3.27 \(\int \frac{\log (1+e^x)}{1+e^{2 x}} \, dx\)

Optimal. Leaf size=102 \[ -\text{PolyLog}\left (2,-e^x\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}+\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac{1}{2} \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]

[Out]

-(Log[(1/2 - I/2)*(I - E^x)]*Log[1 + E^x])/2 - (Log[(-1/2 - I/2)*(I + E^x)]*Log[1 + E^x])/2 - PolyLog[2, -E^x]
- PolyLog[2, (1/2 - I/2)*(1 + E^x)]/2 - PolyLog[2, (1/2 + I/2)*(1 + E^x)]/2

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Rubi [A]  time = 0.146247, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {2282, 266, 36, 29, 31, 2416, 2391, 260, 2394, 2393} \[ -\text{PolyLog}\left (2,-e^x\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}+\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac{1}{2} \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[1 + E^x]/(1 + E^(2*x)),x]

[Out]

-(Log[(1/2 - I/2)*(I - E^x)]*Log[1 + E^x])/2 - (Log[(-1/2 - I/2)*(I + E^x)]*Log[1 + E^x])/2 - PolyLog[2, -E^x]
- PolyLog[2, (1/2 - I/2)*(1 + E^x)]/2 - PolyLog[2, (1/2 + I/2)*(1 + E^x)]/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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

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

Rule 29

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

Rule 31

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 260

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

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rubi steps

\begin{align*} \int \frac{\log \left (1+e^x\right )}{1+e^{2 x}} \, dx &=\operatorname{Subst}\left (\int \frac{\log (1+x)}{x \left (1+x^2\right )} \, dx,x,e^x\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{\log (1+x)}{x}-\frac{x \log (1+x)}{1+x^2}\right ) \, dx,x,e^x\right )\\ &=\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^x\right )-\operatorname{Subst}\left (\int \frac{x \log (1+x)}{1+x^2} \, dx,x,e^x\right )\\ &=-\text{Li}_2\left (-e^x\right )-\operatorname{Subst}\left (\int \left (-\frac{\log (1+x)}{2 (i-x)}+\frac{\log (1+x)}{2 (i+x)}\right ) \, dx,x,e^x\right )\\ &=-\text{Li}_2\left (-e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log (1+x)}{i-x} \, dx,x,e^x\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log (1+x)}{i+x} \, dx,x,e^x\right )\\ &=-\frac{1}{2} \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac{1}{2} \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\text{Li}_2\left (-e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) (i-x)\right )}{1+x} \, dx,x,e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) (i+x)\right )}{1+x} \, dx,x,e^x\right )\\ &=-\frac{1}{2} \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac{1}{2} \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\text{Li}_2\left (-e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log \left (1-\left (\frac{1}{2}+\frac{i}{2}\right ) x\right )}{x} \, dx,x,1+e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log \left (1-\left (\frac{1}{2}-\frac{i}{2}\right ) x\right )}{x} \, dx,x,1+e^x\right )\\ &=-\frac{1}{2} \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac{1}{2} \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\text{Li}_2\left (-e^x\right )-\frac{1}{2} \text{Li}_2\left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (1+e^x\right )\right )-\frac{1}{2} \text{Li}_2\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \left (1+e^x\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0171325, size = 102, normalized size = 1. \[ -\text{PolyLog}\left (2,-e^x\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \text{PolyLog}\left (2,\left (\frac{1}{2}+\frac{i}{2}\right ) \left (e^x+1\right )\right )-\frac{1}{2} \log \left (\left (\frac{1}{2}-\frac{i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac{1}{2} \log \left (\left (-\frac{1}{2}-\frac{i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[(1/2 - I/2)*(I - E^x)]*Log[1 + E^x])/2 - (Log[(-1/2 - I/2)*(I + E^x)]*Log[1 + E^x])/2 - PolyLog[2, -E^x]
- PolyLog[2, (1/2 - I/2)*(1 + E^x)]/2 - PolyLog[2, (1/2 + I/2)*(1 + E^x)]/2

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Maple [A]  time = 0.028, size = 83, normalized size = 0.8 \begin{align*} -{\frac{\ln \left ( 1+{{\rm e}^{x}} \right ) }{2}\ln \left ({\frac{1}{2}}-{\frac{{{\rm e}^{x}}}{2}}+{\frac{i}{2}} \left ( 1+{{\rm e}^{x}} \right ) \right ) }-{\frac{\ln \left ( 1+{{\rm e}^{x}} \right ) }{2}\ln \left ({\frac{1}{2}}-{\frac{{{\rm e}^{x}}}{2}}-{\frac{i}{2}} \left ( 1+{{\rm e}^{x}} \right ) \right ) }-{\frac{1}{2}{\it dilog} \left ({\frac{1}{2}}-{\frac{{{\rm e}^{x}}}{2}}+{\frac{i}{2}} \left ( 1+{{\rm e}^{x}} \right ) \right ) }-{\frac{1}{2}{\it dilog} \left ({\frac{1}{2}}-{\frac{{{\rm e}^{x}}}{2}}-{\frac{i}{2}} \left ( 1+{{\rm e}^{x}} \right ) \right ) }-{\it dilog} \left ( 1+{{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*ln(1+exp(x))*ln(1/2-1/2*exp(x)+1/2*I*(1+exp(x)))-1/2*ln(1+exp(x))*ln(1/2-1/2*exp(x)-1/2*I*(1+exp(x)))-1/2
*dilog(1/2-1/2*exp(x)+1/2*I*(1+exp(x)))-1/2*dilog(1/2-1/2*exp(x)-1/2*I*(1+exp(x)))-dilog(1+exp(x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(e^x + 1)/(e^(2*x) + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(e^x + 1)/(e^(2*x) + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (e^{x} + 1 \right )}}{e^{2 x} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(exp(x) + 1)/(exp(2*x) + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(e^x + 1)/(e^(2*x) + 1), x)