Integrand size = 16, antiderivative size = 34 \[ \int \frac {x^2}{2+e^{-x}+e^x} \, dx=x^2-\frac {x^2}{1+e^x}-2 x \log \left (1+e^x\right )-2 \operatorname {PolyLog}\left (2,-e^x\right ) \] Output:
x^2-x^2/(1+exp(x))-2*x*ln(1+exp(x))-2*polylog(2,-exp(x))
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {x^2}{2+e^{-x}+e^x} \, dx=x \left (\frac {e^x x}{1+e^x}-2 \log \left (1+e^x\right )\right )-2 \operatorname {PolyLog}\left (2,-e^x\right ) \] Input:
Integrate[x^2/(2 + E^(-x) + E^x),x]
Output:
x*((E^x*x)/(1 + E^x) - 2*Log[1 + E^x]) - 2*PolyLog[2, -E^x]
Time = 0.82 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {2697, 7239, 2621, 2615, 2620, 2715, 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.
\(\displaystyle \int \frac {x^2}{e^{-x}+e^x+2} \, dx\) |
\(\Big \downarrow \) 2697 |
\(\displaystyle \int \frac {e^x x^2}{2 e^x+e^{2 x}+1}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^x x^2}{\left (e^x+1\right )^2}dx\) |
\(\Big \downarrow \) 2621 |
\(\displaystyle 2 \int \frac {x}{1+e^x}dx-\frac {x^2}{e^x+1}\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle 2 \left (\frac {x^2}{2}-\int \frac {e^x x}{1+e^x}dx\right )-\frac {x^2}{e^x+1}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle 2 \left (\int \log \left (1+e^x\right )dx+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )-\frac {x^2}{e^x+1}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle 2 \left (\int e^{-x} \log \left (1+e^x\right )de^x+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )-\frac {x^2}{e^x+1}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle 2 \left (-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {x^2}{2}-x \log \left (e^x+1\right )\right )-\frac {x^2}{e^x+1}\) |
Input:
Int[x^2/(2 + E^(-x) + E^x),x]
Output:
-(x^2/(1 + E^x)) + 2*(x^2/2 - x*Log[1 + E^x] - PolyLog[2, -E^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[(((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[(u_)/((a_) + (b_.)*(F_)^(v_) + (c_.)*(F_)^(w_)), x_Symbol] :> Int[u*(F^ v/(c + a*F^v + b*F^(2*v))), x] /; FreeQ[{F, a, b, c}, x] && EqQ[w, -v] && L inearQ[v, x] && If[RationalQ[D[v, x]], GtQ[D[v, x], 0], LtQ[LeafCount[v], L eafCount[w]]]
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[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 = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94
method | result | size |
risch | \(x^{2}-\frac {x^{2}}{1+{\mathrm e}^{x}}-2 x \ln \left (1+{\mathrm e}^{x}\right )-2 \operatorname {polylog}\left (2, -{\mathrm e}^{x}\right )\) | \(32\) |
Input:
int(x^2/(2+exp(-x)+exp(x)),x,method=_RETURNVERBOSE)
Output:
x^2-x^2/(1+exp(x))-2*x*ln(1+exp(x))-2*polylog(2,-exp(x))
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{2+e^{-x}+e^x} \, dx=\frac {x^{2} e^{x} - 2 \, {\left (e^{x} + 1\right )} {\rm Li}_2\left (-e^{x}\right ) - 2 \, {\left (x e^{x} + x\right )} \log \left (e^{x} + 1\right )}{e^{x} + 1} \] Input:
integrate(x^2/(2+exp(-x)+exp(x)),x, algorithm="fricas")
Output:
(x^2*e^x - 2*(e^x + 1)*dilog(-e^x) - 2*(x*e^x + x)*log(e^x + 1))/(e^x + 1)
\[ \int \frac {x^2}{2+e^{-x}+e^x} \, dx=- \frac {x^{2}}{e^{x} + 1} + 2 \int \frac {x}{e^{x} + 1}\, dx \] Input:
integrate(x**2/(2+exp(-x)+exp(x)),x)
Output:
-x**2/(exp(x) + 1) + 2*Integral(x/(exp(x) + 1), x)
Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{2+e^{-x}+e^x} \, dx=x^{2} - 2 \, x \log \left (e^{x} + 1\right ) - \frac {x^{2}}{e^{x} + 1} - 2 \, {\rm Li}_2\left (-e^{x}\right ) \] Input:
integrate(x^2/(2+exp(-x)+exp(x)),x, algorithm="maxima")
Output:
x^2 - 2*x*log(e^x + 1) - x^2/(e^x + 1) - 2*dilog(-e^x)
\[ \int \frac {x^2}{2+e^{-x}+e^x} \, dx=\int { \frac {x^{2}}{e^{\left (-x\right )} + e^{x} + 2} \,d x } \] Input:
integrate(x^2/(2+exp(-x)+exp(x)),x, algorithm="giac")
Output:
integrate(x^2/(e^(-x) + e^x + 2), x)
Timed out. \[ \int \frac {x^2}{2+e^{-x}+e^x} \, dx=\int \frac {x^2}{{\mathrm {e}}^{-x}+{\mathrm {e}}^x+2} \,d x \] Input:
int(x^2/(exp(-x) + exp(x) + 2),x)
Output:
int(x^2/(exp(-x) + exp(x) + 2), x)
\[ \int \frac {x^2}{2+e^{-x}+e^x} \, dx=\frac {2 e^{x} \left (\int \frac {x}{e^{2 x}+2 e^{x}+1}d x \right )-2 e^{x} \mathrm {log}\left (e^{x}+1\right )+2 e^{x} x +2 \left (\int \frac {x}{e^{2 x}+2 e^{x}+1}d x \right )-2 \,\mathrm {log}\left (e^{x}+1\right )-x^{2}}{e^{x}+1} \] Input:
int(x^2/(2+exp(-x)+exp(x)),x)
Output:
(2*e**x*int(x/(e**(2*x) + 2*e**x + 1),x) - 2*e**x*log(e**x + 1) + 2*e**x*x + 2*int(x/(e**(2*x) + 2*e**x + 1),x) - 2*log(e**x + 1) - x**2)/(e**x + 1)