3.99.68 \(\int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log (3 x^2)}{-1+x}} (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log (3 x^2))}{x-2 x^2+x^3} \, dx\)

Optimal. Leaf size=24 \[ e^{e^3 \left (5-\frac {3 \left (\log (3)+\log \left (3 x^2\right )\right )}{-1+x}\right )} \]

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Rubi [F]  time = 3.27, 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 {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}\right ) \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^3*(-5 + 5*x) - 3*E^3*Log[3] - 3*E^3*Log[3*x^2])/(-1 + x))*(E^3*(6 - 6*x) + 3*E^3*x*Log[3] + 3*E^3*x
*Log[3*x^2]))/(x - 2*x^2 + x^3),x]

[Out]

3*Log[9]*Defer[Int][E^(3 + (E^3*(5*x - 5*(1 + (2*Log[27])/5) - 3*Log[x^2]))/(-1 + x))/(-1 + x)^2, x] - 6*Defer
[Int][E^(3 + (E^3*(5*x - 5*(1 + (2*Log[27])/5) - 3*Log[x^2]))/(-1 + x))/(-1 + x), x] + 6*Defer[Int][E^(3 + (E^
3*(5*x - 5*(1 + (2*Log[27])/5) - 3*Log[x^2]))/(-1 + x))/x, x] + 3*Defer[Int][(E^(3 + (E^3*(5*x - 5*(1 + (2*Log
[27])/5) - 3*Log[x^2]))/(-1 + x))*Log[x^2])/(-1 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}\right ) \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x \left (1-2 x+x^2\right )} \, dx\\ &=\int \frac {\exp \left (\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}\right ) \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{(-1+x)^2 x} \, dx\\ &=\int \frac {3 \exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \left (2-2 x (1-\log (3))+x \log \left (x^2\right )\right )}{(1-x)^2 x} \, dx\\ &=3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \left (2-2 x (1-\log (3))+x \log \left (x^2\right )\right )}{(1-x)^2 x} \, dx\\ &=3 \int \left (\frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) (2-x (2-\log (9)))}{(1-x)^2 x}+\frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log \left (x^2\right )}{(-1+x)^2}\right ) \, dx\\ &=3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) (2-x (2-\log (9)))}{(1-x)^2 x} \, dx+3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log \left (x^2\right )}{(-1+x)^2} \, dx\\ &=3 \int \left (-\frac {2 \exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{-1+x}+\frac {2 \exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{x}+\frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log (9)}{(-1+x)^2}\right ) \, dx+3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log \left (x^2\right )}{(-1+x)^2} \, dx\\ &=3 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right ) \log \left (x^2\right )}{(-1+x)^2} \, dx-6 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{-1+x} \, dx+6 \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{x} \, dx+(3 \log (9)) \int \frac {\exp \left (3+\frac {e^3 \left (5 x-5 \left (1+\frac {2 \log (27)}{5}\right )-3 \log \left (x^2\right )\right )}{-1+x}\right )}{(-1+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.79, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((E^3*(-5 + 5*x) - 3*E^3*Log[3] - 3*E^3*Log[3*x^2])/(-1 + x))*(E^3*(6 - 6*x) + 3*E^3*x*Log[3] + 3
*E^3*x*Log[3*x^2]))/(x - 2*x^2 + x^3),x]

[Out]

Integrate[(E^((E^3*(-5 + 5*x) - 3*E^3*Log[3] - 3*E^3*Log[3*x^2])/(-1 + x))*(E^3*(6 - 6*x) + 3*E^3*x*Log[3] + 3
*E^3*x*Log[3*x^2]))/(x - 2*x^2 + x^3), x]

