3.76.44 \(\int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log (e^3 x)}{-3+x}} (9-6 x+x^2+\frac {e^{-1-x} (-3-x+x^2)}{x})}{9 x-6 x^2+x^3} \, dx\)

Optimal. Leaf size=23 \[ e^{3-\frac {e^{-1-x}}{(-3+x) x}} x \]

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Rubi [F]  time = 6.39, 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 {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-(E^(-1 - x)/x) + (-3 + x)*Log[E^3*x])/(-3 + x))*(9 - 6*x + x^2 + (E^(-1 - x)*(-3 - x + x^2))/x))/(9*
x - 6*x^2 + x^3),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{x \left (9-6 x+x^2\right )} \, dx\\ &=\int \frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{(-3+x)^2 x} \, dx\\ &=\int \left (\frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{x}+\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (-3-x+x^2\right )}{(-3+x)^2 x^2}\right ) \, dx\\ &=\int \frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{x} \, dx+\int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (-3-x+x^2\right )}{(-3+x)^2 x^2} \, dx\\ &=\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx+\int \left (\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{3 (-3+x)^2}+\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{3 (-3+x)}-\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{3 x^2}-\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{3 x}\right ) \, dx\\ &=\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{(-3+x)^2} \, dx+\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{-3+x} \, dx-\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{x^2} \, dx-\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{x} \, dx+\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx\\ &=-\left (\frac {1}{3} \int e^{2-\frac {e^{-1-x}}{(-3+x) x}-x} \, dx\right )+\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{(-3+x)^2} \, dx-\frac {1}{3} \int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{x} \, dx+\frac {1}{3} \int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x} x}{-3+x} \, dx+\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx\\ &=-\left (\frac {1}{3} \int e^{2-\frac {e^{-1-x}}{(-3+x) x}-x} \, dx\right )+\frac {1}{3} \int \left (e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}+\frac {3 e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{-3+x}\right ) \, dx+\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{(-3+x)^2} \, dx-\frac {1}{3} \int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{x} \, dx+\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx\\ &=\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{(-3+x)^2} \, dx-\frac {1}{3} \int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{x} \, dx+\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx+\int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{-3+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.81, size = 23, normalized size = 1.00 \begin {gather*} e^{3-\frac {e^{-1-x}}{(-3+x) x}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-(E^(-1 - x)/x) + (-3 + x)*Log[E^3*x])/(-3 + x))*(9 - 6*x + x^2 + (E^(-1 - x)*(-3 - x + x^2))/x
))/(9*x - 6*x^2 + x^3),x]

[Out]

E^(3 - E^(-1 - x)/((-3 + x)*x))*x

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fricas [A]  time = 1.00, size = 30, normalized size = 1.30 \begin {gather*} e^{\left (\frac {{\left (x - 3\right )} \log \relax (x) + 3 \, x - e^{\left (-x - \log \relax (x) - 1\right )} - 9}{x - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-x-3)*exp(-log(x)-x-1)+x^2-6*x+9)*exp((-exp(-log(x)-x-1)+(x-3)*log(x*exp(3)))/(x-3))/(x^3-6*x^2
+9*x),x, algorithm="fricas")

[Out]

e^(((x - 3)*log(x) + 3*x - e^(-x - log(x) - 1) - 9)/(x - 3))

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giac [A]  time = 0.43, size = 43, normalized size = 1.87 \begin {gather*} e^{\left (\frac {x \log \left (x e^{3}\right )}{x - 3} - \frac {e^{\left (-x - \log \relax (x) - 1\right )}}{x - 3} - \frac {3 \, \log \left (x e^{3}\right )}{x - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-x-3)*exp(-log(x)-x-1)+x^2-6*x+9)*exp((-exp(-log(x)-x-1)+(x-3)*log(x*exp(3)))/(x-3))/(x^3-6*x^2
+9*x),x, algorithm="giac")

[Out]

e^(x*log(x*e^3)/(x - 3) - e^(-x - log(x) - 1)/(x - 3) - 3*log(x*e^3)/(x - 3))

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maple [A]  time = 0.06, size = 39, normalized size = 1.70

method	result	size
risch	\({\mathrm e}^{\frac {x^{2} \ln \relax (x )-3 x \ln \relax (x )+3 x^{2}-{\mathrm e}^{-x -1}-9 x}{x \left (x -3\right )}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-x-3)*exp(-ln(x)-x-1)+x^2-6*x+9)*exp((-exp(-ln(x)-x-1)+(x-3)*ln(x*exp(3)))/(x-3))/(x^3-6*x^2+9*x),x,m
ethod=_RETURNVERBOSE)

[Out]

exp((x^2*ln(x)-3*x*ln(x)+3*x^2-exp(-x-1)-9*x)/x/(x-3))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} - 6 \, x + \frac {{\left (x^{2} - x - 3\right )} e^{\left (-x - 1\right )}}{x} + 9\right )} e^{\left (\frac {{\left (x - 3\right )} \log \left (x e^{3}\right ) - \frac {e^{\left (-x - 1\right )}}{x}}{x - 3}\right )}}{x^{3} - 6 \, x^{2} + 9 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-x-3)*exp(-log(x)-x-1)+x^2-6*x+9)*exp((-exp(-log(x)-x-1)+(x-3)*log(x*exp(3)))/(x-3))/(x^3-6*x^2
+9*x),x, algorithm="maxima")

[Out]

integrate((x^2 - 6*x + (x^2 - x - 3)*e^(-x - 1)/x + 9)*e^(((x - 3)*log(x*e^3) - e^(-x - 1)/x)/(x - 3))/(x^3 -
6*x^2 + 9*x), x)

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mupad [B]  time = 4.66, size = 38, normalized size = 1.65 \begin {gather*} x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-1}}{3\,x-x^2}}\,{\mathrm {e}}^{\frac {3\,x}{x-3}}\,{\mathrm {e}}^{-\frac {9}{x-3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(exp(- x - log(x) - 1) - log(x*exp(3))*(x - 3))/(x - 3))*(6*x + exp(- x - log(x) - 1)*(x - x^2 + 3)
- x^2 - 9))/(9*x - 6*x^2 + x^3),x)

[Out]

x*exp((exp(-x)*exp(-1))/(3*x - x^2))*exp((3*x)/(x - 3))*exp(-9/(x - 3))

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sympy [A]  time = 0.61, size = 22, normalized size = 0.96 \begin {gather*} e^{\frac {\left (x - 3\right ) \log {\left (x e^{3} \right )} - \frac {e^{- x - 1}}{x}}{x - 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-x-3)*exp(-ln(x)-x-1)+x**2-6*x+9)*exp((-exp(-ln(x)-x-1)+(x-3)*ln(x*exp(3)))/(x-3))/(x**3-6*x**
2+9*x),x)

[Out]

exp(((x - 3)*log(x*exp(3)) - exp(-x - 1)/x)/(x - 3))

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