3.62.94 \(\int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} (10+40 x+20 x^2))}{x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {10 e^{e^{-4-\frac {e^{2+(-4-x) x}}{x}}}}{x} \]

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Rubi [B]  time = 0.27, antiderivative size = 97, normalized size of antiderivative = 3.59, number of steps used = 1, number of rules used = 1, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {2288} \begin {gather*} \frac {10 \left (2 x^2+4 x+1\right ) \exp \left (-x^2+e^{-\frac {e^{-x^2-4 x+2}+4 x}{x}}-4 x+2\right )}{x^3 \left (\frac {e^{-x^2-4 x+2}+4 x}{x^2}-\frac {2 \left (2-e^{-x^2-4 x+2} (x+2)\right )}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^((-E^(2 - 4*x - x^2) - 4*x)/x)*(-10*x + E^(2 + (-E^(2 - 4*x - x^2) - 4*x)/x - 4*x - x^2)*(10 + 40*x +
20*x^2)))/x^3,x]

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {10 e^{2+e^{-\frac {e^{2-4 x-x^2}+4 x}{x}}-4 x-x^2} \left (1+4 x+2 x^2\right )}{x^3 \left (\frac {e^{2-4 x-x^2}+4 x}{x^2}-\frac {2 \left (2-e^{2-4 x-x^2} (2+x)\right )}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 28, normalized size = 1.04 \begin {gather*} \frac {10 e^{e^{-4-\frac {e^{2-4 x-x^2}}{x}}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^((-E^(2 - 4*x - x^2) - 4*x)/x)*(-10*x + E^(2 + (-E^(2 - 4*x - x^2) - 4*x)/x - 4*x - x^2)*(10 +
40*x + 20*x^2)))/x^3,x]

[Out]

(10*E^E^(-4 - E^(2 - 4*x - x^2)/x))/x

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fricas [A]  time = 0.60, size = 29, normalized size = 1.07 \begin {gather*} 10 \, e^{\left (e^{\left (-\frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x}\right )} - \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)-10*x)*exp(exp((-exp(-x^2-4*x+2)-4*x)
/x)-log(x))/x^2,x, algorithm="fricas")

[Out]

10*e^(e^(-(4*x + e^(-x^2 - 4*x + 2))/x) - log(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {10 \, {\left ({\left (2 \, x^{2} + 4 \, x + 1\right )} e^{\left (-x^{2} - 4 \, x - \frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x} + 2\right )} - x\right )} e^{\left (e^{\left (-\frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x}\right )} - \log \relax (x)\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)-10*x)*exp(exp((-exp(-x^2-4*x+2)-4*x)
/x)-log(x))/x^2,x, algorithm="giac")

[Out]

integrate(10*((2*x^2 + 4*x + 1)*e^(-x^2 - 4*x - (4*x + e^(-x^2 - 4*x + 2))/x + 2) - x)*e^(e^(-(4*x + e^(-x^2 -
4*x + 2))/x) - log(x))/x^2, x)

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maple [A]  time = 0.13, size = 28, normalized size = 1.04

method	result	size
risch	\(\frac {10 \,{\mathrm e}^{{\mathrm e}^{-\frac {{\mathrm e}^{-x^{2}-4 x +2}+4 x}{x}}}}{x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)-10*x)*exp(exp((-exp(-x^2-4*x+2)-4*x)/x)-ln
(x))/x^2,x,method=_RETURNVERBOSE)

[Out]

10/x*exp(exp(-(exp(-x^2-4*x+2)+4*x)/x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 10 \, \int \frac {{\left ({\left (2 \, x^{2} + 4 \, x + 1\right )} e^{\left (-x^{2} - 4 \, x - \frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x} + 2\right )} - x\right )} e^{\left (e^{\left (-\frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x}\right )}\right )}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)-10*x)*exp(exp((-exp(-x^2-4*x+2)-4*x)
/x)-log(x))/x^2,x, algorithm="maxima")

[Out]

10*integrate(((2*x^2 + 4*x + 1)*e^(-x^2 - 4*x - (4*x + e^(-x^2 - 4*x + 2))/x + 2) - x)*e^(e^(-(4*x + e^(-x^2 -
4*x + 2))/x))/x^3, x)

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mupad [B]  time = 4.32, size = 27, normalized size = 1.00 \begin {gather*} \frac {10\,{\mathrm {e}}^{{\mathrm {e}}^{-4}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{-x^2}}{x}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(-(4*x + exp(2 - x^2 - 4*x))/x) - log(x))*(10*x - exp(-(4*x + exp(2 - x^2 - 4*x))/x)*exp(2 - x^2
- 4*x)*(40*x + 20*x^2 + 10)))/x^2,x)

[Out]

(10*exp(exp(-4)*exp(-(exp(-4*x)*exp(2)*exp(-x^2))/x)))/x

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sympy [A]  time = 0.73, size = 22, normalized size = 0.81 \begin {gather*} \frac {10 e^{e^{\frac {- 4 x - e^{- x^{2} - 4 x + 2}}{x}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x**2+40*x+10)*exp(-x**2-4*x+2)*exp((-exp(-x**2-4*x+2)-4*x)/x)-10*x)*exp(exp((-exp(-x**2-4*x+2)-
4*x)/x)-ln(x))/x**2,x)

[Out]

10*exp(exp((-4*x - exp(-x**2 - 4*x + 2))/x))/x

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