3.52.88 \(\int \frac {e^{\frac {2 (e^{3-x}-x^2)}{x}} (2 x^2-4 x^3+e^{3-x} (-4 x-4 x^2))+x^{.\frac {1}{2}/x} (25-25 \log (x))}{50 x^2} \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{25} e^{\frac {2 e^{3-x}}{x}-2 x} x+x^{\left .\frac {1}{2}\right /x} \]

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Rubi [F]  time = 1.34, 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 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{50 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(E^(3 - x) - x^2))/x)*(2*x^2 - 4*x^3 + E^(3 - x)*(-4*x - 4*x^2)) + x^(1/(2*x))*(25 - 25*Log[x]))/(5
0*x^2),x]

[Out]

(-2*Defer[Int][E^(3 + (2*E^(3 - x))/x - 3*x), x])/25 + Defer[Int][E^((2*E^(3 - x))/x - 2*x), x]/25 - (2*Defer[
Int][E^(3 + (2*E^(3 - x))/x - 3*x)/x, x])/25 - (2*Defer[Int][E^((2*E^(3 - x))/x - 2*x)*x, x])/25 + Defer[Int][
x^(-2 + 1/(2*x)), x]/2 - (Log[x]*Defer[Int][x^(-2 + 1/(2*x)), x])/2 + Defer[Int][Defer[Int][x^(-2 + 1/(2*x)),
x]/x, x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{50} \int \frac {e^{\frac {2 \left (e^{3-x}-x^2\right )}{x}} \left (2 x^2-4 x^3+e^{3-x} \left (-4 x-4 x^2\right )\right )+x^{\left .\frac {1}{2}\right /x} (25-25 \log (x))}{x^2} \, dx\\ &=\frac {1}{50} \int \left (-\frac {2 e^{\frac {2 e^{3-x}}{x}-3 x} \left (2 e^3+2 e^3 x-e^x x+2 e^x x^2\right )}{x}-25 x^{-2+\frac {1}{2 x}} (-1+\log (x))\right ) \, dx\\ &=-\left (\frac {1}{25} \int \frac {e^{\frac {2 e^{3-x}}{x}-3 x} \left (2 e^3+2 e^3 x-e^x x+2 e^x x^2\right )}{x} \, dx\right )-\frac {1}{2} \int x^{-2+\frac {1}{2 x}} (-1+\log (x)) \, dx\\ &=-\left (\frac {1}{25} \int \frac {e^{\frac {2 e^{3-x}}{x}-3 x} \left (2 e^3 (1+x)+e^x x (-1+2 x)\right )}{x} \, dx\right )-\frac {1}{2} \int \left (-x^{-2+\frac {1}{2 x}}+x^{-2+\frac {1}{2 x}} \log (x)\right ) \, dx\\ &=-\left (\frac {1}{25} \int \left (\frac {2 e^{3+\frac {2 e^{3-x}}{x}-3 x} (1+x)}{x}+e^{\frac {2 e^{3-x}}{x}-2 x} (-1+2 x)\right ) \, dx\right )+\frac {1}{2} \int x^{-2+\frac {1}{2 x}} \, dx-\frac {1}{2} \int x^{-2+\frac {1}{2 x}} \log (x) \, dx\\ &=-\left (\frac {1}{25} \int e^{\frac {2 e^{3-x}}{x}-2 x} (-1+2 x) \, dx\right )-\frac {2}{25} \int \frac {e^{3+\frac {2 e^{3-x}}{x}-3 x} (1+x)}{x} \, dx+\frac {1}{2} \int x^{-2+\frac {1}{2 x}} \, dx+\frac {1}{2} \int \frac {\int x^{-2+\frac {1}{2 x}} \, dx}{x} \, dx-\frac {1}{2} \log (x) \int x^{-2+\frac {1}{2 x}} \, dx\\ &=-\left (\frac {1}{25} \int \left (-e^{\frac {2 e^{3-x}}{x}-2 x}+2 e^{\frac {2 e^{3-x}}{x}-2 x} x\right ) \, dx\right )-\frac {2}{25} \int \left (e^{3+\frac {2 e^{3-x}}{x}-3 x}+\frac {e^{3+\frac {2 e^{3-x}}{x}-3 x}}{x}\right ) \, dx+\frac {1}{2} \int x^{-2+\frac {1}{2 x}} \, dx+\frac {1}{2} \int \frac {\int x^{-2+\frac {1}{2 x}} \, dx}{x} \, dx-\frac {1}{2} \log (x) \int x^{-2+\frac {1}{2 x}} \, dx\\ &=\frac {1}{25} \int e^{\frac {2 e^{3-x}}{x}-2 x} \, dx-\frac {2}{25} \int e^{3+\frac {2 e^{3-x}}{x}-3 x} \, dx-\frac {2}{25} \int \frac {e^{3+\frac {2 e^{3-x}}{x}-3 x}}{x} \, dx-\frac {2}{25} \int e^{\frac {2 e^{3-x}}{x}-2 x} x \, dx+\frac {1}{2} \int x^{-2+\frac {1}{2 x}} \, dx+\frac {1}{2} \int \frac {\int x^{-2+\frac {1}{2 x}} \, dx}{x} \, dx-\frac {1}{2} \log (x) \int x^{-2+\frac {1}{2 x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.65, size = 37, normalized size = 1.12 \begin {gather*} \frac {1}{50} \left (2 e^{\frac {2 e^{3-x}}{x}-2 x} x+50 x^{\left .\frac {1}{2}\right /x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(E^(3 - x) - x^2))/x)*(2*x^2 - 4*x^3 + E^(3 - x)*(-4*x - 4*x^2)) + x^(1/(2*x))*(25 - 25*Log[x
]))/(50*x^2),x]

