3.73.86 \(\int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x))}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx\)

Optimal. Leaf size=22 \[ e^{\frac {4 \left (\frac {2}{3}+x+\log (x)\right )}{x (-x+\log (3))}} \]

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Rubi [F]  time = 5.46, 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 {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^4-6 x^3 \log (3)+3 x^2 \log ^2(3)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((8 + 12*x + 12*Log[x])/(-3*x^2 + 3*x*Log[3]))*(4*x + 12*x^2 + 4*Log[3] + (24*x - 12*Log[3])*Log[x]))/(
3*x^4 - 6*x^3*Log[3] + 3*x^2*Log[3]^2),x]

[Out]

(4*Defer[Int][E^((4*(2 + 3*x + 3*Log[x]))/(x*(-3*x + Log[27])))/x^2, x])/(3*Log[3]) + (4*Defer[Int][E^((4*(2 +
3*x + 3*Log[x]))/(x*(-3*x + Log[27])))/x, x])/Log[3]^2 + (4*(2 + Log[27])*Defer[Int][E^((4*(2 + 3*x + 3*Log[x
]))/(x*(-3*x + Log[27])))/(x - Log[3])^2, x])/(3*Log[3]) - (4*Defer[Int][E^((4*(2 + 3*x + 3*Log[x]))/(x*(-3*x
+ Log[27])))/(x - Log[3]), x])/Log[3]^2 - (4*Defer[Int][(E^((4*(2 + 3*x + 3*Log[x]))/(x*(-3*x + Log[27])))*Log
[x])/x^2, x])/Log[3] + (4*Defer[Int][(E^((4*(2 + 3*x + 3*Log[x]))/(x*(-3*x + Log[27])))*Log[x])/(x - Log[3])^2
, x])/Log[3]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{x^2 \left (3 x^2-6 x \log (3)+3 \log ^2(3)\right )} \, dx\\ &=\int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{3 x^2 (x-\log (3))^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {8+12 x+12 \log (x)}{-3 x^2+3 x \log (3)}} \left (4 x+12 x^2+4 \log (3)+(24 x-12 \log (3)) \log (x)\right )}{x^2 (x-\log (3))^2} \, dx\\ &=\frac {1}{3} \int \frac {4 \exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) \left (x+3 x^2+\log (3)+6 x \log (x)-3 \log (3) \log (x)\right )}{x^2 (x-\log (3))^2} \, dx\\ &=\frac {4}{3} \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) \left (x+3 x^2+\log (3)+6 x \log (x)-3 \log (3) \log (x)\right )}{x^2 (x-\log (3))^2} \, dx\\ &=\frac {4}{3} \int \left (\frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) \left (x+3 x^2+\log (3)\right )}{x^2 (x-\log (3))^2}+\frac {3 \exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) (2 x-\log (3)) \log (x)}{x^2 (x-\log (3))^2}\right ) \, dx\\ &=\frac {4}{3} \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) \left (x+3 x^2+\log (3)\right )}{x^2 (x-\log (3))^2} \, dx+4 \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) (2 x-\log (3)) \log (x)}{x^2 (x-\log (3))^2} \, dx\\ &=\frac {4}{3} \int \left (\frac {3 \exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right )}{x \log ^2(3)}-\frac {3 \exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right )}{(x-\log (3)) \log ^2(3)}+\frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right )}{x^2 \log (3)}+\frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) (2+\log (27))}{(x-\log (3))^2 \log (3)}\right ) \, dx+4 \int \left (-\frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) \log (x)}{x^2 \log (3)}+\frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) \log (x)}{(x-\log (3))^2 \log (3)}\right ) \, dx\\ &=\frac {4 \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right )}{x} \, dx}{\log ^2(3)}-\frac {4 \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right )}{x-\log (3)} \, dx}{\log ^2(3)}+\frac {4 \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right )}{x^2} \, dx}{3 \log (3)}-\frac {4 \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) \log (x)}{x^2} \, dx}{\log (3)}+\frac {4 \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right ) \log (x)}{(x-\log (3))^2} \, dx}{\log (3)}+\frac {(4 (2+\log (27))) \int \frac {\exp \left (\frac {4 (2+3 x+3 \log (x))}{x (-3 x+\log (27))}\right )}{(x-\log (3))^2} \, dx}{3 \log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.29, size = 38, normalized size = 1.73 \begin {gather*} e^{-\frac {8+12 x}{3 x^2-x \log (27)}} x^{-\frac {4}{x^2-x \log (3)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((8 + 12*x + 12*Log[x])/(-3*x^2 + 3*x*Log[3]))*(4*x + 12*x^2 + 4*Log[3] + (24*x - 12*Log[3])*Log[
x]))/(3*x^4 - 6*x^3*Log[3] + 3*x^2*Log[3]^2),x]

[Out]

1/(E^((8 + 12*x)/(3*x^2 - x*Log[27]))*x^(4/(x^2 - x*Log[3])))

