3.78.60 \(\int \frac {e^{3-x} (81 x^2+63 x^3-17 x^4+x^5)+e^{-\frac {e^{-2+x}}{-9 x+x^2}} (e (-9+11 x-x^2)+e^{3-x} (81 x^2-18 x^3+x^4))}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} (81 x^2-18 x^3+x^4)+e^{3-x} (81 x^3-18 x^4+x^5)} \, dx\)

Optimal. Leaf size=23 \[ x+\log \left (e^{-\frac {e^{-2+x}}{-9 x+x^2}}+x\right ) \]

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Rubi [F]  time = 9.43, 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^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(3 - x)*(81*x^2 + 63*x^3 - 17*x^4 + x^5) + (E*(-9 + 11*x - x^2) + E^(3 - x)*(81*x^2 - 18*x^3 + x^4))/E^
(E^(-2 + x)/(-9*x + x^2)))/(E^(3 - x - E^(-2 + x)/(-9*x + x^2))*(81*x^2 - 18*x^3 + x^4) + E^(3 - x)*(81*x^3 -
18*x^4 + x^5)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 (-9+x)^2 x^2+e^{2+\frac {e^{-2+x}}{(-9+x) x}} (-9+x)^2 x^2 (1+x)-e^x \left (9-11 x+x^2\right )}{e^2 (9-x)^2 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx\\ &=\frac {\int \frac {e^2 (-9+x)^2 x^2+e^{2+\frac {e^{-2+x}}{(-9+x) x}} (-9+x)^2 x^2 (1+x)-e^x \left (9-11 x+x^2\right )}{(9-x)^2 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{e^2}\\ &=\frac {\int \left (\frac {e^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}}+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}{1+e^{\frac {e^{-2+x}}{(-9+x) x}} x}-\frac {e^x \left (9-11 x+x^2\right )}{(-9+x)^2 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}\right ) \, dx}{e^2}\\ &=-\frac {\int \frac {e^x \left (9-11 x+x^2\right )}{(-9+x)^2 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{e^2}+\int \frac {1+e^{\frac {e^{-2+x}}{(-9+x) x}}+e^{\frac {e^{-2+x}}{(-9+x) x}} x}{1+e^{\frac {e^{-2+x}}{(-9+x) x}} x} \, dx\\ &=-\frac {\int \left (-\frac {e^x}{9 (-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}+\frac {e^x}{9 (-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}+\frac {e^x}{9 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}-\frac {e^x}{9 x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}\right ) \, dx}{e^2}+\int \frac {1+e^{-\frac {e^{-2+x}}{(-9+x) x}}+x}{e^{-\frac {e^{-2+x}}{(-9+x) x}}+x} \, dx\\ &=\frac {\int \frac {e^x}{(-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{(-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\frac {\int \frac {e^x}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\int \left (\frac {1+x}{x}-\frac {1}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}\right ) \, dx\\ &=\frac {\int \frac {e^x}{(-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{(-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\frac {\int \frac {e^x}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\int \frac {1+x}{x} \, dx-\int \frac {1}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx\\ &=\frac {\int \frac {e^x}{(-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{(-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\frac {\int \frac {e^x}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\int \left (1+\frac {1}{x}\right ) \, dx-\int \frac {1}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx\\ &=x+\log (x)+\frac {\int \frac {e^x}{(-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{(-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\frac {\int \frac {e^x}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\int \frac {1}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.32, size = 50, normalized size = 2.17 \begin {gather*} \frac {e^2 x+\frac {e^x}{9 x-x^2}+e^2 \log \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 - x)*(81*x^2 + 63*x^3 - 17*x^4 + x^5) + (E*(-9 + 11*x - x^2) + E^(3 - x)*(81*x^2 - 18*x^3 + x^
4))/E^(E^(-2 + x)/(-9*x + x^2)))/(E^(3 - x - E^(-2 + x)/(-9*x + x^2))*(81*x^2 - 18*x^3 + x^4) + E^(3 - x)*(81*
x^3 - 18*x^4 + x^5)),x]

[Out]

(E^2*x + E^x/(9*x - x^2) + E^2*Log[1 + E^(E^(-2 + x)/((-9 + x)*x))*x])/E^2

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fricas [A]  time = 1.02, size = 21, normalized size = 0.91 \begin {gather*} x + \log \left (x + e^{\left (-\frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4-18*x^3+81*x^2)*exp(3-x)+(-x^2+11*x-9)*exp(1))*exp(-exp(1)/(x^2-9*x)/exp(3-x))+(x^5-17*x^4+63*
x^3+81*x^2)*exp(3-x))/((x^4-18*x^3+81*x^2)*exp(3-x)*exp(-exp(1)/(x^2-9*x)/exp(3-x))+(x^5-18*x^4+81*x^3)*exp(3-
x)),x, algorithm="fricas")

