3.12.80 \(\int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx\)

Optimal. Leaf size=36 \[ 5 \left (e^4-e^{5 e^5}-e^{\frac {-e^{2/x}+x}{x}}\right )+x \]

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Rubi [F]  time = 0.44, 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 {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2/x + (-E^(2/x) + x)/x)*(-10 - 5*x) + x^3)/x^3,x]

[Out]

x - 10*Defer[Int][E^(1 + 2/x - E^(2/x)/x)/x^3, x] + 5*Defer[Subst][Defer[Int][E^(1 + 2*x - E^(2*x)*x), x], x,
x^(-1)]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-\frac {5 e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}} (2+x)}{x^3}\right ) \, dx\\ &=x-5 \int \frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}} (2+x)}{x^3} \, dx\\ &=x-5 \int \left (\frac {2 e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^3}+\frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^2}\right ) \, dx\\ &=x-5 \int \frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^2} \, dx-10 \int \frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^3} \, dx\\ &=x+5 \operatorname {Subst}\left (\int e^{1+2 x-e^{2 x} x} \, dx,x,\frac {1}{x}\right )-10 \int \frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 20, normalized size = 0.56 \begin {gather*} -5 e^{1-\frac {e^{2/x}}{x}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2/x + (-E^(2/x) + x)/x)*(-10 - 5*x) + x^3)/x^3,x]

[Out]

-5*E^(1 - E^(2/x)/x) + x

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fricas [A]  time = 0.98, size = 34, normalized size = 0.94 \begin {gather*} {\left (x e^{\frac {2}{x}} - 5 \, e^{\left (\frac {x - e^{\frac {2}{x}} + 2}{x}\right )}\right )} e^{\left (-\frac {2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x^3)/x^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.87, size = 34, normalized size = 0.94 \begin {gather*} {\left (x e^{\frac {2}{x}} - 5 \, e^{\left (\frac {x - e^{\frac {2}{x}} + 2}{x}\right )}\right )} e^{\left (-\frac {2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x^3)/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 20, normalized size = 0.56

method	result	size
risch	\(x -5 \,{\mathrm e}^{\frac {-{\mathrm e}^{\frac {2}{x}}+x}{x}}\)	\(20\)
norman	\(\frac {x^{3}-5 x^{2} {\mathrm e}^{\frac {-{\mathrm e}^{\frac {2}{x}}+x}{x}}}{x^{2}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.71, size = 18, normalized size = 0.50 \begin {gather*} x - 5 \, e^{\left (-\frac {e^{\frac {2}{x}}}{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x^3)/x^3,x, algorithm="maxima")

[Out]

x - 5*e^(-e^(2/x)/x + 1)

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mupad [B]  time = 0.89, size = 18, normalized size = 0.50 \begin {gather*} x-5\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{2/x}}{x}}\,\mathrm {e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 - exp((x - exp(2/x))/x)*exp(2/x)*(5*x + 10))/x^3,x)

[Out]

x - 5*exp(-exp(2/x)/x)*exp(1)

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sympy [A]  time = 0.30, size = 12, normalized size = 0.33 \begin {gather*} x - 5 e^{\frac {x - e^{\frac {2}{x}}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x**3)/x**3,x)

[Out]

x - 5*exp((x - exp(2/x))/x)

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