3.64.69 \(\int \frac {e^{\frac {-e^{x^2} x+e^x (-5+x+x^4)}{x}} (-2 e^{x^2} x^3+e^x (5-5 x+x^2+3 x^4+x^5))}{x^2} \, dx\)

Optimal. Leaf size=30 \[ e^{e^x+\frac {e^x \left (-5-e^{-x+x^2} x+x^4\right )}{x}} \]

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Rubi [F]  time = 3.74, 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^{\frac {-e^{x^2} x+e^x \left (-5+x+x^4\right )}{x}} \left (-2 e^{x^2} x^3+e^x \left (5-5 x+x^2+3 x^4+x^5\right )\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-(E^x^2*x) + E^x*(-5 + x + x^4))/x)*(-2*E^x^2*x^3 + E^x*(5 - 5*x + x^2 + 3*x^4 + x^5)))/x^2,x]

[Out]

Defer[Int][E^(-E^x^2 + x + (E^x*(-5 + x + x^4))/x), x] + 5*Defer[Int][E^(-E^x^2 + x + (E^x*(-5 + x + x^4))/x)/
x^2, x] - 5*Defer[Int][E^(-E^x^2 + x + (E^x*(-5 + x + x^4))/x)/x, x] - 2*Defer[Int][E^(-E^x^2 + x^2 + (E^x*(-5
+ x + x^4))/x)*x, x] + 3*Defer[Int][E^(-E^x^2 + x + (E^x*(-5 + x + x^4))/x)*x^2, x] + Defer[Int][E^(-E^x^2 +
x + (E^x*(-5 + x + x^4))/x)*x^3, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-e^{x^2}+\frac {e^x \left (-5+x+x^4\right )}{x}} \left (-2 e^{x^2} x^3+e^x \left (5-5 x+x^2+3 x^4+x^5\right )\right )}{x^2} \, dx\\ &=\int \left (-2 e^{-e^{x^2}+x^2+\frac {e^x \left (-5+x+x^4\right )}{x}} x+\frac {e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}} \left (5-5 x+x^2+3 x^4+x^5\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int e^{-e^{x^2}+x^2+\frac {e^x \left (-5+x+x^4\right )}{x}} x \, dx\right )+\int \frac {e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}} \left (5-5 x+x^2+3 x^4+x^5\right )}{x^2} \, dx\\ &=-\left (2 \int e^{-e^{x^2}+x^2+\frac {e^x \left (-5+x+x^4\right )}{x}} x \, dx\right )+\int \left (e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}}+\frac {5 e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}}}{x^2}-\frac {5 e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}}}{x}+3 e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}} x^2+e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}} x^3\right ) \, dx\\ &=-\left (2 \int e^{-e^{x^2}+x^2+\frac {e^x \left (-5+x+x^4\right )}{x}} x \, dx\right )+3 \int e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}} x^2 \, dx+5 \int \frac {e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}}}{x^2} \, dx-5 \int \frac {e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}}}{x} \, dx+\int e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}} \, dx+\int e^{-e^{x^2}+x+\frac {e^x \left (-5+x+x^4\right )}{x}} x^3 \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.09, size = 23, normalized size = 0.77 \begin {gather*} e^{-e^{x^2}+\frac {e^x \left (-5+x+x^4\right )}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-(E^x^2*x) + E^x*(-5 + x + x^4))/x)*(-2*E^x^2*x^3 + E^x*(5 - 5*x + x^2 + 3*x^4 + x^5)))/x^2,x]

[Out]

E^(-E^x^2 + (E^x*(-5 + x + x^4))/x)

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fricas [A]  time = 0.83, size = 23, normalized size = 0.77 \begin {gather*} e^{\left (-\frac {x e^{\left (x^{2}\right )} - {\left (x^{4} + x - 5\right )} e^{x}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3*exp(x)*exp(x^2-x)+(x^5+3*x^4+x^2-5*x+5)*exp(x))*exp((-x*exp(x)*exp(x^2-x)+(x^4+x-5)*exp(x))/
x)/x^2,x, algorithm="fricas")

[Out]

e^(-(x*e^(x^2) - (x^4 + x - 5)*e^x)/x)

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giac [A]  time = 0.21, size = 23, normalized size = 0.77 \begin {gather*} e^{\left (x^{3} e^{x} - \frac {5 \, e^{x}}{x} - e^{\left (x^{2}\right )} + e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3*exp(x)*exp(x^2-x)+(x^5+3*x^4+x^2-5*x+5)*exp(x))*exp((-x*exp(x)*exp(x^2-x)+(x^4+x-5)*exp(x))/
x)/x^2,x, algorithm="giac")

[Out]

e^(x^3*e^x - 5*e^x/x - e^(x^2) + e^x)

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maple [A]  time = 0.08, size = 28, normalized size = 0.93

method	result	size
risch	\({\mathrm e}^{\frac {-{\mathrm e}^{x^{2}} x +{\mathrm e}^{x} x^{4}+{\mathrm e}^{x} x -5 \,{\mathrm e}^{x}}{x}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^3*exp(x)*exp(x^2-x)+(x^5+3*x^4+x^2-5*x+5)*exp(x))*exp((-x*exp(x)*exp(x^2-x)+(x^4+x-5)*exp(x))/x)/x^2
,x,method=_RETURNVERBOSE)

[Out]

exp((-exp(x^2)*x+exp(x)*x^4+exp(x)*x-5*exp(x))/x)

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maxima [A]  time = 0.68, size = 23, normalized size = 0.77 \begin {gather*} e^{\left (x^{3} e^{x} - \frac {5 \, e^{x}}{x} - e^{\left (x^{2}\right )} + e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3*exp(x)*exp(x^2-x)+(x^5+3*x^4+x^2-5*x+5)*exp(x))*exp((-x*exp(x)*exp(x^2-x)+(x^4+x-5)*exp(x))/
x)/x^2,x, algorithm="maxima")

[Out]

e^(x^3*e^x - 5*e^x/x - e^(x^2) + e^x)

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mupad [B]  time = 4.39, size = 26, normalized size = 0.87 \begin {gather*} {\mathrm {e}}^{-{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {5\,{\mathrm {e}}^x}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(x)*(x + x^4 - 5) - x*exp(x^2 - x)*exp(x))/x)*(exp(x)*(x^2 - 5*x + 3*x^4 + x^5 + 5) - 2*x^3*exp(x
^2 - x)*exp(x)))/x^2,x)

[Out]

exp(-exp(x^2))*exp(exp(x))*exp(x^3*exp(x))*exp(-(5*exp(x))/x)

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sympy [A]  time = 0.50, size = 24, normalized size = 0.80 \begin {gather*} e^{\frac {- x e^{x} e^{x^{2} - x} + \left (x^{4} + x - 5\right ) e^{x}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**3*exp(x)*exp(x**2-x)+(x**5+3*x**4+x**2-5*x+5)*exp(x))*exp((-x*exp(x)*exp(x**2-x)+(x**4+x-5)*e
xp(x))/x)/x**2,x)

[Out]

exp((-x*exp(x)*exp(x**2 - x) + (x**4 + x - 5)*exp(x))/x)

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