3.41.79 $\int \frac{e^{-x^3}((12-4x)\log (3-x)+(4x+(-36x^3+12x^4)\log (3-x))\log (x)+(e^{x^3}(3-x)\log (3-x)+e^{x^3}x\log (x))\log (\log (4)))}{(-3x+x^2){\log }^2(3-x)}dx$

Optimal. Leaf size=28 $$-4+\frac{\log (x)(-4e^{-x^3}-\log (\log (4)))}{\log (3-x)}$$

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Rubi [F]  time = 3.00, 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 = {} $$\begin{array}{c}\int \frac{e^{-x^3}((12-4x)\log (3-x)+(4x+(-36x^3+12x^4)\log (3-x))\log (x)+(e^{x^3}(3-x)\log (3-x)+e^{x^3}x\log (x))\log (\log (4)))}{(-3x+x^2){\log }^2(3-x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[((12 - 4*x)*Log[3 - x] + (4*x + (-36*x^3 + 12*x^4)*Log[3 - x])*Log[x] + (E^x^3*(3 - x)*Log[3 - x] + E^x^3*
x*Log[x])*Log[Log[4]])/(E^x^3*(-3*x + x^2)*Log[3 - x]^2),x]

[Out]

(-4*(3*x^3*Log[3 - x]*Log[x] - x^4*Log[3 - x]*Log[x]))/(E^x^3*(3 - x)*x^3*Log[3 - x]^2) - Log[Log[4]]*Defer[In
t][1/(x*Log[3 - x]), x] + Log[Log[4]]*Defer[Subst][Defer[Int][Log[3 - x]/(x*Log[x]^2), x], x, 3 - x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{-x^3}((12-4x)\log (3-x)+(4x+(-36x^3+12x^4)\log (3-x))\log (x)+(e^{x^3}(3-x)\log (3-x)+e^{x^3}x\log (x))\log (\log (4)))}{(-3+x)x{\log }^2(3-x)}dx\\ & =\int \frac{e^{-x^3}(-((-3+x)\log (3-x)(-4+12x^3\log (x)-e^{x^3}\log (\log (4))))-x\log (x)(4+e^{x^3}\log (\log (4))))}{(3-x)x{\log }^2(3-x)}dx\\ & =\int (\frac{4e^{-x^3}(3\log (3-x)-x\log (3-x)+x\log (x)-9x^3\log (3-x)\log (x)+3x^4\log (3-x)\log (x))}{(-3+x)x{\log }^2(3-x)}-\frac{(-3\log (3-x)+x\log (3-x)-x\log (x))\log (\log (4))}{(-3+x)x{\log }^2(3-x)})dx\\ & =4\int \frac{e^{-x^3}(3\log (3-x)-x\log (3-x)+x\log (x)-9x^3\log (3-x)\log (x)+3x^4\log (3-x)\log (x))}{(-3+x)x{\log }^2(3-x)}dx-\log (\log (4))\int \frac{-3\log (3-x)+x\log (3-x)-x\log (x)}{(-3+x)x{\log }^2(3-x)}dx\\ & =-\frac{4e^{-x^3}(3x^3\log (3-x)\log (x)-x^4\log (3-x)\log (x))}{(3-x)x^3{\log }^2(3-x)}-\log (\log (4))\int \frac{\frac{\log (3-x)}{x}-\frac{\log (x)}{-3+x}}{{\log }^2(3-x)}dx\\ & =-\frac{4e^{-x^3}(3x^3\log (3-x)\log (x)-x^4\log (3-x)\log (x))}{(3-x)x^3{\log }^2(3-x)}-\log (\log (4))\int (\frac{1}{x\log (3-x)}-\frac{\log (x)}{(-3+x){\log }^2(3-x)})dx\\ & =-\frac{4e^{-x^3}(3x^3\log (3-x)\log (x)-x^4\log (3-x)\log (x))}{(3-x)x^3{\log }^2(3-x)}-\log (\log (4))\int \frac{1}{x\log (3-x)}dx+\log (\log (4))\int \frac{\log (x)}{(-3+x){\log }^2(3-x)}dx\\ & =-\frac{4e^{-x^3}(3x^3\log (3-x)\log (x)-x^4\log (3-x)\log (x))}{(3-x)x^3{\log }^2(3-x)}-\log (\log (4))\int \frac{1}{x\log (3-x)}dx+\log (\log (4))\mathrm{Subst}(\int \frac{\log (3-x)}{x{\log }^2(x)}dx,x,3-x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.34, size = 25, normalized size = 0.89 $$\begin{array}{c}-\frac{\log (x)(4e^{-x^3}+\log (\log (4)))}{\log (3-x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[((12 - 4*x)*Log[3 - x] + (4*x + (-36*x^3 + 12*x^4)*Log[3 - x])*Log[x] + (E^x^3*(3 - x)*Log[3 - x] +
E^x^3*x*Log[x])*Log[Log[4]])/(E^x^3*(-3*x + x^2)*Log[3 - x]^2),x]

