\(\int e^{-e^{\frac {1}{5} (2-80 x^2-10 x^3+10 x^4)}+x} (1+e^{\frac {1}{5} (2-80 x^2-10 x^3+10 x^4)} (32 x+6 x^2-8 x^3)) \, dx\) [1446]

Optimal result

Integrand size = 70, antiderivative size = 28 \[ \int e^{-e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )}+x} \left (1+e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )} \left (32 x+6 x^2-8 x^3\right )\right ) \, dx=1+e^{-e^{\frac {2}{5}+2 x^2 (x+(-4+x) (2+x))}+x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int e^{-e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )}+x} \left (1+e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )} \left (32 x+6 x^2-8 x^3\right )\right ) \, dx=e^{x-e^{\frac {2}{5} \left (5 x^4-5 x^3-40 x^2+1\right )}} \]

[In]

Int[E^(-E^((2 - 80*x^2 - 10*x^3 + 10*x^4)/5) + x)*(1 + E^((2 - 80*x^2 - 10*x^3 + 10*x^4)/5)*(32*x + 6*x^2 - 8*
x^3)),x]

[Out]

E^(-E^((2*(1 - 40*x^2 - 5*x^3 + 5*x^4))/5) + 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}& = e^{-e^{\frac {2}{5} \left (1-40 x^2-5 x^3+5 x^4\right )}+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.83 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int e^{-e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )}+x} \left (1+e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )} \left (32 x+6 x^2-8 x^3\right )\right ) \, dx=e^{-e^{\frac {2}{5}-16 x^2-2 x^3+2 x^4}+x} \]

[In]

Integrate[E^(-E^((2 - 80*x^2 - 10*x^3 + 10*x^4)/5) + x)*(1 + E^((2 - 80*x^2 - 10*x^3 + 10*x^4)/5)*(32*x + 6*x^
2 - 8*x^3)),x]

[Out]

E^(-E^(2/5 - 16*x^2 - 2*x^3 + 2*x^4) + x)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
norman	\({\mathrm e}^{-{\mathrm e}^{2 x^{4}-2 x^{3}-16 x^{2}+\frac {2}{5}}+x}\)	\(24\)
risch	\({\mathrm e}^{-{\mathrm e}^{2 x^{4}-2 x^{3}-16 x^{2}+\frac {2}{5}}+x}\)	\(24\)
parallelrisch	\({\mathrm e}^{-{\mathrm e}^{2 x^{4}-2 x^{3}-16 x^{2}+\frac {2}{5}}+x}\)	\(24\)

[In]

int(((-8*x^3+6*x^2+32*x)*exp(2*x^4-2*x^3-16*x^2+2/5)+1)*exp(-exp(2*x^4-2*x^3-16*x^2+2/5)+x),x,method=_RETURNVE
RBOSE)

[Out]

exp(-exp(2*x^4-2*x^3-16*x^2+2/5)+x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int e^{-e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )}+x} \left (1+e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )} \left (32 x+6 x^2-8 x^3\right )\right ) \, dx=e^{\left (x - e^{\left (2 \, x^{4} - 2 \, x^{3} - 16 \, x^{2} + \frac {2}{5}\right )}\right )} \]

[In]

integrate(((-8*x^3+6*x^2+32*x)*exp(2*x^4-2*x^3-16*x^2+2/5)+1)*exp(-exp(2*x^4-2*x^3-16*x^2+2/5)+x),x, algorithm
="fricas")

[Out]

e^(x - e^(2*x^4 - 2*x^3 - 16*x^2 + 2/5))

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int e^{-e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )}+x} \left (1+e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )} \left (32 x+6 x^2-8 x^3\right )\right ) \, dx=e^{x - e^{2 x^{4} - 2 x^{3} - 16 x^{2} + \frac {2}{5}}} \]

[In]

integrate(((-8*x**3+6*x**2+32*x)*exp(2*x**4-2*x**3-16*x**2+2/5)+1)*exp(-exp(2*x**4-2*x**3-16*x**2+2/5)+x),x)

[Out]

exp(x - exp(2*x**4 - 2*x**3 - 16*x**2 + 2/5))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int e^{-e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )}+x} \left (1+e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )} \left (32 x+6 x^2-8 x^3\right )\right ) \, dx=e^{\left (x - e^{\left (2 \, x^{4} - 2 \, x^{3} - 16 \, x^{2} + \frac {2}{5}\right )}\right )} \]

[In]

integrate(((-8*x^3+6*x^2+32*x)*exp(2*x^4-2*x^3-16*x^2+2/5)+1)*exp(-exp(2*x^4-2*x^3-16*x^2+2/5)+x),x, algorithm
="maxima")

[Out]

e^(x - e^(2*x^4 - 2*x^3 - 16*x^2 + 2/5))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int e^{-e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )}+x} \left (1+e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )} \left (32 x+6 x^2-8 x^3\right )\right ) \, dx=e^{\left (x - e^{\left (2 \, x^{4} - 2 \, x^{3} - 16 \, x^{2} + \frac {2}{5}\right )}\right )} \]

[In]

integrate(((-8*x^3+6*x^2+32*x)*exp(2*x^4-2*x^3-16*x^2+2/5)+1)*exp(-exp(2*x^4-2*x^3-16*x^2+2/5)+x),x, algorithm
="giac")

[Out]

e^(x - e^(2*x^4 - 2*x^3 - 16*x^2 + 2/5))

Mupad [B] (verification not implemented)

Time = 7.75 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int e^{-e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )}+x} \left (1+e^{\frac {1}{5} \left (2-80 x^2-10 x^3+10 x^4\right )} \left (32 x+6 x^2-8 x^3\right )\right ) \, dx={\mathrm {e}}^x\,{\mathrm {e}}^{-{\mathrm {e}}^{2/5}\,{\mathrm {e}}^{-2\,x^3}\,{\mathrm {e}}^{2\,x^4}\,{\mathrm {e}}^{-16\,x^2}} \]

[In]

int(exp(x - exp(2*x^4 - 2*x^3 - 16*x^2 + 2/5))*(exp(2*x^4 - 2*x^3 - 16*x^2 + 2/5)*(32*x + 6*x^2 - 8*x^3) + 1),
x)

[Out]

exp(x)*exp(-exp(2/5)*exp(-2*x^3)*exp(2*x^4)*exp(-16*x^2))