\(\int \frac {\log (1+e^x)}{1+e^{2 x}} \, dx\) [27]

Optimal result

Integrand size = 16, antiderivative size = 102 \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2720, 2863, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2720

\(\displaystyle \int \frac {e^{-x} \log \left (e^x+1\right )}{e^{2 x}+1}de^x\)

\(\Big \downarrow \) 2863

\(\displaystyle \int \left (e^{-x} \log \left (e^x+1\right )-\frac {e^x \log \left (e^x+1\right )}{e^{2 x}+1}\right )de^x\)

\(\Big \downarrow \) 2009

\(\displaystyle -\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 2863

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]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81

method	result	size
default	\(-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\operatorname {dilog}\left (1+{\mathrm e}^{x}\right )\)	\(83\)
risch	\(-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\operatorname {dilog}\left (1+{\mathrm e}^{x}\right )\)	\(83\)

Input:

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

Output:

-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))
 

Fricas [F]

\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int \frac {\log {\left (e^{x} + 1 \right )}}{e^{2 x} + 1}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int \frac {\ln \left ({\mathrm {e}}^x+1\right )}{{\mathrm {e}}^{2\,x}+1} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int \frac {\mathrm {log}\left (e^{x}+1\right )}{e^{2 x}+1}d x \] Input:

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

Output:

int(log(e**x + 1)/(e**(2*x) + 1),x)