3.8.80 \(\int \frac {1}{10} e^{-\frac {e^{16+x}}{2}} (2+8 x+e^{e^x} (2+2 e^x x-e^{16+x} x)+e^{16+x} (-x-2 x^2)) \, dx\)

Optimal. Leaf size=30 \[ -3+\frac {1}{5} e^{-\frac {e^{16+x}}{2}} \left (x+x \left (e^{e^x}+2 x\right )\right ) \]

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Rubi [F]  time = 0.54, 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}{10} e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 + 8*x + E^E^x*(2 + 2*E^x*x - E^(16 + x)*x) + E^(16 + x)*(-x - 2*x^2))/(10*E^(E^(16 + x)/2)),x]

[Out]

(E^(E^x - E^(16 + x)/2 + x)*(2 - E^16)*x)/(5*(2*E^x - E^(16 + x))) + ExpIntegralEi[-1/2*E^(16 + x)]/5 + (4*Def
er[Int][x/E^(E^(16 + x)/2), x])/5 - Defer[Int][E^(16 - E^(16 + x)/2 + x)*x, x]/10 - Defer[Int][E^(16 - E^(16 +
x)/2 + x)*x^2, x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int e^{-\frac {e^{16+x}}{2}} \left (2+8 x+e^{e^x} \left (2+2 e^x x-e^{16+x} x\right )+e^{16+x} \left (-x-2 x^2\right )\right ) \, dx\\ &=\frac {1}{10} \int \left (2 e^{-\frac {e^{16+x}}{2}}+8 e^{-\frac {e^{16+x}}{2}} x-e^{16-\frac {e^{16+x}}{2}+x} x (1+2 x)+e^{e^x-\frac {e^{16+x}}{2}} \left (2+2 e^x \left (1-\frac {e^{16}}{2}\right ) x\right )\right ) \, dx\\ &=-\left (\frac {1}{10} \int e^{16-\frac {e^{16+x}}{2}+x} x (1+2 x) \, dx\right )+\frac {1}{10} \int e^{e^x-\frac {e^{16+x}}{2}} \left (2+2 e^x \left (1-\frac {e^{16}}{2}\right ) x\right ) \, dx+\frac {1}{5} \int e^{-\frac {e^{16+x}}{2}} \, dx+\frac {4}{5} \int e^{-\frac {e^{16+x}}{2}} x \, dx\\ &=\frac {e^{e^x-\frac {e^{16+x}}{2}+x} \left (2-e^{16}\right ) x}{5 \left (2 e^x-e^{16+x}\right )}-\frac {1}{10} \int \left (e^{16-\frac {e^{16+x}}{2}+x} x+2 e^{16-\frac {e^{16+x}}{2}+x} x^2\right ) \, dx+\frac {1}{5} \operatorname {Subst}\left (\int \frac {e^{-x/2}}{x} \, dx,x,e^{16+x}\right )+\frac {4}{5} \int e^{-\frac {e^{16+x}}{2}} x \, dx\\ &=\frac {e^{e^x-\frac {e^{16+x}}{2}+x} \left (2-e^{16}\right ) x}{5 \left (2 e^x-e^{16+x}\right )}+\frac {1}{5} \text {Ei}\left (-\frac {e^{16+x}}{2}\right )-\frac {1}{10} \int e^{16-\frac {e^{16+x}}{2}+x} x \, dx-\frac {1}{5} \int e^{16-\frac {e^{16+x}}{2}+x} x^2 \, dx+\frac {4}{5} \int e^{-\frac {e^{16+x}}{2}} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 26, normalized size = 0.87 \begin {gather*} \frac {1}{5} e^{-\frac {e^{16+x}}{2}} x \left (1+e^{e^x}+2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 8*x + E^E^x*(2 + 2*E^x*x - E^(16 + x)*x) + E^(16 + x)*(-x - 2*x^2))/(10*E^(E^(16 + x)/2)),x]

[Out]

(x*(1 + E^E^x + 2*x))/(5*E^(E^(16 + x)/2))

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fricas [A]  time = 0.95, size = 30, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, {\left (2 \, x^{2} + x\right )} e^{\left (-\frac {1}{2} \, e^{\left (x + 16\right )}\right )} + \frac {1}{5} \, x e^{\left (-\frac {1}{2} \, e^{\left (x + 16\right )} + e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+16)+8*x+2)/exp(1/2*exp(x+16)),x, algo
rithm="fricas")

[Out]

