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

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

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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 = {} $$\begin{array}{c}\int \frac{e^{\frac{8+12x+12\log (x)}{-3x^2+3x\log (3)}}(4x+12x^2+4\log (3)+(24x-12\log (3))\log (x))}{3x^4-6x^3\log (3)+3x^2{\log }^2(3)}dx\end{array}$$

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

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

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 $$\begin{array}{c}{e}^{(-\frac{4(3x+3\log (x)+2)}{3(x^2-x\log (3))})}\end{array}$$

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 $$\begin{array}{c}{e}^{(-\frac{4x}{x^2-x\log (3)}-\frac{4\log (x)}{x^2-x\log (3)}-\frac{8}{3(x^2-x\log (3))})}\end{array}$$

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	${\mathrm{e}}^{\frac{4\ln (x)+4x+\frac{8}{3}}{x(\ln (3)-x)}}$	$24$
norman	$\frac{x\ln (3){\mathrm{e}}^{\frac{12\ln (x)+12x+8}{3x\ln (3)-3x^2}}-x^2{\mathrm{e}}^{\frac{12\ln (x)+12x+8}{3x\ln (3)-3x^2}}}{x(\ln (3)-x)}$	$71$

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 $$\begin{array}{c}{e}^{(-\frac{4\log (x)}{x\log (3)-\log (3{)}^2}-\frac{8}{3(x\log (3)-\log (3{)}^2)}-\frac{4}{x-\log (3)}+\frac{4\log (x)}{x\log (3)}+\frac{8}{3x\log (3)})}\end{array}$$

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 $$\begin{array}{c}{x}^{\frac{4}{x\ln (3)-x^2}}{\mathrm{e}}^{-\frac{4}{x-\ln (3)}}{\mathrm{e}}^{\frac{8}{3x\ln (3)-3x^2}}\end{array}$$

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 $$\begin{array}{c}{e}^{\frac{12x+12\log (x)+8}{-3x^2+3x\log (3)}}\end{array}$$

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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