm e}^{2 x}}}\right ) {\mathrm e}^{x}}{2 \sqrt {\left ({\mathrm e}^{x}+1\right ) {\mathrm e}^{x}}} \]

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*((exp(x)+1)/exp(x))^(1/2)*exp(x)/((exp(x)+1)*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 = 1.21, size = 36, normalized size = 1.44 \[ \frac {1}{2} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \sqrt {e^{\left (-x\right )} + 1}}{\sqrt {2} + \sqrt {e^{\left (-x\right )} + 1}}\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ -\int \frac {\sqrt {{\mathrm {e}}^{-x}+1}}{{\mathrm {e}}^{-x}-{\mathrm {e}}^x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.97, size = 65, normalized size = 2.60 \[ 2 \left (\begin {cases} - \frac {\sqrt {2} \operatorname {acoth}{\left (\frac {\sqrt {2} \sqrt {1 + e^{- x}}}{2} \right )}}{2} & \text {for}\: 1 + e^{- x} > 2 \\- \frac {\sqrt {2} \operatorname {atanh}{\left (\frac {\sqrt {2} \sqrt {1 + e^{- x}}}{2} \right )}}{2} & \text {for}\: 1 + e^{- x} < 2 \end {cases}\right ) \]

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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