3.82.76 \(\int e^{x^2+e^{\frac {1}{2} (5+e^{3+e^x})} x^2-2 x^3+x^4+e^{\frac {1}{4} (5+e^{3+e^x})} (-2 x^2+2 x^3)} (4 x-12 x^2+8 x^3+e^{\frac {1}{2} (5+e^{3+e^x})} (4 x+e^{3+e^x+x} x^2)+e^{\frac {1}{4} (5+e^{3+e^x})} (-8 x+12 x^2+e^{3+e^x+x} (-x^2+x^3))) \, dx\)

Optimal. Leaf size=30 \[ 2 e^{\left (-x+e^{\frac {1}{4} \left (5+e^{3+e^x}\right )} x+x^2\right )^2} \]

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Rubi [A]  time = 6.34, antiderivative size = 32, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6688, 6706} \begin {gather*} 2 e^{\left (-x-e^{\frac {1}{4} \left (e^{e^x+3}+5\right )}+1\right )^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(x^2 + E^((5 + E^(3 + E^x))/2)*x^2 - 2*x^3 + x^4 + E^((5 + E^(3 + E^x))/4)*(-2*x^2 + 2*x^3))*(4*x - 12*x
^2 + 8*x^3 + E^((5 + E^(3 + E^x))/2)*(4*x + E^(3 + E^x + x)*x^2) + E^((5 + E^(3 + E^x))/4)*(-8*x + 12*x^2 + E^
(3 + E^x + x)*(-x^2 + x^3))),x]

[Out]

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

Rule 6688

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

Rule 6706

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 {gather*} \begin {aligned} \text {integral} &=\int e^{x^2 \left (-1+e^{\frac {1}{4} \left (5+e^{3+e^x}\right )}+x\right )^2} \left (1-e^{\frac {1}{4} \left (5+e^{3+e^x}\right )}-x\right ) x \left (4-4 e^{\frac {1}{4} \left (5+e^{3+e^x}\right )}-8 x-e^{\frac {17}{4}+\frac {e^{3+e^x}}{4}+e^x+x} x\right ) \, dx\\ &=2 e^{\left (1-e^{\frac {1}{4} \left (5+e^{3+e^x}\right )}-x\right )^2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 28, normalized size = 0.93 \begin {gather*} 2 e^{x^2 \left (-1+e^{\frac {1}{4} \left (5+e^{3+e^x}\right )}+x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(x^2 + E^((5 + E^(3 + E^x))/2)*x^2 - 2*x^3 + x^4 + E^((5 + E^(3 + E^x))/4)*(-2*x^2 + 2*x^3))*(4*x
- 12*x^2 + 8*x^3 + E^((5 + E^(3 + E^x))/2)*(4*x + E^(3 + E^x + x)*x^2) + E^((5 + E^(3 + E^x))/4)*(-8*x + 12*x^
2 + E^(3 + E^x + x)*(-x^2 + x^3))),x]

[Out]

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

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fricas [B]  time = 0.62, size = 66, normalized size = 2.20 \begin {gather*} 2 \, e^{\left (x^{4} - 2 \, x^{3} + x^{2} e^{\left (\frac {1}{2} \, {\left (e^{\left (x + e^{x} + 3\right )} + 5 \, e^{x}\right )} e^{\left (-x\right )}\right )} + x^{2} + 2 \, {\left (x^{3} - x^{2}\right )} e^{\left (\frac {1}{4} \, {\left (e^{\left (x + e^{x} + 3\right )} + 5 \, e^{x}\right )} e^{\left (-x\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(x)*exp(3+exp(x))+4*x)*exp(1/4*exp(3+exp(x))+5/4)^2+((x^3-x^2)*exp(x)*exp(3+exp(x))+12*x^2-
8*x)*exp(1/4*exp(3+exp(x))+5/4)+8*x^3-12*x^2+4*x)*exp(x^2*exp(1/4*exp(3+exp(x))+5/4)^2+(2*x^3-2*x^2)*exp(1/4*e
xp(3+exp(x))+5/4)+x^4-2*x^3+x^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (8 \, x^{3} - 12 \, x^{2} + {\left (x^{2} e^{\left (x + e^{x} + 3\right )} + 4 \, x\right )} e^{\left (\frac {1}{2} \, e^{\left (e^{x} + 3\right )} + \frac {5}{2}\right )} + {\left (12 \, x^{2} + {\left (x^{3} - x^{2}\right )} e^{\left (x + e^{x} + 3\right )} - 8 \, x\right )} e^{\left (\frac {1}{4} \, e^{\left (e^{x} + 3\right )} + \frac {5}{4}\right )} + 4 \, x\right )} e^{\left (x^{4} - 2 \, x^{3} + x^{2} e^{\left (\frac {1}{2} \, e^{\left (e^{x} + 3\right )} + \frac {5}{2}\right )} + x^{2} + 2 \, {\left (x^{3} - x^{2}\right )} e^{\left (\frac {1}{4} \, e^{\left (e^{x} + 3\right )} + \frac {5}{4}\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(x)*exp(3+exp(x))+4*x)*exp(1/4*exp(3+exp(x))+5/4)^2+((x^3-x^2)*exp(x)*exp(3+exp(x))+12*x^2-
8*x)*exp(1/4*exp(3+exp(x))+5/4)+8*x^3-12*x^2+4*x)*exp(x^2*exp(1/4*exp(3+exp(x))+5/4)^2+(2*x^3-2*x^2)*exp(1/4*e
xp(3+exp(x))+5/4)+x^4-2*x^3+x^2),x, algorithm="giac")

