3.95.93 \(\int \frac {e^{-x} (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} (80+64 x-16 x^2+2 e^x x^2)+e^x (10 x^2-2 x^3)))}{x^2} \, dx\)

Optimal. Leaf size=31 \[ 3+e^{1-\left (5+e^{\frac {8 e^{-x}}{x}}-x\right )^2}-\log (x) \]

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Rubi [F]  time = 11.98, 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 {e^{-x} \left (-e^x x+\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)\right ) \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^x*x) + E^(-24 - E^(16/(E^x*x)) + 10*x - x^2 + E^(8/(E^x*x))*(-10 + 2*x))*(E^(16/(E^x*x))*(16 + 16*x)
+ E^(8/(E^x*x))*(80 + 64*x - 16*x^2 + 2*E^x*x^2) + E^x*(10*x^2 - 2*x^3)))/(E^x*x^2),x]

[Out]

-Log[x] - 16*Defer[Int][E^(-24 - E^(16/(E^x*x)) + 2*E^(8/(E^x*x))*(-5 + x) + 8/(E^x*x) + 9*x - x^2), x] + 10*D
efer[Int][E^(-24 - E^(16/(E^x*x)) + 2*E^(8/(E^x*x))*(-5 + x) + 10*x - x^2), x] + 2*Defer[Int][E^(-24 - E^(16/(
E^x*x)) + 2*E^(8/(E^x*x))*(-5 + x) + 8/(E^x*x) + 10*x - x^2), x] + 80*Defer[Int][E^(-24 - E^(16/(E^x*x)) + 2*E
^(8/(E^x*x))*(-5 + x) + 8/(E^x*x) + 9*x - x^2)/x^2, x] + 16*Defer[Int][E^(-24 - E^(16/(E^x*x)) + 2*E^(8/(E^x*x
))*(-5 + x) + 16/(E^x*x) + 9*x - x^2)/x^2, x] + 64*Defer[Int][E^(-24 - E^(16/(E^x*x)) + 2*E^(8/(E^x*x))*(-5 +
x) + 8/(E^x*x) + 9*x - x^2)/x, x] + 16*Defer[Int][E^(-24 - E^(16/(E^x*x)) + 2*E^(8/(E^x*x))*(-5 + x) + 16/(E^x
*x) + 9*x - x^2)/x, x] - 2*Defer[Int][E^(-24 - E^(16/(E^x*x)) + 2*E^(8/(E^x*x))*(-5 + x) + 10*x - x^2)*x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x+2 \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+9 x-x^2\right ) \left (5+e^{\frac {8 e^{-x}}{x}}-x\right ) \left (e^x x^2+8 e^{\frac {8 e^{-x}}{x}} (1+x)\right )}{x^2} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {2 \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+9 x-x^2\right ) \left (5+e^{\frac {8 e^{-x}}{x}}-x\right ) \left (8 e^{\frac {8 e^{-x}}{x}}+8 e^{\frac {8 e^{-x}}{x}} x+e^x x^2\right )}{x^2}\right ) \, dx\\ &=-\log (x)+2 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+9 x-x^2\right ) \left (5+e^{\frac {8 e^{-x}}{x}}-x\right ) \left (8 e^{\frac {8 e^{-x}}{x}}+8 e^{\frac {8 e^{-x}}{x}} x+e^x x^2\right )}{x^2} \, dx\\ &=-\log (x)+2 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+9 x-x^2\right ) \left (5+e^{\frac {8 e^{-x}}{x}}-x\right ) \left (e^x x^2+8 e^{\frac {8 e^{-x}}{x}} (1+x)\right )}{x^2} \, dx\\ &=-\log (x)+2 \int \left (\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) \left (5+e^{\frac {8 e^{-x}}{x}}-x\right )+\frac {8 \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right ) \left (5+e^{\frac {8 e^{-x}}{x}}-x\right ) (1+x)}{x^2}\right ) \, dx\\ &=-\log (x)+2 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) \left (5+e^{\frac {8 e^{-x}}{x}}-x\right ) \, dx+16 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right ) \left (5+e^{\frac {8 e^{-x}}{x}}-x\right ) (1+x)}{x^2} \, dx\\ &=-\log (x)+2 \int \left (5 \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right )+\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+10 x-x^2\right )-\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) x\right ) \, dx+16 \int \left (\frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {16 e^{-x}}{x}+9 x-x^2\right ) (1+x)}{x^2}-\frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right ) (-5+x) (1+x)}{x^2}\right ) \, dx\\ &=-\log (x)+2 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+10 x-x^2\right ) \, dx-2 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) x \, dx+10 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) \, dx+16 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {16 e^{-x}}{x}+9 x-x^2\right ) (1+x)}{x^2} \, dx-16 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right ) (-5+x) (1+x)}{x^2} \, dx\\ &=-\log (x)+2 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+10 x-x^2\right ) \, dx-2 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) x \, dx+10 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) \, dx-16 \int \left (\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right )-\frac {5 \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right )}{x^2}-\frac {4 \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right )}{x}\right ) \, dx+16 \int \left (\frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {16 e^{-x}}{x}+9 x-x^2\right )}{x^2}+\frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {16 e^{-x}}{x}+9 x-x^2\right )}{x}\right ) \, dx\\ &=-\log (x)+2 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+10 x-x^2\right ) \, dx-2 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) x \, dx+10 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2\right ) \, dx-16 \int \exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right ) \, dx+16 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {16 e^{-x}}{x}+9 x-x^2\right )}{x^2} \, dx+16 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {16 e^{-x}}{x}+9 x-x^2\right )}{x} \, dx+64 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right )}{x} \, dx+80 \int \frac {\exp \left (-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+\frac {8 e^{-x}}{x}+9 x-x^2\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 48, normalized size = 1.55 \begin {gather*} e^{-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^x*x) + E^(-24 - E^(16/(E^x*x)) + 10*x - x^2 + E^(8/(E^x*x))*(-10 + 2*x))*(E^(16/(E^x*x))*(16 +
16*x) + E^(8/(E^x*x))*(80 + 64*x - 16*x^2 + 2*E^x*x^2) + E^x*(10*x^2 - 2*x^3)))/(E^x*x^2),x]

