Integrand size = 16, antiderivative size = 44 \[ \int \frac {x}{1+2 e^x+e^{2 x}} \, dx=-x+\frac {x}{1+e^x}+\frac {x^2}{2}+\log \left (1+e^x\right )-x \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right ) \] Output:
-x+x/(1+exp(x))+1/2*x^2+ln(1+exp(x))-x*ln(1+exp(x))-polylog(2,-exp(x))
Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {x}{1+2 e^x+e^{2 x}} \, dx=\frac {1}{2} x \left (-2+\frac {2}{1+e^x}+x\right )-(-1+x) \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right ) \] Input:
Integrate[x/(1 + 2*E^x + E^(2*x)),x]
Output:
(x*(-2 + 2/(1 + E^x) + x))/2 - (-1 + x)*Log[1 + E^x] - PolyLog[2, -E^x]
Time = 0.82 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {7239, 2616, 2615, 2620, 2621, 2715, 2720, 47, 14, 16, 2838}
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.
\(\displaystyle \int \frac {x}{2 e^x+e^{2 x}+1} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x}{\left (e^x+1\right )^2}dx\) |
\(\Big \downarrow \) 2616 |
\(\displaystyle \int \frac {x}{1+e^x}dx-\int \frac {e^x x}{\left (1+e^x\right )^2}dx\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle -\int \frac {e^x x}{\left (1+e^x\right )^2}dx-\int \frac {e^x x}{1+e^x}dx+\frac {x^2}{2}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -\int \frac {e^x x}{\left (1+e^x\right )^2}dx+\int \log \left (1+e^x\right )dx+\frac {x^2}{2}-x \log \left (e^x+1\right )\) |
\(\Big \downarrow \) 2621 |
\(\displaystyle -\int \frac {1}{1+e^x}dx+\int \log \left (1+e^x\right )dx+\frac {x^2}{2}+\frac {x}{e^x+1}-x \log \left (e^x+1\right )\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -\int \frac {1}{1+e^x}dx+\int e^{-x} \log \left (1+e^x\right )de^x+\frac {x^2}{2}+\frac {x}{e^x+1}-x \log \left (e^x+1\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -\int \frac {e^{-x}}{1+e^x}de^x+\int e^{-x} \log \left (1+e^x\right )de^x+\frac {x^2}{2}+\frac {x}{e^x+1}-x \log \left (e^x+1\right )\) |
\(\Big \downarrow \) 47 |
\(\displaystyle -\int e^{-x}de^x+\int \frac {1}{1+e^x}de^x+\int e^{-x} \log \left (1+e^x\right )de^x+\frac {x^2}{2}+\frac {x}{e^x+1}-x \log \left (e^x+1\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \int \frac {1}{1+e^x}de^x+\int e^{-x} \log \left (1+e^x\right )de^x+\frac {x^2}{2}+\frac {x}{e^x+1}-x \log \left (e^x+1\right )-\log \left (e^x\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \int e^{-x} \log \left (1+e^x\right )de^x+\frac {x^2}{2}+\frac {x}{e^x+1}-x \log \left (e^x+1\right )-\log \left (e^x\right )+\log \left (e^x+1\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -\operatorname {PolyLog}\left (2,-e^x\right )+\frac {x^2}{2}+\frac {x}{e^x+1}-x \log \left (e^x+1\right )-\log \left (e^x\right )+\log \left (e^x+1\right )\) |
Input:
Int[x/(1 + 2*E^x + E^(2*x)),x]
Output:
x/(1 + E^x) + x^2/2 - Log[E^x] + Log[1 + E^x] - x*Log[1 + E^x] - PolyLog[2 , -E^x]