[Out]

integrate((8*x^3 - 12*x^2 + (x^2*e^(x + e^x + 3) + 4*x)*e^(1/2*e^(e^x + 3) + 5/2) + (12*x^2 + (x^3 - x^2)*e^(x
+ e^x + 3) - 8*x)*e^(1/4*e^(e^x + 3) + 5/4) + 4*x)*e^(x^4 - 2*x^3 + x^2*e^(1/2*e^(e^x + 3) + 5/2) + x^2 + 2*(
x^3 - x^2)*e^(1/4*e^(e^x + 3) + 5/4)), x)

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maple [B]  time = 0.15, size = 51, normalized size = 1.70

method	result	size
risch	\(2 \,{\mathrm e}^{x^{2} \left (2 x \,{\mathrm e}^{\frac {{\mathrm e}^{3+{\mathrm e}^{x}}}{4}+\frac {5}{4}}+x^{2}-2 \,{\mathrm e}^{\frac {{\mathrm e}^{3+{\mathrm e}^{x}}}{4}+\frac {5}{4}}+{\mathrm e}^{\frac {{\mathrm e}^{3+{\mathrm e}^{x}}}{2}+\frac {5}{2}}-2 x +1\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*exp(x)*exp(3+exp(x))+4*x)*exp(1/4*exp(3+exp(x))+5/4)^2+((x^3-x^2)*exp(x)*exp(3+exp(x))+12*x^2-8*x)*e
xp(1/4*exp(3+exp(x))+5/4)+8*x^3-12*x^2+4*x)*exp(x^2*exp(1/4*exp(3+exp(x))+5/4)^2+(2*x^3-2*x^2)*exp(1/4*exp(3+e
xp(x))+5/4)+x^4-2*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (8 \, x^{3} - 12 \, x^{2} + {\left (x^{2} e^{\left (x + e^{x} + 3\right )} + 4 \, x\right )} e^{\left (\frac {1}{2} \, e^{\left (e^{x} + 3\right )} + \frac {5}{2}\right )} + {\left (12 \, x^{2} + {\left (x^{3} - x^{2}\right )} e^{\left (x + e^{x} + 3\right )} - 8 \, x\right )} e^{\left (\frac {1}{4} \, e^{\left (e^{x} + 3\right )} + \frac {5}{4}\right )} + 4 \, x\right )} e^{\left (x^{4} - 2 \, x^{3} + x^{2} e^{\left (\frac {1}{2} \, e^{\left (e^{x} + 3\right )} + \frac {5}{2}\right )} + x^{2} + 2 \, {\left (x^{3} - x^{2}\right )} e^{\left (\frac {1}{4} \, e^{\left (e^{x} + 3\right )} + \frac {5}{4}\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(x)*exp(3+exp(x))+4*x)*exp(1/4*exp(3+exp(x))+5/4)^2+((x^3-x^2)*exp(x)*exp(3+exp(x))+12*x^2-
8*x)*exp(1/4*exp(3+exp(x))+5/4)+8*x^3-12*x^2+4*x)*exp(x^2*exp(1/4*exp(3+exp(x))+5/4)^2+(2*x^3-2*x^2)*exp(1/4*e
xp(3+exp(x))+5/4)+x^4-2*x^3+x^2),x, algorithm="maxima")

[Out]

integrate((8*x^3 - 12*x^2 + (x^2*e^(x + e^x + 3) + 4*x)*e^(1/2*e^(e^x + 3) + 5/2) + (12*x^2 + (x^3 - x^2)*e^(x
+ e^x + 3) - 8*x)*e^(1/4*e^(e^x + 3) + 5/4) + 4*x)*e^(x^4 - 2*x^3 + x^2*e^(1/2*e^(e^x + 3) + 5/2) + x^2 + 2*(
x^3 - x^2)*e^(1/4*e^(e^x + 3) + 5/4)), x)

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mupad [B]  time = 6.00, size = 63, normalized size = 2.10 \begin {gather*} 2\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{5/2}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^3}{2}}}\,{\mathrm {e}}^{-2\,x^2\,{\mathrm {e}}^{5/4}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^3}{4}}}\,{\mathrm {e}}^{2\,x^3\,{\mathrm {e}}^{5/4}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^3}{4}}}\,{\mathrm {e}}^{-2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2 - 2*x^3 + x^4 - exp(exp(exp(x) + 3)/4 + 5/4)*(2*x^2 - 2*x^3) + x^2*exp(exp(exp(x) + 3)/2 + 5/2))*(
4*x + exp(exp(exp(x) + 3)/2 + 5/2)*(4*x + x^2*exp(exp(x) + 3)*exp(x)) - 12*x^2 + 8*x^3 - exp(exp(exp(x) + 3)/4
+ 5/4)*(8*x - 12*x^2 + exp(exp(x) + 3)*exp(x)*(x^2 - x^3))),x)

[Out]

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

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sympy [B]  time = 6.64, size = 54, normalized size = 1.80 \begin {gather*} 2 e^{x^{4} - 2 x^{3} + x^{2} e^{\frac {e^{e^{x} + 3}}{2} + \frac {5}{2}} + x^{2} + \left (2 x^{3} - 2 x^{2}\right ) e^{\frac {e^{e^{x} + 3}}{4} + \frac {5}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2*exp(x)*exp(3+exp(x))+4*x)*exp(1/4*exp(3+exp(x))+5/4)**2+((x**3-x**2)*exp(x)*exp(3+exp(x))+12*
x**2-8*x)*exp(1/4*exp(3+exp(x))+5/4)+8*x**3-12*x**2+4*x)*exp(x**2*exp(1/4*exp(3+exp(x))+5/4)**2+(2*x**3-2*x**2
)*exp(1/4*exp(3+exp(x))+5/4)+x**4-2*x**3+x**2),x)

[Out]

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

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