3.55.81 \(\int \frac {5 x^3+e^{e^{\frac {-25+5 x-5 x^2}{x}}} (-x+e^{\frac {-25+5 x-5 x^2}{x}} (25-5 x^2))}{5 x^3} \, dx\)

Optimal. Leaf size=30 \[ 4+x+\frac {\frac {1}{5} e^{e^{\frac {5 \left (-5+x-x^2\right )}{x}}}+x}{x} \]

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Rubi [F]  time = 0.78, 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 {5 x^3+e^{e^{\frac {-25+5 x-5 x^2}{x}}} \left (-x+e^{\frac {-25+5 x-5 x^2}{x}} \left (25-5 x^2\right )\right )}{5 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x^3 + E^E^((-25 + 5*x - 5*x^2)/x)*(-x + E^((-25 + 5*x - 5*x^2)/x)*(25 - 5*x^2)))/(5*x^3),x]

[Out]

x + 5*Defer[Int][E^(5 + E^(5 - 25/x - 5*x) - 25/x - 5*x)/x^3, x] - Defer[Int][E^E^(5 - 25/x - 5*x)/x^2, x]/5 -
Defer[Int][E^(5 + E^(5 - 25/x - 5*x) - 25/x - 5*x)/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {5 x^3+e^{e^{\frac {-25+5 x-5 x^2}{x}}} \left (-x+e^{\frac {-25+5 x-5 x^2}{x}} \left (25-5 x^2\right )\right )}{x^3} \, dx\\ &=\frac {1}{5} \int \left (-\frac {5 e^{5+e^{5-\frac {25}{x}-5 x}-\frac {25}{x}-5 x} \left (-5+x^2\right )}{x^3}+\frac {-e^{e^{5-\frac {25}{x}-5 x}}+5 x^2}{x^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-e^{e^{5-\frac {25}{x}-5 x}}+5 x^2}{x^2} \, dx-\int \frac {e^{5+e^{5-\frac {25}{x}-5 x}-\frac {25}{x}-5 x} \left (-5+x^2\right )}{x^3} \, dx\\ &=\frac {1}{5} \int \left (5-\frac {e^{e^{5-\frac {25}{x}-5 x}}}{x^2}\right ) \, dx-\int \left (-\frac {5 e^{5+e^{5-\frac {25}{x}-5 x}-\frac {25}{x}-5 x}}{x^3}+\frac {e^{5+e^{5-\frac {25}{x}-5 x}-\frac {25}{x}-5 x}}{x}\right ) \, dx\\ &=x-\frac {1}{5} \int \frac {e^{e^{5-\frac {25}{x}-5 x}}}{x^2} \, dx+5 \int \frac {e^{5+e^{5-\frac {25}{x}-5 x}-\frac {25}{x}-5 x}}{x^3} \, dx-\int \frac {e^{5+e^{5-\frac {25}{x}-5 x}-\frac {25}{x}-5 x}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.61, size = 23, normalized size = 0.77 \begin {gather*} \frac {e^{e^{5-\frac {25}{x}-5 x}}}{5 x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^3 + E^E^((-25 + 5*x - 5*x^2)/x)*(-x + E^((-25 + 5*x - 5*x^2)/x)*(25 - 5*x^2)))/(5*x^3),x]

[Out]

E^E^(5 - 25/x - 5*x)/(5*x) + x

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fricas [A]  time = 1.62, size = 26, normalized size = 0.87 \begin {gather*} \frac {5 \, x^{2} + e^{\left (e^{\left (-\frac {5 \, {\left (x^{2} - x + 5\right )}}{x}\right )}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((-5*x^2+25)*exp((-5*x^2+5*x-25)/x)-x)*exp(exp((-5*x^2+5*x-25)/x))+5*x^3)/x^3,x, algorithm="fri
cas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((-5*x^2+25)*exp((-5*x^2+5*x-25)/x)-x)*exp(exp((-5*x^2+5*x-25)/x))+5*x^3)/x^3,x, algorithm="gia
c")

[Out]

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

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maple [A]  time = 0.23, size = 23, normalized size = 0.77

method	result	size
risch	\(x +\frac {{\mathrm e}^{{\mathrm e}^{-\frac {5 \left (x^{2}-x +5\right )}{x}}}}{5 x}\)	\(23\)
norman	\(\frac {x^{3}+\frac {x \,{\mathrm e}^{{\mathrm e}^{\frac {-5 x^{2}+5 x -25}{x}}}}{5}}{x^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(((-5*x^2+25)*exp((-5*x^2+5*x-25)/x)-x)*exp(exp((-5*x^2+5*x-25)/x))+5*x^3)/x^3,x,method=_RETURNVERBOSE
)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x - \frac {1}{5} \, \int \frac {{\left (5 \, x^{2} e^{5} + x e^{\left (5 \, x + \frac {25}{x}\right )} - 25 \, e^{5}\right )} e^{\left (-5 \, x - \frac {25}{x} + e^{\left (-5 \, x - \frac {25}{x} + 5\right )}\right )}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((-5*x^2+25)*exp((-5*x^2+5*x-25)/x)-x)*exp(exp((-5*x^2+5*x-25)/x))+5*x^3)/x^3,x, algorithm="max
ima")

[Out]

x - 1/5*integrate((5*x^2*e^5 + x*e^(5*x + 25/x) - 25*e^5)*e^(-5*x - 25/x + e^(-5*x - 25/x + 5))/x^3, x)

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mupad [B]  time = 3.51, size = 21, normalized size = 0.70 \begin {gather*} x+\frac {{\mathrm {e}}^{{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {25}{x}}}}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(exp(-(5*x^2 - 5*x + 25)/x))*(x + exp(-(5*x^2 - 5*x + 25)/x)*(5*x^2 - 25)))/5 - x^3)/x^3,x)

[Out]

x + exp(exp(-5*x)*exp(5)*exp(-25/x))/(5*x)

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sympy [A]  time = 0.29, size = 19, normalized size = 0.63 \begin {gather*} x + \frac {e^{e^{\frac {- 5 x^{2} + 5 x - 25}{x}}}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((-5*x**2+25)*exp((-5*x**2+5*x-25)/x)-x)*exp(exp((-5*x**2+5*x-25)/x))+5*x**3)/x**3,x)

[Out]

x + exp(exp((-5*x**2 + 5*x - 25)/x))/(5*x)

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