[Out]

-((Log[x]*(4/E^x^3 + Log[Log[4]]))/Log[3 - x])

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fricas [A]  time = 0.75, size = 33, normalized size = 1.18 $$\begin{array}{c}-\frac{(e^{\left(x^3\right)}\log (x)\log (2\log (2))+4\log (x))e^{(-x^3)}}{\log (-x+3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x^3)*log(x)+(3-x)*exp(x^3)*log(3-x))*log(2*log(2))+((12*x^4-36*x^3)*log(3-x)+4*x)*log(x)+(-4
*x+12)*log(3-x))/(x^2-3*x)/exp(x^3)/log(3-x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 32, normalized size = 1.14 $$\begin{array}{c}-\frac{4e^{(-x^3)}\log (x)+\log (2)\log (x)+\log (x)\log (\log (2))}{\log (-x+3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x^3)*log(x)+(3-x)*exp(x^3)*log(3-x))*log(2*log(2))+((12*x^4-36*x^3)*log(3-x)+4*x)*log(x)+(-4
*x+12)*log(3-x))/(x^2-3*x)/exp(x^3)/log(3-x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 36, normalized size = 1.29

method	result	size
risch	$-\frac{\ln (x)({\mathrm{e}}^{x^3}\ln (2)+{\mathrm{e}}^{x^3}\ln (\ln (2))+4){\mathrm{e}}^{-x^3}}{\ln (3-x)}$	$36$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(x^3)*ln(x)+(3-x)*exp(x^3)*ln(3-x))*ln(2*ln(2))+((12*x^4-36*x^3)*ln(3-x)+4*x)*ln(x)+(-4*x+12)*ln(3-
x))/(x^2-3*x)/exp(x^3)/ln(3-x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 34, normalized size = 1.21 $$\begin{array}{c}-\frac{((\log (2)+\log (\log (2)))e^{\left(x^3\right)}\log (x)+4\log (x))e^{(-x^3)}}{\log (-x+3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x^3)*log(x)+(3-x)*exp(x^3)*log(3-x))*log(2*log(2))+((12*x^4-36*x^3)*log(3-x)+4*x)*log(x)+(-4
*x+12)*log(3-x))/(x^2-3*x)/exp(x^3)/log(3-x)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}-\int -\frac{{\mathrm{e}}^{-x^3}(\ln (2\ln (2))({\mathrm{e}}^{x^3}\ln (3-x)(x-3)-x{\mathrm{e}}^{x^3}\ln (x))+\ln (3-x)(4x-12)-\ln (x)(4x-\ln (3-x)(36x^3-12x^4)))}{{\ln (3-x)}^2(3x-x^2)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x^3)*(log(2*log(2))*(exp(x^3)*log(3 - x)*(x - 3) - x*exp(x^3)*log(x)) + log(3 - x)*(4*x - 12) - log(
x)*(4*x - log(3 - x)*(36*x^3 - 12*x^4))))/(log(3 - x)^2*(3*x - x^2)),x)

[Out]

-int(-(exp(-x^3)*(log(2*log(2))*(exp(x^3)*log(3 - x)*(x - 3) - x*exp(x^3)*log(x)) + log(3 - x)*(4*x - 12) - lo
g(x)*(4*x - log(3 - x)*(36*x^3 - 12*x^4))))/(log(3 - x)^2*(3*x - x^2)), x)

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sympy [A]  time = 0.43, size = 36, normalized size = 1.29 $$\begin{array}{c}\frac{-\log (2)\log (x)-\log (x)\log (\log (2))}{\log (3-x)}-\frac{4e^{-x^3}\log (x)}{\log (3-x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x**3)*ln(x)+(3-x)*exp(x**3)*ln(3-x))*ln(2*ln(2))+((12*x**4-36*x**3)*ln(3-x)+4*x)*ln(x)+(-4*x
+12)*ln(3-x))/(x**2-3*x)/exp(x**3)/ln(3-x)**2,x)

[Out]

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

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