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

Optimal. Leaf size=23 \[ x \left (3-x-e^{-e^{\frac {x^2}{8}}} x\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 0.31, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{4} e^{-e^{\frac {x^2}{8}}} \left (e^{e^{\frac {x^2}{8}}} (12-8 x)-8 x+e^{\frac {x^2}{8}} x^3\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{-e^{\frac {x^2}{8}}} \left (e^{e^{\frac {x^2}{8}}} (12-8 x)-8 x+e^{\frac {x^2}{8}} x^3\right ) \, dx\\ &=\frac {1}{4} \int \left (-8 e^{-e^{\frac {x^2}{8}}} x+e^{-e^{\frac {x^2}{8}}+\frac {x^2}{8}} x^3-4 (-3+2 x)\right ) \, dx\\ &=-\frac {1}{4} (3-2 x)^2+\frac {1}{4} \int e^{-e^{\frac {x^2}{8}}+\frac {x^2}{8}} x^3 \, dx-2 \int e^{-e^{\frac {x^2}{8}}} x \, dx\\ &=-\frac {1}{4} (3-2 x)^2+\frac {1}{8} \operatorname {Subst}\left (\int e^{\frac {1}{8} \left (-8 e^{x/8}+x\right )} x \, dx,x,x^2\right )-\operatorname {Subst}\left (\int e^{-e^{x/8}} \, dx,x,x^2\right )\\ &=-\frac {1}{4} (3-2 x)^2+\frac {1}{8} \operatorname {Subst}\left (\int e^{\frac {1}{8} \left (-8 e^{x/8}+x\right )} x \, dx,x,x^2\right )-8 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,e^{\frac {x^2}{8}}\right )\\ &=-\frac {1}{4} (3-2 x)^2-8 \text {Ei}\left (-e^{\frac {x^2}{8}}\right )+\frac {1}{8} \operatorname {Subst}\left (\int e^{\frac {1}{8} \left (-8 e^{x/8}+x\right )} x \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 23, normalized size = 1.00 \begin {gather*} x \left (3+\left (-1-e^{-e^{\frac {x^2}{8}}}\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*(3 + (-1 - E^(-E^(x^2/8)))*x)

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 30, normalized size = 1.30 \begin {gather*} -{\left (x^{2} + {\left (x^{2} - 3 \, x\right )} e^{\left (e^{\left (\frac {1}{8} \, x^{2}\right )}\right )}\right )} e^{\left (-e^{\left (\frac {1}{8} \, x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8*x+12)*exp(exp(1/16*x^2)^2)+x^3*exp(1/16*x^2)^2-8*x)/exp(exp(1/16*x^2)^2),x, algorithm="fric
as")

[Out]

-(x^2 + (x^2 - 3*x)*e^(e^(1/8*x^2)))*e^(-e^(1/8*x^2))

________________________________________________________________________________________

giac [A]  time = 0.28, size = 23, normalized size = 1.00 \begin {gather*} -x^{2} e^{\left (-e^{\left (\frac {1}{8} \, x^{2}\right )}\right )} - x^{2} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8*x+12)*exp(exp(1/16*x^2)^2)+x^3*exp(1/16*x^2)^2-8*x)/exp(exp(1/16*x^2)^2),x, algorithm="giac
")

[Out]

-x^2*e^(-e^(1/8*x^2)) - x^2 + 3*x

________________________________________________________________________________________

maple [A]  time = 0.05, size = 24, normalized size = 1.04

method	result	size
risch	\(-x^{2}+3 x -x^{2} {\mathrm e}^{-{\mathrm e}^{\frac {x^{2}}{8}}}\)	\(24\)
norman	\(\left (-x^{2}+3 x \,{\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{8}}}-x^{2} {\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{8}}}\right ) {\mathrm e}^{-{\mathrm e}^{\frac {x^{2}}{8}}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-8*x+12)*exp(exp(1/16*x^2)^2)+x^3*exp(1/16*x^2)^2-8*x)/exp(exp(1/16*x^2)^2),x,method=_RETURNVERBOSE)

[Out]

-x^2+3*x-x^2*exp(-exp(1/8*x^2))

________________________________________________________________________________________

maxima [A]  time = 0.84, size = 23, normalized size = 1.00 \begin {gather*} -x^{2} e^{\left (-e^{\left (\frac {1}{8} \, x^{2}\right )}\right )} - x^{2} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8*x+12)*exp(exp(1/16*x^2)^2)+x^3*exp(1/16*x^2)^2-8*x)/exp(exp(1/16*x^2)^2),x, algorithm="maxi
ma")

[Out]

-x^2*e^(-e^(1/8*x^2)) - x^2 + 3*x

________________________________________________________________________________________

mupad [B]  time = 1.09, size = 17, normalized size = 0.74 \begin {gather*} -x\,\left (x+x\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {x^2}{8}}}-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*(x + x*exp(-exp(x^2/8)) - 3)

________________________________________________________________________________________

sympy [A]  time = 0.29, size = 17, normalized size = 0.74 \begin {gather*} - x^{2} - x^{2} e^{- e^{\frac {x^{2}}{8}}} + 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8*x+12)*exp(exp(1/16*x**2)**2)+x**3*exp(1/16*x**2)**2-8*x)/exp(exp(1/16*x**2)**2),x)

[Out]

-x**2 - x**2*exp(-exp(x**2/8)) + 3*x

________________________________________________________________________________________