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fricas [A]  time = 0.69, size = 23, normalized size = 1.05 \begin {gather*} e^{\left (-\frac {4 \, {\left (3 \, x + 3 \, \log \relax (x) + 2\right )}}{3 \, {\left (x^{2} - x \log \relax (3)\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(3)+24*x)*log(x)+4*log(3)+12*x^2+4*x)*exp((12*log(x)+12*x+8)/(3*x*log(3)-3*x^2))/(3*x^2*log
(3)^2-6*x^3*log(3)+3*x^4),x, algorithm="fricas")

[Out]

e^(-4/3*(3*x + 3*log(x) + 2)/(x^2 - x*log(3)))

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giac [A]  time = 0.25, size = 44, normalized size = 2.00 \begin {gather*} e^{\left (-\frac {4 \, x}{x^{2} - x \log \relax (3)} - \frac {4 \, \log \relax (x)}{x^{2} - x \log \relax (3)} - \frac {8}{3 \, {\left (x^{2} - x \log \relax (3)\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(3)+24*x)*log(x)+4*log(3)+12*x^2+4*x)*exp((12*log(x)+12*x+8)/(3*x*log(3)-3*x^2))/(3*x^2*log
(3)^2-6*x^3*log(3)+3*x^4),x, algorithm="giac")

[Out]

e^(-4*x/(x^2 - x*log(3)) - 4*log(x)/(x^2 - x*log(3)) - 8/3/(x^2 - x*log(3)))

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maple [A]  time = 0.16, size = 24, normalized size = 1.09

method	result	size
risch	\({\mathrm e}^{\frac {4 \ln \relax (x )+4 x +\frac {8}{3}}{x \left (\ln \relax (3)-x \right )}}\)	\(24\)
norman	\(\frac {x \ln \relax (3) {\mathrm e}^{\frac {12 \ln \relax (x )+12 x +8}{3 x \ln \relax (3)-3 x^{2}}}-x^{2} {\mathrm e}^{\frac {12 \ln \relax (x )+12 x +8}{3 x \ln \relax (3)-3 x^{2}}}}{x \left (\ln \relax (3)-x \right )}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*ln(3)+24*x)*ln(x)+4*ln(3)+12*x^2+4*x)*exp((12*ln(x)+12*x+8)/(3*x*ln(3)-3*x^2))/(3*x^2*ln(3)^2-6*x^3*
ln(3)+3*x^4),x,method=_RETURNVERBOSE)

[Out]

exp(4/3*(3*ln(x)+3*x+2)/x/(ln(3)-x))

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maxima [B]  time = 0.59, size = 64, normalized size = 2.91 \begin {gather*} e^{\left (-\frac {4 \, \log \relax (x)}{x \log \relax (3) - \log \relax (3)^{2}} - \frac {8}{3 \, {\left (x \log \relax (3) - \log \relax (3)^{2}\right )}} - \frac {4}{x - \log \relax (3)} + \frac {4 \, \log \relax (x)}{x \log \relax (3)} + \frac {8}{3 \, x \log \relax (3)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(3)+24*x)*log(x)+4*log(3)+12*x^2+4*x)*exp((12*log(x)+12*x+8)/(3*x*log(3)-3*x^2))/(3*x^2*log
(3)^2-6*x^3*log(3)+3*x^4),x, algorithm="maxima")

[Out]

e^(-4*log(x)/(x*log(3) - log(3)^2) - 8/3/(x*log(3) - log(3)^2) - 4/(x - log(3)) + 4*log(x)/(x*log(3)) + 8/3/(x
*log(3)))

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mupad [B]  time = 4.82, size = 44, normalized size = 2.00 \begin {gather*} x^{\frac {4}{x\,\ln \relax (3)-x^2}}\,{\mathrm {e}}^{-\frac {4}{x-\ln \relax (3)}}\,{\mathrm {e}}^{\frac {8}{3\,x\,\ln \relax (3)-3\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((12*x + 12*log(x) + 8)/(3*x*log(3) - 3*x^2))*(4*x + 4*log(3) + log(x)*(24*x - 12*log(3)) + 12*x^2))/(
3*x^2*log(3)^2 - 6*x^3*log(3) + 3*x^4),x)

[Out]

x^(4/(x*log(3) - x^2))*exp(-4/(x - log(3)))*exp(8/(3*x*log(3) - 3*x^2))

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sympy [A]  time = 0.76, size = 22, normalized size = 1.00 \begin {gather*} e^{\frac {12 x + 12 \log {\relax (x )} + 8}{- 3 x^{2} + 3 x \log {\relax (3 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*ln(3)+24*x)*ln(x)+4*ln(3)+12*x**2+4*x)*exp((12*ln(x)+12*x+8)/(3*x*ln(3)-3*x**2))/(3*x**2*ln(3)
**2-6*x**3*ln(3)+3*x**4),x)

[Out]

exp((12*x + 12*log(x) + 8)/(-3*x**2 + 3*x*log(3)))

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