[Out]

E^(-24 - E^(16/(E^x*x)) + 2*E^(8/(E^x*x))*(-5 + x) + 10*x - x^2) - Log[x]

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fricas [A]  time = 0.55, size = 43, normalized size = 1.39 \begin {gather*} e^{\left (-x^{2} + 2 \, {\left (x - 5\right )} e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )} + 10 \, x - e^{\left (\frac {16 \, e^{\left (-x\right )}}{x}\right )} - 24\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp(4/exp(x)/x)^2+(-2*x^3+10*x^2)*exp(x)
)*exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp(4/exp(x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x, algorithm="fricas")

[Out]

e^(-x^2 + 2*(x - 5)*e^(8*e^(-x)/x) + 10*x - e^(16*e^(-x)/x) - 24) - log(x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, {\left ({\left (x^{3} - 5 \, x^{2}\right )} e^{x} - 8 \, {\left (x + 1\right )} e^{\left (\frac {16 \, e^{\left (-x\right )}}{x}\right )} - {\left (x^{2} e^{x} - 8 \, x^{2} + 32 \, x + 40\right )} e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )}\right )} e^{\left (-x^{2} + 2 \, {\left (x - 5\right )} e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )} + 10 \, x - e^{\left (\frac {16 \, e^{\left (-x\right )}}{x}\right )} - 24\right )} + x e^{x}\right )} e^{\left (-x\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp(4/exp(x)/x)^2+(-2*x^3+10*x^2)*exp(x)
)*exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp(4/exp(x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x, algorithm="giac")

