Integrand size = 16, antiderivative size = 102 \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \]
-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)))
Time = 0.01 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i-e^x\right )\right ) \log \left (1+e^x\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+e^x\right )\right ) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right ) \]
-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
Time = 0.31 (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 |
\(\displaystyle -\operatorname {PolyLog}\left (2,-e^x\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+e^x\right )\right )-\frac {1}{2} \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-e^x+i\right )\right ) \log \left (e^x+1\right )-\frac {1}{2} \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (e^x+i\right )\right ) \log \left (e^x+1\right )\) |
-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
3.1.27.3.1 Defintions of rubi rules used
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]]]
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]
Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\operatorname {dilog}\left (1+{\mathrm e}^{x}\right )-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\ln \left (1+{\mathrm e}^{x}\right ) \ln \left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {1}{2}-\frac {{\mathrm e}^{x}}{2}-\frac {i \left (1+{\mathrm e}^{x}\right )}{2}\right )}{2}\) | \(83\) |
-dilog(1+exp(x))-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)))
\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \]
\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int \frac {\log {\left (e^{x} + 1 \right )}}{e^{2 x} + 1}\, dx \]
\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \]
\[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int { \frac {\log \left (e^{x} + 1\right )}{e^{\left (2 \, x\right )} + 1} \,d x } \]
Timed out. \[ \int \frac {\log \left (1+e^x\right )}{1+e^{2 x}} \, dx=\int \frac {\ln \left ({\mathrm {e}}^x+1\right )}{{\mathrm {e}}^{2\,x}+1} \,d x \]