3.39.88 \(\int \frac {e^{-25+\frac {e^{-25+10 x-x^2}}{5 x}+10 x-x^2} (-1+10 x-2 x^2+(1-10 x+2 x^2) \log (3))}{5 x^2} \, dx\)

Optimal. Leaf size=25 \[ e^{\frac {e^{-(-5+x)^2}}{5 x}} (1-\log (3)) \]

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Rubi [F]  time = 1.89, 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^{-25+\frac {e^{-25+10 x-x^2}}{5 x}+10 x-x^2} \left (-1+10 x-2 x^2+\left (1-10 x+2 x^2\right ) \log (3)\right )}{5 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-25 + E^(-25 + 10*x - x^2)/(5*x) + 10*x - x^2)*(-1 + 10*x - 2*x^2 + (1 - 10*x + 2*x^2)*Log[3]))/(5*x^2
),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{-25+\frac {e^{-25+10 x-x^2}}{5 x}+10 x-x^2} \left (-1+10 x-2 x^2+\left (1-10 x+2 x^2\right ) \log (3)\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}} \left (1-10 x+2 x^2\right ) (-1+\log (3))}{x^2} \, dx\\ &=\frac {1}{5} (-1+\log (3)) \int \frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}} \left (1-10 x+2 x^2\right )}{x^2} \, dx\\ &=\frac {1}{5} (-1+\log (3)) \int \left (2 e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}+\frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}}{x^2}-\frac {10 e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}}{x}\right ) \, dx\\ &=-\left (\frac {1}{5} (2 (1-\log (3))) \int e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}} \, dx\right )+(2 (1-\log (3))) \int \frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}}{x} \, dx+\frac {1}{5} (-1+\log (3)) \int \frac {e^{-(-5+x)^2+\frac {e^{-(-5+x)^2}}{5 x}}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 24, normalized size = 0.96 \begin {gather*} -e^{\frac {e^{-(-5+x)^2}}{5 x}} (-1+\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-25 + E^(-25 + 10*x - x^2)/(5*x) + 10*x - x^2)*(-1 + 10*x - 2*x^2 + (1 - 10*x + 2*x^2)*Log[3]))/
(5*x^2),x]

[Out]

-(E^(1/(5*E^(-5 + x)^2*x))*(-1 + Log[3]))

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fricas [B]  time = 0.91, size = 47, normalized size = 1.88 \begin {gather*} -{\left (\log \relax (3) - 1\right )} e^{\left (x^{2} - 10 \, x - \frac {5 \, x^{3} - 50 \, x^{2} + 125 \, x - e^{\left (-x^{2} + 10 \, x - 25\right )}}{5 \, x} + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x^2-10*x+1)*log(3)-2*x^2+10*x-1)*exp(1/5/x/exp(x^2-10*x+25))/x^2/exp(x^2-10*x+25),x, algorit
hm="fricas")

[Out]

-(log(3) - 1)*e^(x^2 - 10*x - 1/5*(5*x^3 - 50*x^2 + 125*x - e^(-x^2 + 10*x - 25))/x + 25)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, x^{2} - {\left (2 \, x^{2} - 10 \, x + 1\right )} \log \relax (3) - 10 \, x + 1\right )} e^{\left (-x^{2} + 10 \, x + \frac {e^{\left (-x^{2} + 10 \, x - 25\right )}}{5 \, x} - 25\right )}}{5 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x^2-10*x+1)*log(3)-2*x^2+10*x-1)*exp(1/5/x/exp(x^2-10*x+25))/x^2/exp(x^2-10*x+25),x, algorit
hm="giac")

[Out]

integrate(-1/5*(2*x^2 - (2*x^2 - 10*x + 1)*log(3) - 10*x + 1)*e^(-x^2 + 10*x + 1/5*e^(-x^2 + 10*x - 25)/x - 25
)/x^2, x)

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maple [A]  time = 0.08, size = 25, normalized size = 1.00

method	result	size
norman	\(\left (1-\ln \relax (3)\right ) {\mathrm e}^{\frac {{\mathrm e}^{-x^{2}+10 x -25}}{5 x}}\)	\(25\)
risch	\(-{\mathrm e}^{\frac {{\mathrm e}^{-\left (x -5\right )^{2}}}{5 x}} \ln \relax (3)+{\mathrm e}^{\frac {{\mathrm e}^{-\left (x -5\right )^{2}}}{5 x}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((2*x^2-10*x+1)*ln(3)-2*x^2+10*x-1)*exp(1/5/x/exp(x^2-10*x+25))/x^2/exp(x^2-10*x+25),x,method=_RETURNV
ERBOSE)

[Out]

(1-ln(3))*exp(1/5/x/exp(x^2-10*x+25))

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maxima [A]  time = 0.62, size = 23, normalized size = 0.92 \begin {gather*} -{\left (\log \relax (3) - 1\right )} e^{\left (\frac {e^{\left (-x^{2} + 10 \, x - 25\right )}}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x^2-10*x+1)*log(3)-2*x^2+10*x-1)*exp(1/5/x/exp(x^2-10*x+25))/x^2/exp(x^2-10*x+25),x, algorit
hm="maxima")

[Out]

-(log(3) - 1)*e^(1/5*e^(-x^2 + 10*x - 25)/x)

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mupad [B]  time = 2.52, size = 24, normalized size = 0.96 \begin {gather*} -{\mathrm {e}}^{\frac {{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{-25}\,{\mathrm {e}}^{-x^2}}{5\,x}}\,\left (\ln \relax (3)-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(10*x - x^2 - 25)/(5*x))*exp(10*x - x^2 - 25)*(10*x + log(3)*(2*x^2 - 10*x + 1) - 2*x^2 - 1))/(5*x
^2),x)

[Out]

-exp((exp(10*x)*exp(-25)*exp(-x^2))/(5*x))*(log(3) - 1)

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sympy [A]  time = 0.66, size = 19, normalized size = 0.76 \begin {gather*} \left (1 - \log {\relax (3 )}\right ) e^{\frac {e^{- x^{2} + 10 x - 25}}{5 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x**2-10*x+1)*ln(3)-2*x**2+10*x-1)*exp(1/5/x/exp(x**2-10*x+25))/x**2/exp(x**2-10*x+25),x)

[Out]

(1 - log(3))*exp(exp(-x**2 + 10*x - 25)/(5*x))

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