[Out]

integrate(-(2*((x^3 - 5*x^2)*e^x - 8*(x + 1)*e^(16*e^(-x)/x) - (x^2*e^x - 8*x^2 + 32*x + 40)*e^(8*e^(-x)/x))*e
^(-x^2 + 2*(x - 5)*e^(8*e^(-x)/x) + 10*x - e^(16*e^(-x)/x) - 24) + x*e^x)*e^(-x)/x^2, x)

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maple [A]  time = 0.10, size = 54, normalized size = 1.74

method	result	size
risch	\(-\ln \relax (x )+{\mathrm e}^{2 \,{\mathrm e}^{\frac {8 \,{\mathrm e}^{-x}}{x}} x -x^{2}-{\mathrm e}^{\frac {16 \,{\mathrm e}^{-x}}{x}}-10 \,{\mathrm e}^{\frac {8 \,{\mathrm e}^{-x}}{x}}+10 x -24}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp(4/exp(x)/x)^2+(-2*x^3+10*x^2)*exp(x))*exp(
-exp(4/exp(x)/x)^4+(2*x-10)*exp(4/exp(x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-ln(x)+exp(2*exp(8*exp(-x)/x)*x-x^2-exp(16*exp(-x)/x)-10*exp(8*exp(-x)/x)+10*x-24)

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maxima [A]  time = 0.62, size = 53, normalized size = 1.71 \begin {gather*} e^{\left (-x^{2} + 2 \, x e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )} + 10 \, x - e^{\left (\frac {16 \, e^{\left (-x\right )}}{x}\right )} - 10 \, e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )} - 24\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp(4/exp(x)/x)^2+(-2*x^3+10*x^2)*exp(x)
)*exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp(4/exp(x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x, algorithm="maxima")

[Out]

e^(-x^2 + 2*x*e^(8*e^(-x)/x) + 10*x - e^(16*e^(-x)/x) - 10*e^(8*e^(-x)/x) - 24) - log(x)

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mupad [B]  time = 6.40, size = 58, normalized size = 1.87 \begin {gather*} {\mathrm {e}}^{2\,x\,{\mathrm {e}}^{\frac {8\,{\mathrm {e}}^{-x}}{x}}}\,{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{-24}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {16\,{\mathrm {e}}^{-x}}{x}}}\,{\mathrm {e}}^{-10\,{\mathrm {e}}^{\frac {8\,{\mathrm {e}}^{-x}}{x}}}-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(exp(10*x - exp((16*exp(-x))/x) + exp((8*exp(-x))/x)*(2*x - 10) - x^2 - 24)*(exp(x)*(10*x^2 - 2*x
^3) + exp((16*exp(-x))/x)*(16*x + 16) + exp((8*exp(-x))/x)*(64*x + 2*x^2*exp(x) - 16*x^2 + 80)) - x*exp(x)))/x
^2,x)

[Out]

exp(2*x*exp((8*exp(-x))/x))*exp(10*x)*exp(-24)*exp(-x^2)*exp(-exp((16*exp(-x))/x))*exp(-10*exp((8*exp(-x))/x))
- log(x)

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sympy [A]  time = 4.75, size = 34, normalized size = 1.10 \begin {gather*} e^{- x^{2} + 10 x + \left (2 x - 10\right ) e^{\frac {8 e^{- x}}{x}} - e^{\frac {16 e^{- x}}{x}} - 24} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+16)*exp(4/exp(x)/x)**4+(2*exp(x)*x**2-16*x**2+64*x+80)*exp(4/exp(x)/x)**2+(-2*x**3+10*x**2)*
exp(x))*exp(-exp(4/exp(x)/x)**4+(2*x-10)*exp(4/exp(x)/x)**2-x**2+10*x-24)-exp(x)*x)/exp(x)/x**2,x)

[Out]

exp(-x**2 + 10*x + (2*x - 10)*exp(8*exp(-x)/x) - exp(16*exp(-x)/x) - 24) - log(x)

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