Integrand size = 18, antiderivative size = 72 \[ \int \frac {x^2}{1+2 e^x+e^{2 x}} \, dx=-x^2+\frac {x^2}{1+e^x}+\frac {x^3}{3}+2 x \log \left (1+e^x\right )-x^2 \log \left (1+e^x\right )+2 \operatorname {PolyLog}\left (2,-e^x\right )-2 x \operatorname {PolyLog}\left (2,-e^x\right )+2 \operatorname {PolyLog}\left (3,-e^x\right ) \]
-x^2+x^2/(1+exp(x))+1/3*x^3+2*x*ln(1+exp(x))-x^2*ln(1+exp(x))+2*polylog(2, -exp(x))-2*x*polylog(2,-exp(x))+2*polylog(3,-exp(x))
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{1+2 e^x+e^{2 x}} \, dx=\frac {x^2 \left (e^x (-3+x)+x\right )}{3 \left (1+e^x\right )}-(-2+x) x \log \left (1+e^x\right )-2 (-1+x) \operatorname {PolyLog}\left (2,-e^x\right )+2 \operatorname {PolyLog}\left (3,-e^x\right ) \]
(x^2*(E^x*(-3 + x) + x))/(3*(1 + E^x)) - (-2 + x)*x*Log[1 + E^x] - 2*(-1 + x)*PolyLog[2, -E^x] + 2*PolyLog[3, -E^x]
Time = 0.83 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {7239, 2616, 2615, 2620, 2621, 2615, 2620, 2715, 2838, 3011, 2720, 7143}
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.
| \(\displaystyle \int \frac {x^2}{2 e^x+e^{2 x}+1} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {x^2}{\left (e^x+1\right )^2}dx\) |
| \(\Big \downarrow \) 2616 |
| \(\displaystyle \int \frac {x^2}{1+e^x}dx-\int \frac {e^x x^2}{\left (1+e^x\right )^2}dx\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle -\int \frac {e^x x^2}{\left (1+e^x\right )^2}dx-\int \frac {e^x x^2}{1+e^x}dx+\frac {x^3}{3}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -\int \frac {e^x x^2}{\left (1+e^x\right )^2}dx+2 \int x \log \left (1+e^x\right )dx+\frac {x^3}{3}-x^2 \log \left (e^x+1\right )\) |
| \(\Big \downarrow \) 2621 |
| \(\displaystyle -2 \int \frac {x}{1+e^x}dx+2 \int x \log \left (1+e^x\right )dx+\frac {x^3}{3}+\frac {x^2}{e^x+1}-x^2 \log \left (e^x+1\right )\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle -2 \left (\frac {x^2}{2}-\int \frac {e^x x}{1+e^x}dx\right )+2 \int x \log \left (1+e^x\right )dx+\frac {x^3}{3}+\frac {x^2}{e^x+1}-x^2 \log \left (e^x+1\right )\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -2 \left (\int \log \left (1+e^x\right )dx+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )+2 \int x \log \left (1+e^x\right )dx+\frac {x^3}{3}+\frac {x^2}{e^x+1}-x^2 \log \left (e^x+1\right )\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle -2 \left (\int e^{-x} \log \left (1+e^x\right )de^x+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )+2 \int x \log \left (1+e^x\right )dx+\frac {x^3}{3}+\frac {x^2}{e^x+1}-x^2 \log \left (e^x+1\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle 2 \int x \log \left (1+e^x\right )dx-2 \left (-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )+\frac {x^3}{3}+\frac {x^2}{e^x+1}-x^2 \log \left (e^x+1\right )\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle 2 \left (\int \operatorname {PolyLog}\left (2,-e^x\right )dx-x \operatorname {PolyLog}\left (2,-e^x\right )\right )-2 \left (-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )+\frac {x^3}{3}+\frac {x^2}{e^x+1}-x^2 \log \left (e^x+1\right )\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle 2 \left (\int e^{-x} \operatorname {PolyLog}\left (2,-e^x\right )de^x-x \operatorname {PolyLog}\left (2,-e^x\right )\right )-2 \left (-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )+\frac {x^3}{3}+\frac {x^2}{e^x+1}-x^2 \log \left (e^x+1\right )\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle -2 \left (-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )+2 \left (\operatorname {PolyLog}\left (3,-e^x\right )-x \operatorname {PolyLog}\left (2,-e^x\right )\right )+\frac {x^3}{3}+\frac {x^2}{e^x+1}-x^2 \log \left (e^x+1\right )\) |
x^2/(1 + E^x) + x^3/3 - x^2*Log[1 + E^x] - 2*(x^2/2 - x*Log[1 + E^x] - Pol yLog[2, -E^x]) + 2*(-(x*PolyLog[2, -E^x]) + PolyLog[3, -E^x])
3.6.15.3.1 Defintions of rubi rules used
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[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90
| method | result | size |
| risch | \(-x^{2}+\frac {x^{2}}{1+{\mathrm e}^{x}}+\frac {x^{3}}{3}+2 x \ln \left (1+{\mathrm e}^{x}\right )-x^{2} \ln \left (1+{\mathrm e}^{x}\right )+2 \,\operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )-2 x \,\operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )+2 \,\operatorname {Li}_{3}\left (-{\mathrm e}^{x}\right )\) | \(65\) |
-x^2+x^2/(1+exp(x))+1/3*x^3+2*x*ln(1+exp(x))-x^2*ln(1+exp(x))+2*polylog(2, -exp(x))-2*x*polylog(2,-exp(x))+2*polylog(3,-exp(x))
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {x^2}{1+2 e^x+e^{2 x}} \, dx=\frac {x^{3} - 6 \, {\left ({\left (x - 1\right )} e^{x} + x - 1\right )} {\rm Li}_2\left (-e^{x}\right ) + {\left (x^{3} - 3 \, x^{2}\right )} e^{x} - 3 \, {\left (x^{2} + {\left (x^{2} - 2 \, x\right )} e^{x} - 2 \, x\right )} \log \left (e^{x} + 1\right ) + 6 \, {\left (e^{x} + 1\right )} {\rm polylog}\left (3, -e^{x}\right )}{3 \, {\left (e^{x} + 1\right )}} \]
1/3*(x^3 - 6*((x - 1)*e^x + x - 1)*dilog(-e^x) + (x^3 - 3*x^2)*e^x - 3*(x^ 2 + (x^2 - 2*x)*e^x - 2*x)*log(e^x + 1) + 6*(e^x + 1)*polylog(3, -e^x))/(e ^x + 1)
\[ \int \frac {x^2}{1+2 e^x+e^{2 x}} \, dx=\frac {x^{2}}{e^{x} + 1} + \int \frac {x \left (x - 2\right )}{e^{x} + 1}\, dx \]
Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{1+2 e^x+e^{2 x}} \, dx=\frac {1}{3} \, x^{3} - x^{2} \log \left (e^{x} + 1\right ) - x^{2} - 2 \, x {\rm Li}_2\left (-e^{x}\right ) + 2 \, x \log \left (e^{x} + 1\right ) + \frac {x^{2}}{e^{x} + 1} + 2 \, {\rm Li}_2\left (-e^{x}\right ) + 2 \, {\rm Li}_{3}(-e^{x}) \]
1/3*x^3 - x^2*log(e^x + 1) - x^2 - 2*x*dilog(-e^x) + 2*x*log(e^x + 1) + x^ 2/(e^x + 1) + 2*dilog(-e^x) + 2*polylog(3, -e^x)
\[ \int \frac {x^2}{1+2 e^x+e^{2 x}} \, dx=\int { \frac {x^{2}}{e^{\left (2 \, x\right )} + 2 \, e^{x} + 1} \,d x } \]
Timed out. \[ \int \frac {x^2}{1+2 e^x+e^{2 x}} \, dx=\int \frac {x^2}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1} \,d x \]