[Out]

(2*E^((2*E^(3 - x))/x - 2*x)*x + 50*x^(1/(2*x)))/50

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fricas [A]  time = 0.60, size = 29, normalized size = 0.88 \begin {gather*} \frac {1}{25} \, x e^{\left (-\frac {2 \, {\left (x^{2} - e^{\left (-x + 3\right )}\right )}}{x}\right )} + x^{\frac {1}{2 \, x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/50*((-25*log(x)+25)*exp(1/2*log(x)/x)+((-4*x^2-4*x)*exp(3-x)-4*x^3+2*x^2)*exp((exp(3-x)-x^2)/x)^2)
/x^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {25 \, x^{\frac {1}{2 \, x}} {\left (\log \relax (x) - 1\right )} + 2 \, {\left (2 \, x^{3} - x^{2} + 2 \, {\left (x^{2} + x\right )} e^{\left (-x + 3\right )}\right )} e^{\left (-\frac {2 \, {\left (x^{2} - e^{\left (-x + 3\right )}\right )}}{x}\right )}}{50 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/50*((-25*log(x)+25)*exp(1/2*log(x)/x)+((-4*x^2-4*x)*exp(3-x)-4*x^3+2*x^2)*exp((exp(3-x)-x^2)/x)^2)
/x^2,x, algorithm="giac")

[Out]

integrate(-1/50*(25*x^(1/2/x)*(log(x) - 1) + 2*(2*x^3 - x^2 + 2*(x^2 + x)*e^(-x + 3))*e^(-2*(x^2 - e^(-x + 3))
/x))/x^2, x)

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maple [A]  time = 0.11, size = 30, normalized size = 0.91

method	result	size
risch	\(x^{\frac {1}{2 x}}+\frac {x \,{\mathrm e}^{-\frac {2 \left (-{\mathrm e}^{3-x}+x^{2}\right )}{x}}}{25}\)	\(30\)
default	\({\mathrm e}^{\frac {\ln \relax (x )}{2 x}}+\frac {x \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{3-x}-2 x^{2}}{x}}}{25}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/50*((-25*ln(x)+25)*exp(1/2*ln(x)/x)+((-4*x^2-4*x)*exp(3-x)-4*x^3+2*x^2)*exp((exp(3-x)-x^2)/x)^2)/x^2,x,m
ethod=_RETURNVERBOSE)

[Out]

x^(1/2/x)+1/25*x*exp(-2*(-exp(3-x)+x^2)/x)

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maxima [A]  time = 0.40, size = 35, normalized size = 1.06 \begin {gather*} \frac {1}{25} \, {\left (x e^{\left (\frac {2 \, e^{\left (-x + 3\right )}}{x}\right )} + 25 \, e^{\left (2 \, x + \frac {\log \relax (x)}{2 \, x}\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/50*((-25*log(x)+25)*exp(1/2*log(x)/x)+((-4*x^2-4*x)*exp(3-x)-4*x^3+2*x^2)*exp((exp(3-x)-x^2)/x)^2)
/x^2,x, algorithm="maxima")

[Out]

1/25*(x*e^(2*e^(-x + 3)/x) + 25*e^(2*x + 1/2*log(x)/x))*e^(-2*x)

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mupad [B]  time = 3.65, size = 27, normalized size = 0.82 \begin {gather*} x^{\frac {1}{2\,x}}+\frac {x\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x}}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp((2*(exp(3 - x) - x^2))/x)*(exp(3 - x)*(4*x + 4*x^2) - 2*x^2 + 4*x^3))/50 + (exp(log(x)/(2*x))*(25*l
og(x) - 25))/50)/x^2,x)

[Out]

x^(1/(2*x)) + (x*exp(-2*x)*exp((2*exp(-x)*exp(3))/x))/25

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/50*((-25*ln(x)+25)*exp(1/2*ln(x)/x)+((-4*x**2-4*x)*exp(3-x)-4*x**3+2*x**2)*exp((exp(3-x)-x**2)/x)*
*2)/x**2,x)

[Out]

Timed out

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