3.46.100 \(\int \frac {e^{-2 x-\frac {1}{5} e^{-2 x} (5 e^{-e^x+x} x+x^2)} (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} (-5 x+5 x^2+5 e^x x^2))}{5 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^{-e^{-2 x} \left (e^{-e^x+x}+\frac {x}{5}\right ) x}}{x} \]

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Rubi [F]  time = 2.96, 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 {\exp \left (-2 x-\frac {1}{5} e^{-2 x} \left (5 e^{-e^x+x} x+x^2\right )\right ) \left (-5 e^{2 x}-2 x^2+2 x^3+e^{-e^x+x} \left (-5 x+5 x^2+5 e^x x^2\right )\right )}{5 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[Int][E^(-E^x - (5*E^(-E^x + x)*x + x^2)/(5*E^(2*x))), x] - (2*Defer[Int][E^(-2*x - (5*E^(-E^x + x)*x + x
^2)/(5*E^(2*x))), x])/5 + Defer[Int][E^(-E^x - x - (5*E^(-E^x + x)*x + x^2)/(5*E^(2*x))), x] - Defer[Int][1/(E
^((5*E^(-E^x + x)*x + x^2)/(5*E^(2*x)))*x^2), x] - Defer[Int][E^(-E^x - x - (5*E^(-E^x + x)*x + x^2)/(5*E^(2*x
)))/x, x] + (2*Defer[Int][E^(-2*x - (5*E^(-E^x + x)*x + x^2)/(5*E^(2*x)))*x, x])/5

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 29, normalized size = 1.00 \begin {gather*} \frac {e^{-\frac {1}{5} e^{-2 x} x \left (5 e^{-e^x+x}+x\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 36, normalized size = 1.24 \begin {gather*} \frac {e^{\left (-\frac {1}{5} \, {\left (x^{2} + 10 \, x e^{\left (2 \, x\right )} + 5 \, x e^{\left (x - e^{x}\right )}\right )} e^{\left (-2 \, x\right )} + 2 \, x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((5*exp(x)*x^2+5*x^2-5*x)*exp(x-exp(x))-5*exp(x)^2+2*x^3-2*x^2)/x^2/exp(x)^2/exp(1/5*(5*x*exp(x-
exp(x))+x^2)/exp(x)^2),x, algorithm="fricas")

[Out]

e^(-1/5*(x^2 + 10*x*e^(2*x) + 5*x*e^(x - e^x))*e^(-2*x) + 2*x)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((5*exp(x)*x^2+5*x^2-5*x)*exp(x-exp(x))-5*exp(x)^2+2*x^3-2*x^2)/x^2/exp(x)^2/exp(1/5*(5*x*exp(x-
exp(x))+x^2)/exp(x)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 24, normalized size = 0.83

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {x \left (5 \,{\mathrm e}^{x -{\mathrm e}^{x}}+x \right ) {\mathrm e}^{-2 x}}{5}}}{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((5*exp(x)*x^2+5*x^2-5*x)*exp(x-exp(x))-5*exp(x)^2+2*x^3-2*x^2)/x^2/exp(x)^2/exp(1/5*(5*x*exp(x-exp(x)
)+x^2)/exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((5*exp(x)*x^2+5*x^2-5*x)*exp(x-exp(x))-5*exp(x)^2+2*x^3-2*x^2)/x^2/exp(x)^2/exp(1/5*(5*x*exp(x-
exp(x))+x^2)/exp(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.49, size = 27, normalized size = 0.93 \begin {gather*} \frac {{\mathrm {e}}^{-x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^{-2\,x}}{5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.37, size = 22, normalized size = 0.76 \begin {gather*} \frac {e^{- \left (\frac {x^{2}}{5} + x e^{x - e^{x}}\right ) e^{- 2 x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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