[Out]

x + log(x + e^(-e^(x - 2)/(x^2 - 9*x)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{5} - 17 \, x^{4} + 63 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x + 3\right )} - {\left ({\left (x^{2} - 11 \, x + 9\right )} e - {\left (x^{4} - 18 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x + 3\right )}\right )} e^{\left (-\frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, x}\right )}}{{\left (x^{4} - 18 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x - \frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, x} + 3\right )} + {\left (x^{5} - 18 \, x^{4} + 81 \, x^{3}\right )} e^{\left (-x + 3\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4-18*x^3+81*x^2)*exp(3-x)+(-x^2+11*x-9)*exp(1))*exp(-exp(1)/(x^2-9*x)/exp(3-x))+(x^5-17*x^4+63*
x^3+81*x^2)*exp(3-x))/((x^4-18*x^3+81*x^2)*exp(3-x)*exp(-exp(1)/(x^2-9*x)/exp(3-x))+(x^5-18*x^4+81*x^3)*exp(3-
x)),x, algorithm="giac")

[Out]

integrate(((x^5 - 17*x^4 + 63*x^3 + 81*x^2)*e^(-x + 3) - ((x^2 - 11*x + 9)*e - (x^4 - 18*x^3 + 81*x^2)*e^(-x +
3))*e^(-e^(x - 2)/(x^2 - 9*x)))/((x^4 - 18*x^3 + 81*x^2)*e^(-x - e^(x - 2)/(x^2 - 9*x) + 3) + (x^5 - 18*x^4 +
81*x^3)*e^(-x + 3)), x)

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maple [A]  time = 0.44, size = 49, normalized size = 2.13

method	result	size
risch	\(x -\frac {{\mathrm e}^{x -2}}{\left (x -9\right ) x}+\frac {{\mathrm e}^{x -2}}{x^{2}-9 x}+\ln \left ({\mathrm e}^{-\frac {{\mathrm e}^{x -2}}{\left (x -9\right ) x}}+x \right )\)	\(49\)
norman	\(\frac {\left (x^{3} {\mathrm e}^{3-x}-81 x \,{\mathrm e}^{3-x}\right ) {\mathrm e}^{x -3}}{x \left (x -9\right )}+\ln \left ({\mathrm e}^{-\frac {{\mathrm e} \,{\mathrm e}^{x -3}}{x^{2}-9 x}}+x \right )\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^4-18*x^3+81*x^2)*exp(3-x)+(-x^2+11*x-9)*exp(1))*exp(-exp(1)/(x^2-9*x)/exp(3-x))+(x^5-17*x^4+63*x^3+81
*x^2)*exp(3-x))/((x^4-18*x^3+81*x^2)*exp(3-x)*exp(-exp(1)/(x^2-9*x)/exp(3-x))+(x^5-18*x^4+81*x^3)*exp(3-x)),x,
method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.49, size = 67, normalized size = 2.91 \begin {gather*} \frac {9 \, x^{2} e^{2} - 81 \, x e^{2} - e^{x}}{9 \, {\left (x e^{2} - 9 \, e^{2}\right )}} + \log \relax (x) + \log \left (\frac {x e^{\left (\frac {e^{x}}{9 \, {\left (x e^{2} - 9 \, e^{2}\right )}}\right )} + e^{\left (\frac {e^{\left (x - 2\right )}}{9 \, x}\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4-18*x^3+81*x^2)*exp(3-x)+(-x^2+11*x-9)*exp(1))*exp(-exp(1)/(x^2-9*x)/exp(3-x))+(x^5-17*x^4+63*
x^3+81*x^2)*exp(3-x))/((x^4-18*x^3+81*x^2)*exp(3-x)*exp(-exp(1)/(x^2-9*x)/exp(3-x))+(x^5-18*x^4+81*x^3)*exp(3-
x)),x, algorithm="maxima")

[Out]

1/9*(9*x^2*e^2 - 81*x*e^2 - e^x)/(x*e^2 - 9*e^2) + log(x) + log((x*e^(1/9*e^x/(x*e^2 - 9*e^2)) + e^(1/9*e^(x -
2)/x))/x)

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mupad [B]  time = 5.82, size = 20, normalized size = 0.87 \begin {gather*} x+\ln \left (x+{\mathrm {e}}^{-\frac {{\mathrm {e}}^{x-2}}{x\,\left (x-9\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(x - 3)*exp(1))/(9*x - x^2))*(exp(3 - x)*(81*x^2 - 18*x^3 + x^4) - exp(1)*(x^2 - 11*x + 9)) + exp
(3 - x)*(81*x^2 + 63*x^3 - 17*x^4 + x^5))/(exp(3 - x)*(81*x^3 - 18*x^4 + x^5) + exp((exp(x - 3)*exp(1))/(9*x -
x^2))*exp(3 - x)*(81*x^2 - 18*x^3 + x^4)),x)

[Out]

x + log(x + exp(-exp(x - 2)/(x*(x - 9))))

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sympy [A]  time = 0.74, size = 22, normalized size = 0.96 \begin {gather*} x + \log {\left (x + e^{- \frac {e e^{x - 3}}{x^{2} - 9 x}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**4-18*x**3+81*x**2)*exp(3-x)+(-x**2+11*x-9)*exp(1))*exp(-exp(1)/(x**2-9*x)/exp(3-x))+(x**5-17*x
**4+63*x**3+81*x**2)*exp(3-x))/((x**4-18*x**3+81*x**2)*exp(3-x)*exp(-exp(1)/(x**2-9*x)/exp(3-x))+(x**5-18*x**4
+81*x**3)*exp(3-x)),x)

[Out]

x + log(x + exp(-E*exp(x - 3)/(x**2 - 9*x)))

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