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x _))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ b/a Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] , x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[1/a Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Simp[b/a Int[(c + d*x)^m*(F^(g*(e + f*x)))^ n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n }, x] && ILtQ[p, 0] && IGtQ[m, 0]
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*( (e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*((a + b*(F^(g*(e + f*x)))^n)^(p + 1)/(b*f*g*n*(p + 1)*Log [F])), x] - Simp[d*(m/(b*f*g*n*(p + 1)*Log[F])) Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
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[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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {x^{2}}{2}+\ln \left (1+{\mathrm e}^{x}\right )-\frac {x \,{\mathrm e}^{x}}{1+{\mathrm e}^{x}}-\operatorname {dilog}\left (1+{\mathrm e}^{x}\right )-x \ln \left (1+{\mathrm e}^{x}\right )\) | \(38\) |
risch | \(\frac {x}{1+{\mathrm e}^{x}}+\frac {x^{2}}{2}-x \ln \left (1+{\mathrm e}^{x}\right )-\operatorname {polylog}\left (2, -{\mathrm e}^{x}\right )-\ln \left ({\mathrm e}^{x}\right )+\ln \left (1+{\mathrm e}^{x}\right )\) | \(41\) |
Input:
int(x/(1+2*exp(x)+exp(2*x)),x,method=_RETURNVERBOSE)
Output:
1/2*x^2+ln(1+exp(x))-x*exp(x)/(1+exp(x))-dilog(1+exp(x))-x*ln(1+exp(x))
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \frac {x}{1+2 e^x+e^{2 x}} \, dx=\frac {x^{2} - 2 \, {\left (e^{x} + 1\right )} {\rm Li}_2\left (-e^{x}\right ) + {\left (x^{2} - 2 \, x\right )} e^{x} - 2 \, {\left ({\left (x - 1\right )} e^{x} + x - 1\right )} \log \left (e^{x} + 1\right )}{2 \, {\left (e^{x} + 1\right )}} \] Input:
integrate(x/(1+2*exp(x)+exp(2*x)),x, algorithm="fricas")
Output:
1/2*(x^2 - 2*(e^x + 1)*dilog(-e^x) + (x^2 - 2*x)*e^x - 2*((x - 1)*e^x + x - 1)*log(e^x + 1))/(e^x + 1)
\[ \int \frac {x}{1+2 e^x+e^{2 x}} \, dx=\frac {x}{e^{x} + 1} + \int \frac {x - 1}{e^{x} + 1}\, dx \] Input:
integrate(x/(1+2*exp(x)+exp(2*x)),x)
Output:
x/(exp(x) + 1) + Integral((x - 1)/(exp(x) + 1), x)
Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {x}{1+2 e^x+e^{2 x}} \, dx=\frac {1}{2} \, x^{2} - x \log \left (e^{x} + 1\right ) - x + \frac {x}{e^{x} + 1} - {\rm Li}_2\left (-e^{x}\right ) + \log \left (e^{x} + 1\right ) \] Input:
integrate(x/(1+2*exp(x)+exp(2*x)),x, algorithm="maxima")
Output:
1/2*x^2 - x*log(e^x + 1) - x + x/(e^x + 1) - dilog(-e^x) + log(e^x + 1)
\[ \int \frac {x}{1+2 e^x+e^{2 x}} \, dx=\int { \frac {x}{e^{\left (2 \, x\right )} + 2 \, e^{x} + 1} \,d x } \] Input:
integrate(x/(1+2*exp(x)+exp(2*x)),x, algorithm="giac")
Output:
integrate(x/(e^(2*x) + 2*e^x + 1), x)
Timed out. \[ \int \frac {x}{1+2 e^x+e^{2 x}} \, dx=\int \frac {x}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1} \,d x \] Input:
int(x/(exp(2*x) + 2*exp(x) + 1),x)
Output:
int(x/(exp(2*x) + 2*exp(x) + 1), x)
\[ \int \frac {x}{1+2 e^x+e^{2 x}} \, dx=\int \frac {x}{e^{2 x}+2 e^{x}+1}d x \] Input:
int(x/(1+2*exp(x)+exp(2*x)),x)
Output:
int(x/(e**(2*x) + 2*e**x + 1),x)