\(\int \frac {e^{-5+e^x+\frac {e^{-5+e^x} (4+x^3)}{3 x+3 x^2}} (-4-8 x+2 x^3+x^4+e^x (4 x+4 x^2+x^4+x^5))}{3 x^2+6 x^3+3 x^4} \, dx\) [5597]

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

Optimal result

Integrand size = 82, antiderivative size = 26 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x (1+x)}} \]

[Out]

exp(1/3*exp(exp(x)-5)/(1+x)/x*(x^3+4))

Rubi [F]

\[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=\int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx \]

[In]

Int[(E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))*(-4 - 8*x + 2*x^3 + x^4 + E^x*(4*x + 4*x^2 + x^4 +
x^5)))/(3*x^2 + 6*x^3 + 3*x^4),x]

[Out]

Defer[Int][E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2)), x]/3 - Defer[Int][E^(-5 + E^x + x + (E^(-5 +
 E^x)*(4 + x^3))/(3*x + 3*x^2)), x]/3 - (8*Defer[Int][E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))/(-
1 - x), x])/3 - (4*Defer[Int][E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))/x^2, x])/3 + (4*Defer[Int]
[E^(-5 + E^x + x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))/x, x])/3 + Defer[Int][E^(-5 + E^x + x + (E^(-5 + E^
x)*(4 + x^3))/(3*x + 3*x^2))*x, x]/3 + Defer[Int][E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))/(1 + x
)^2, x] - (8*Defer[Int][E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))/(1 + x), x])/3 - Defer[Int][E^(-
5 + E^x + x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))/(1 + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{x^2 \left (3+6 x+3 x^2\right )} \, dx \\ & = \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2 (1+x)^2} \, dx \\ & = \frac {1}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{x^2 (1+x)^2} \, dx \\ & = \frac {1}{3} \int \left (-\frac {4 \exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x^2 (1+x)^2}-\frac {8 \exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x (1+x)^2}+\frac {2 \exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) x}{(1+x)^2}+\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) x^2}{(1+x)^2}+\frac {\exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) \left (4+x^3\right )}{x (1+x)}\right ) \, dx \\ & = \frac {1}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) x^2}{(1+x)^2} \, dx+\frac {1}{3} \int \frac {\exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) \left (4+x^3\right )}{x (1+x)} \, dx+\frac {2}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) x}{(1+x)^2} \, dx-\frac {4}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x^2 (1+x)^2} \, dx-\frac {8}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x (1+x)^2} \, dx \\ & = \frac {1}{3} \int \left (\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )+\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{(1+x)^2}-\frac {2 \exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{1+x}\right ) \, dx+\frac {1}{3} \int \left (-\exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )+\frac {4 \exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x}+\exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) x-\frac {3 \exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{1+x}\right ) \, dx+\frac {2}{3} \int \left (-\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{(1+x)^2}+\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{1+x}\right ) \, dx-\frac {4}{3} \int \left (\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x^2}-\frac {2 \exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x}+\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{(1+x)^2}+\frac {2 \exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{1+x}\right ) \, dx-\frac {8}{3} \int \left (\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{-1-x}+\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x}-\frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{(1+x)^2}\right ) \, dx \\ & = \frac {1}{3} \int \exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) \, dx-\frac {1}{3} \int \exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) \, dx+\frac {1}{3} \int \exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right ) x \, dx+\frac {1}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{(1+x)^2} \, dx-\frac {2}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{(1+x)^2} \, dx-\frac {4}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x^2} \, dx+\frac {4}{3} \int \frac {\exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{x} \, dx-\frac {4}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{(1+x)^2} \, dx-\frac {8}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{-1-x} \, dx+\frac {8}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{(1+x)^2} \, dx-\frac {8}{3} \int \frac {\exp \left (-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{1+x} \, dx-\int \frac {\exp \left (-5+e^x+x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}\right )}{1+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x (1+x)}} \]

[In]

Integrate[(E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))*(-4 - 8*x + 2*x^3 + x^4 + E^x*(4*x + 4*x^2 +
x^4 + x^5)))/(3*x^2 + 6*x^3 + 3*x^4),x]

[Out]

E^((E^(-5 + E^x)*(4 + x^3))/(3*x*(1 + x)))

Maple [A] (verified)

Time = 20.70 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85

method result size
risch \({\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x}-5} \left (x^{3}+4\right )}{3 \left (1+x \right ) x}}\) \(22\)
parallelrisch \({\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x}-5} \left (x^{3}+4\right )}{3 \left (1+x \right ) x}}\) \(22\)

[In]

int(((x^5+x^4+4*x^2+4*x)*exp(x)+x^4+2*x^3-8*x-4)*exp(exp(x)-5)*exp((x^3+4)*exp(exp(x)-5)/(3*x^2+3*x))/(3*x^4+6
*x^3+3*x^2),x,method=_RETURNVERBOSE)

[Out]

exp(1/3*exp(exp(x)-5)/(1+x)/x*(x^3+4))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (21) = 42\).

Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\left (-\frac {15 \, x^{2} - 3 \, {\left (x^{2} + x\right )} e^{x} - {\left (x^{3} + 4\right )} e^{\left (e^{x} - 5\right )} + 15 \, x}{3 \, {\left (x^{2} + x\right )}} - e^{x} + 5\right )} \]

[In]

integrate(((x^5+x^4+4*x^2+4*x)*exp(x)+x^4+2*x^3-8*x-4)*exp(exp(x)-5)*exp((x^3+4)*exp(exp(x)-5)/(3*x^2+3*x))/(3
*x^4+6*x^3+3*x^2),x, algorithm="fricas")

[Out]

e^(-1/3*(15*x^2 - 3*(x^2 + x)*e^x - (x^3 + 4)*e^(e^x - 5) + 15*x)/(x^2 + x) - e^x + 5)

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\frac {\left (x^{3} + 4\right ) e^{e^{x} - 5}}{3 x^{2} + 3 x}} \]

[In]

integrate(((x**5+x**4+4*x**2+4*x)*exp(x)+x**4+2*x**3-8*x-4)*exp(exp(x)-5)*exp((x**3+4)*exp(exp(x)-5)/(3*x**2+3
*x))/(3*x**4+6*x**3+3*x**2),x)

[Out]

exp((x**3 + 4)*exp(exp(x) - 5)/(3*x**2 + 3*x))

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\left (\frac {1}{3} \, x e^{\left (e^{x} - 5\right )} + \frac {4 \, e^{\left (e^{x} - 5\right )}}{3 \, x} - \frac {e^{\left (e^{x}\right )}}{x e^{5} + e^{5}} - \frac {1}{3} \, e^{\left (e^{x} - 5\right )}\right )} \]

[In]

integrate(((x^5+x^4+4*x^2+4*x)*exp(x)+x^4+2*x^3-8*x-4)*exp(exp(x)-5)*exp((x^3+4)*exp(exp(x)-5)/(3*x^2+3*x))/(3
*x^4+6*x^3+3*x^2),x, algorithm="maxima")

[Out]

e^(1/3*x*e^(e^x - 5) + 4/3*e^(e^x - 5)/x - e^(e^x)/(x*e^5 + e^5) - 1/3*e^(e^x - 5))

Giac [F]

\[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=\int { \frac {{\left (x^{4} + 2 \, x^{3} + {\left (x^{5} + x^{4} + 4 \, x^{2} + 4 \, x\right )} e^{x} - 8 \, x - 4\right )} e^{\left (\frac {{\left (x^{3} + 4\right )} e^{\left (e^{x} - 5\right )}}{3 \, {\left (x^{2} + x\right )}} + e^{x} - 5\right )}}{3 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}} \,d x } \]

[In]

integrate(((x^5+x^4+4*x^2+4*x)*exp(x)+x^4+2*x^3-8*x-4)*exp(exp(x)-5)*exp((x^3+4)*exp(exp(x)-5)/(3*x^2+3*x))/(3
*x^4+6*x^3+3*x^2),x, algorithm="giac")

[Out]

integrate(1/3*(x^4 + 2*x^3 + (x^5 + x^4 + 4*x^2 + 4*x)*e^x - 8*x - 4)*e^(1/3*(x^3 + 4)*e^(e^x - 5)/(x^2 + x) +
 e^x - 5)/(x^4 + 2*x^3 + x^2), x)

Mupad [B] (verification not implemented)

Time = 11.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx={\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-5}}{3\,x^2+3\,x}}\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-5}}{3\,x^2+3\,x}} \]

[In]

int((exp((exp(exp(x) - 5)*(x^3 + 4))/(3*x + 3*x^2))*exp(exp(x) - 5)*(exp(x)*(4*x + 4*x^2 + x^4 + x^5) - 8*x +
2*x^3 + x^4 - 4))/(3*x^2 + 6*x^3 + 3*x^4),x)

[Out]

exp((x^3*exp(exp(x))*exp(-5))/(3*x + 3*x^2))*exp((4*exp(exp(x))*exp(-5))/(3*x + 3*x^2))