3.507 \(\int \frac{1}{-1+e^x+e^{2 x}} \, dx\)

Optimal. Leaf size=56 \[ -x+\frac{1}{10} \left (5+\sqrt{5}\right ) \log \left (2 e^x+1-\sqrt{5}\right )+\frac{1}{10} \left (5-\sqrt{5}\right ) \log \left (2 e^x+1+\sqrt{5}\right ) \]

[Out]

-x + ((5 + Sqrt[5])*Log[1 - Sqrt[5] + 2*E^x])/10 + ((5 - Sqrt[5])*Log[1 + Sqrt[5] + 2*E^x])/10

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Rubi [A]  time = 0.0316725, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2282, 705, 29, 632, 31} \[ -x+\frac{1}{10} \left (5+\sqrt{5}\right ) \log \left (2 e^x+1-\sqrt{5}\right )+\frac{1}{10} \left (5-\sqrt{5}\right ) \log \left (2 e^x+1+\sqrt{5}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + E^x + E^(2*x))^(-1),x]

[Out]

-x + ((5 + Sqrt[5])*Log[1 - Sqrt[5] + 2*E^x])/10 + ((5 - Sqrt[5])*Log[1 + Sqrt[5] + 2*E^x])/10

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 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{-1+e^x+e^{2 x}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{x \left (-1+x+x^2\right )} \, dx,x,e^x\right )\\ &=-\operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,e^x\right )-\operatorname{Subst}\left (\int \frac{-1-x}{-1+x+x^2} \, dx,x,e^x\right )\\ &=-x+\frac{1}{10} \left (5-\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}+\frac{\sqrt{5}}{2}+x} \, dx,x,e^x\right )+\frac{1}{10} \left (5+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}-\frac{\sqrt{5}}{2}+x} \, dx,x,e^x\right )\\ &=-x+\frac{1}{10} \left (5+\sqrt{5}\right ) \log \left (1-\sqrt{5}+2 e^x\right )+\frac{1}{10} \left (5-\sqrt{5}\right ) \log \left (1+\sqrt{5}+2 e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.0249438, size = 44, normalized size = 0.79 \[ -x+\frac{1}{2} \log \left (-e^x-e^{2 x}+1\right )-\frac{\tanh ^{-1}\left (\frac{2 e^x+1}{\sqrt{5}}\right )}{\sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^x + E^(2*x))^(-1),x]

[Out]

-x - ArcTanh[(1 + 2*E^x)/Sqrt[5]]/Sqrt[5] + Log[1 - E^x - E^(2*x)]/2

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Maple [A]  time = 0.006, size = 35, normalized size = 0.6 \begin{align*} -\ln \left ({{\rm e}^{x}} \right ) +{\frac{\ln \left ( -1+{{\rm e}^{x}}+ \left ({{\rm e}^{x}} \right ) ^{2} \right ) }{2}}-{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{ \left ( 1+2\,{{\rm e}^{x}} \right ) \sqrt{5}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(exp(x))+1/2*ln(-1+exp(x)+exp(x)^2)-1/5*5^(1/2)*arctanh(1/5*(1+2*exp(x))*5^(1/2))

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Maxima [A]  time = 1.47232, size = 58, normalized size = 1.04 \begin{align*} \frac{1}{10} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - 2 \, e^{x} - 1}{\sqrt{5} + 2 \, e^{x} + 1}\right ) - x + \frac{1}{2} \, \log \left (e^{\left (2 \, x\right )} + e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*sqrt(5)*log(-(sqrt(5) - 2*e^x - 1)/(sqrt(5) + 2*e^x + 1)) - x + 1/2*log(e^(2*x) + e^x - 1)

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Fricas [A]  time = 1.52817, size = 163, normalized size = 2.91 \begin{align*} \frac{1}{10} \, \sqrt{5} \log \left (-\frac{2 \,{\left (\sqrt{5} - 1\right )} e^{x} + \sqrt{5} - 2 \, e^{\left (2 \, x\right )} - 3}{e^{\left (2 \, x\right )} + e^{x} - 1}\right ) - x + \frac{1}{2} \, \log \left (e^{\left (2 \, x\right )} + e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*sqrt(5)*log(-(2*(sqrt(5) - 1)*e^x + sqrt(5) - 2*e^(2*x) - 3)/(e^(2*x) + e^x - 1)) - x + 1/2*log(e^(2*x) +
e^x - 1)

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Sympy [A]  time = 0.122494, size = 22, normalized size = 0.39 \begin{align*} - x + \operatorname{RootSum}{\left (5 z^{2} - 5 z + 1, \left ( i \mapsto i \log{\left (- 5 i + e^{x} + 3 \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + RootSum(5*_z**2 - 5*_z + 1, Lambda(_i, _i*log(-5*_i + exp(x) + 3)))

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Giac [A]  time = 1.29618, size = 62, normalized size = 1.11 \begin{align*} \frac{1}{10} \, \sqrt{5} \log \left (\frac{{\left | -\sqrt{5} + 2 \, e^{x} + 1 \right |}}{\sqrt{5} + 2 \, e^{x} + 1}\right ) - x + \frac{1}{2} \, \log \left ({\left | e^{\left (2 \, x\right )} + e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*sqrt(5)*log(abs(-sqrt(5) + 2*e^x + 1)/(sqrt(5) + 2*e^x + 1)) - x + 1/2*log(abs(e^(2*x) + e^x - 1))