\(\int \frac {e^{\frac {x^2}{1+x+x^2}} (2 x+x^2)}{1+2 x+3 x^2+2 x^3+x^4} \, dx\) [3371]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 42, antiderivative size = 17 \[ \int \frac {e^{\frac {x^2}{1+x+x^2}} \left (2 x+x^2\right )}{1+2 x+3 x^2+2 x^3+x^4} \, dx=e^{\frac {x}{x+\frac {x+x^2}{x^2}}} \]

[Out]

exp(x/(x+(x^2+x)/x^2))

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1607, 6820, 6838} \[ \int \frac {e^{\frac {x^2}{1+x+x^2}} \left (2 x+x^2\right )}{1+2 x+3 x^2+2 x^3+x^4} \, dx=e^{\frac {x^2}{x^2+x+1}} \]

[In]

Int[(E^(x^2/(1 + x + x^2))*(2*x + x^2))/(1 + 2*x + 3*x^2 + 2*x^3 + x^4),x]

[Out]

E^(x^2/(1 + x + x^2))

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {x^2}{1+x+x^2}} x (2+x)}{1+2 x+3 x^2+2 x^3+x^4} \, dx \\ & = \int \frac {e^{\frac {x^2}{1+x+x^2}} x (2+x)}{\left (1+x+x^2\right )^2} \, dx \\ & = e^{\frac {x^2}{1+x+x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {x^2}{1+x+x^2}} \left (2 x+x^2\right )}{1+2 x+3 x^2+2 x^3+x^4} \, dx=e^{\frac {x^2}{1+x+x^2}} \]

[In]

Integrate[(E^(x^2/(1 + x + x^2))*(2*x + x^2))/(1 + 2*x + 3*x^2 + 2*x^3 + x^4),x]

[Out]

E^(x^2/(1 + x + x^2))

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method result size
gosper \({\mathrm e}^{\frac {x^{2}}{x^{2}+x +1}}\) \(14\)
risch \({\mathrm e}^{\frac {x^{2}}{x^{2}+x +1}}\) \(14\)
parallelrisch \({\mathrm e}^{\frac {x^{2}}{x^{2}+x +1}}\) \(14\)
norman \(\frac {x \,{\mathrm e}^{\frac {x^{2}}{x^{2}+x +1}}+{\mathrm e}^{\frac {x^{2}}{x^{2}+x +1}} x^{2}+{\mathrm e}^{\frac {x^{2}}{x^{2}+x +1}}}{x^{2}+x +1}\) \(56\)

[In]

int((x^2+2*x)*exp(x^2/(x^2+x+1))/(x^4+2*x^3+3*x^2+2*x+1),x,method=_RETURNVERBOSE)

[Out]

exp(x^2/(x^2+x+1))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {x^2}{1+x+x^2}} \left (2 x+x^2\right )}{1+2 x+3 x^2+2 x^3+x^4} \, dx=e^{\left (\frac {x^{2}}{x^{2} + x + 1}\right )} \]

[In]

integrate((x^2+2*x)*exp(x^2/(x^2+x+1))/(x^4+2*x^3+3*x^2+2*x+1),x, algorithm="fricas")

[Out]

e^(x^2/(x^2 + x + 1))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {e^{\frac {x^2}{1+x+x^2}} \left (2 x+x^2\right )}{1+2 x+3 x^2+2 x^3+x^4} \, dx=e^{\frac {x^{2}}{x^{2} + x + 1}} \]

[In]

integrate((x**2+2*x)*exp(x**2/(x**2+x+1))/(x**4+2*x**3+3*x**2+2*x+1),x)

[Out]

exp(x**2/(x**2 + x + 1))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\frac {x^2}{1+x+x^2}} \left (2 x+x^2\right )}{1+2 x+3 x^2+2 x^3+x^4} \, dx=e^{\left (-\frac {x}{x^{2} + x + 1} - \frac {1}{x^{2} + x + 1} + 1\right )} \]

[In]

integrate((x^2+2*x)*exp(x^2/(x^2+x+1))/(x^4+2*x^3+3*x^2+2*x+1),x, algorithm="maxima")

[Out]

e^(-x/(x^2 + x + 1) - 1/(x^2 + x + 1) + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {x^2}{1+x+x^2}} \left (2 x+x^2\right )}{1+2 x+3 x^2+2 x^3+x^4} \, dx=e^{\left (\frac {x^{2}}{x^{2} + x + 1}\right )} \]

[In]

integrate((x^2+2*x)*exp(x^2/(x^2+x+1))/(x^4+2*x^3+3*x^2+2*x+1),x, algorithm="giac")

[Out]

e^(x^2/(x^2 + x + 1))

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {x^2}{1+x+x^2}} \left (2 x+x^2\right )}{1+2 x+3 x^2+2 x^3+x^4} \, dx={\mathrm {e}}^{\frac {x^2}{x^2+x+1}} \]

[In]

int((exp(x^2/(x + x^2 + 1))*(2*x + x^2))/(2*x + 3*x^2 + 2*x^3 + x^4 + 1),x)

[Out]

exp(x^2/(x + x^2 + 1))