1/5*(2*x^2 + x)*e^(-1/2*e^(x + 16)) + 1/5*x*e^(-1/2*e^(x + 16) + e^x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {1}{10} \, {\left ({\left (2 \, x^{2} + x\right )} e^{\left (x + 16\right )} + {\left (x e^{\left (x + 16\right )} - 2 \, x e^{x} - 2\right )} e^{\left (e^{x}\right )} - 8 \, x - 2\right )} e^{\left (-\frac {1}{2} \, e^{\left (x + 16\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+16)+8*x+2)/exp(1/2*exp(x+16)),x, algo
rithm="giac")

[Out]

integrate(-1/10*((2*x^2 + x)*e^(x + 16) + (x*e^(x + 16) - 2*x*e^x - 2)*e^(e^x) - 8*x - 2)*e^(-1/2*e^(x + 16)),
x)

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maple [A]  time = 0.13, size = 19, normalized size = 0.63

method	result	size
risch	\(\frac {\left (2 x +1+{\mathrm e}^{{\mathrm e}^{x}}\right ) x \,{\mathrm e}^{-\frac {{\mathrm e}^{x +16}}{2}}}{5}\)	\(19\)
norman	\(\left (\frac {x}{5}+\frac {2 x^{2}}{5}+\frac {x \,{\mathrm e}^{{\mathrm e}^{x}}}{5}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{x +16}}{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+16)+8*x+2)/exp(1/2*exp(x+16)),x,method=_RET
URNVERBOSE)

[Out]

1/5*(2*x+1+exp(exp(x)))*x*exp(-1/2*exp(x+16))

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maxima [A]  time = 0.96, size = 21, normalized size = 0.70 \begin {gather*} \frac {1}{5} \, {\left (2 \, x^{2} + x e^{\left (e^{x}\right )} + x\right )} e^{\left (-\frac {1}{2} \, e^{\left (x + 16\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x^2-x)*exp(x+16)+8*x+2)/exp(1/2*exp(x+16)),x, algo
rithm="maxima")

[Out]

1/5*(2*x^2 + x*e^(e^x) + x)*e^(-1/2*e^(x + 16))

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mupad [B]  time = 0.14, size = 18, normalized size = 0.60 \begin {gather*} \frac {x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{16}\,{\mathrm {e}}^x}{2}}\,\left (2\,x+{\mathrm {e}}^{{\mathrm {e}}^x}+1\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-exp(x + 16)/2)*((4*x)/5 - (exp(x + 16)*(x + 2*x^2))/10 + (exp(exp(x))*(2*x*exp(x) - x*exp(x + 16) + 2
))/10 + 1/5),x)

[Out]

(x*exp(-(exp(16)*exp(x))/2)*(2*x + exp(exp(x)) + 1))/5

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sympy [B]  time = 38.64, size = 139, normalized size = 4.63 \begin {gather*} \frac {2 x^{2} e^{- \frac {e^{16} e^{x}}{2}}}{5} - \frac {2 x e^{- \frac {e^{16} e^{x}}{2} + e^{x}}}{-10 + 5 e^{16}} + \frac {x e^{16} e^{- \frac {e^{16} e^{x}}{2} + e^{x}}}{-10 + 5 e^{16}} + \frac {x e^{- \frac {e^{16} e^{x}}{2}}}{5} - \frac {e^{16} \operatorname {Ei}{\left (- \frac {e^{16} e^{x}}{2} + e^{x} \right )}}{-10 + 5 e^{16}} + \frac {2 \operatorname {Ei}{\left (- \frac {e^{16} e^{x}}{2} + e^{x} \right )}}{-10 + 5 e^{16}} + \frac {\operatorname {Ei}{\left (- \frac {e^{16} e^{x}}{2} + e^{x} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x*exp(x+16)+2*exp(x)*x+2)*exp(exp(x))+(-2*x**2-x)*exp(x+16)+8*x+2)/exp(1/2*exp(x+16)),x)

[Out]

2*x**2*exp(-exp(16)*exp(x)/2)/5 - 2*x*exp(-exp(16)*exp(x)/2 + exp(x))/(-10 + 5*exp(16)) + x*exp(16)*exp(-exp(1
6)*exp(x)/2 + exp(x))/(-10 + 5*exp(16)) + x*exp(-exp(16)*exp(x)/2)/5 - exp(16)*Ei(-exp(16)*exp(x)/2 + exp(x))/
(-10 + 5*exp(16)) + 2*Ei(-exp(16)*exp(x)/2 + exp(x))/(-10 + 5*exp(16)) + Ei(-exp(16)*exp(x)/2 + exp(x))/5

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