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fricas [A]  time = 0.81, size = 31, normalized size = 1.29 \begin {gather*} e^{\left (\frac {5 \, {\left (x - 1\right )} e^{3} - 3 \, e^{3} \log \relax (3) - 3 \, e^{3} \log \left (3 \, x^{2}\right )}{x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x*exp(3)*log(3*x^2)+3*x*exp(3)*log(3)+(6-6*x)*exp(3))*exp((-3*exp(3)*log(3*x^2)-3*exp(3)*log(3)+(
5*x-5)*exp(3))/(x-1))/(x^3-2*x^2+x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.26, size = 47, normalized size = 1.96 \begin {gather*} e^{\left (\frac {5 \, x e^{3}}{x - 1} - \frac {3 \, e^{3} \log \relax (3)}{x - 1} - \frac {3 \, e^{3} \log \left (3 \, x^{2}\right )}{x - 1} - \frac {5 \, e^{3}}{x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x*exp(3)*log(3*x^2)+3*x*exp(3)*log(3)+(6-6*x)*exp(3))*exp((-3*exp(3)*log(3*x^2)-3*exp(3)*log(3)+(
5*x-5)*exp(3))/(x-1))/(x^3-2*x^2+x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.30, size = 28, normalized size = 1.17

method	result	size
risch	\({\mathrm e}^{-\frac {{\mathrm e}^{3} \left (3 \ln \left (3 x^{2}\right )+3 \ln \relax (3)-5 x +5\right )}{x -1}}\)	\(28\)
default	\(\frac {x \,{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \relax (3)+\left (5 x -5\right ) {\mathrm e}^{3}}{x -1}}-{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \relax (3)+\left (5 x -5\right ) {\mathrm e}^{3}}{x -1}}}{x -1}\)	\(76\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \relax (3)+\left (5 x -5\right ) {\mathrm e}^{3}}{x -1}}-{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \relax (3)+\left (5 x -5\right ) {\mathrm e}^{3}}{x -1}}}{x -1}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x*exp(3)*ln(3*x^2)+3*x*exp(3)*ln(3)+(6-6*x)*exp(3))*exp((-3*exp(3)*ln(3*x^2)-3*exp(3)*ln(3)+(5*x-5)*exp
(3))/(x-1))/(x^3-2*x^2+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.62, size = 28, normalized size = 1.17 \begin {gather*} e^{\left (-\frac {6 \, e^{3} \log \relax (3)}{x - 1} - \frac {6 \, e^{3} \log \relax (x)}{x - 1} + 5 \, e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x*exp(3)*log(3*x^2)+3*x*exp(3)*log(3)+(6-6*x)*exp(3))*exp((-3*exp(3)*log(3*x^2)-3*exp(3)*log(3)+(
5*x-5)*exp(3))/(x-1))/(x^3-2*x^2+x),x, algorithm="maxima")

[Out]

e^(-6*e^3*log(3)/(x - 1) - 6*e^3*log(x)/(x - 1) + 5*e^3)

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mupad [B]  time = 6.09, size = 36, normalized size = 1.50 \begin {gather*} {\mathrm {e}}^{-\frac {5\,{\mathrm {e}}^3}{x-1}}\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^3}{x-1}}\,{\left (\frac {1}{729\,x^6}\right )}^{\frac {{\mathrm {e}}^3}{x-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(3*exp(3)*log(3) + 3*exp(3)*log(3*x^2) - exp(3)*(5*x - 5))/(x - 1))*(3*x*exp(3)*log(3) - exp(3)*(6*x
- 6) + 3*x*exp(3)*log(3*x^2)))/(x - 2*x^2 + x^3),x)

[Out]

exp(-(5*exp(3))/(x - 1))*exp((5*x*exp(3))/(x - 1))*(1/(729*x^6))^(exp(3)/(x - 1))

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sympy [A]  time = 0.53, size = 32, normalized size = 1.33 \begin {gather*} e^{\frac {\left (5 x - 5\right ) e^{3} - 3 e^{3} \log {\left (3 x^{2} \right )} - 3 e^{3} \log {\relax (3 )}}{x - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x*exp(3)*ln(3*x**2)+3*x*exp(3)*ln(3)+(6-6*x)*exp(3))*exp((-3*exp(3)*ln(3*x**2)-3*exp(3)*ln(3)+(5*
x-5)*exp(3))/(x-1))/(x**3-2*x**2+x),x)

[Out]

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

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