3.5.14 $\int \frac{e^{2e^x}(6-3e^4)+12x^2-3e^4{x}^2-3x^2\log (2)+(-16x+6e^{4}x+2x\log (2))\log (3)+(6-3e^4){\log }^2(3)+e^{e^x}(-16x+6e^{4}x+2x\log (2)+e^x(-2x^2+x^2\log (2))+(12-6e^4)\log (3))}{e^{2e^x}{x}^4+x^6-2x^5\log (3)+x^4{\log }^2(3)+e^{e^x}(-2x^5+2x^4\log (3))}dx$

Optimal. Leaf size=30 $$\frac{-2+e^4+\frac{x(2-\log (2))}{e^{e^x}-x+\log (3)}}{x^3}$$

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Rubi [F]  time = 8.31, 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 = {} $$\begin{array}{c}\int \frac{e^{2e^x}(6-3e^4)+12x^2-3e^4{x}^2-3x^2\log (2)+(-16x+6e^{4}x+2x\log (2))\log (3)+(6-3e^4){\log }^2(3)+e^{e^x}(-16x+6e^{4}x+2x\log (2)+e^x(-2x^2+x^2\log (2))+(12-6e^4)\log (3))}{e^{2e^x}{x}^4+x^6-2x^5\log (3)+x^4{\log }^2(3)+e^{e^x}(-2x^5+2x^4\log (3))}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^(2*E^x)*(6 - 3*E^4) + 12*x^2 - 3*E^4*x^2 - 3*x^2*Log[2] + (-16*x + 6*E^4*x + 2*x*Log[2])*Log[3] + (6 -
3*E^4)*Log[3]^2 + E^E^x*(-16*x + 6*E^4*x + 2*x*Log[2] + E^x*(-2*x^2 + x^2*Log[2]) + (12 - 6*E^4)*Log[3]))/(E^(
2*E^x)*x^4 + x^6 - 2*x^5*Log[3] + x^4*Log[3]^2 + E^E^x*(-2*x^5 + 2*x^4*Log[3])),x]

[Out]

6*Log[3]^2*Defer[Int][1/(x^4*(-E^E^x + x - Log[3])^2), x] - 3*E^4*Log[3]^2*Defer[Int][1/(x^4*(-E^E^x + x - Log
[3])^2), x] + 12*Log[3]*Defer[Int][E^E^x/(x^4*(-E^E^x + x - Log[3])^2), x] - 6*Log[3]*Defer[Int][E^(4 + E^x)/(
x^4*(-E^E^x + x - Log[3])^2), x] - Log[3]*(16 - Log[4])*Defer[Int][1/(x^3*(-E^E^x + x - Log[3])^2), x] + 3*E^4
*Log[9]*Defer[Int][1/(x^3*(-E^E^x + x - Log[3])^2), x] - (16 - Log[4])*Defer[Int][E^E^x/(x^3*(-E^E^x + x - Log
[3])^2), x] + 6*Defer[Int][E^(4 + E^x)/(x^3*(-E^E^x + x - Log[3])^2), x] - 3*E^4*Defer[Int][1/(x^2*(-E^E^x + x
- Log[3])^2), x] + (12 - Log[8])*Defer[Int][1/(x^2*(-E^E^x + x - Log[3])^2), x] + 3*(2 - E^4)*Defer[Int][E^(2
*E^x)/(x^4*(E^E^x - x + Log[3])^2), x] - (2 - Log[2])*Defer[Int][E^(E^x + x)/(x^2*(E^E^x - x + Log[3])^2), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{2e^x}(6-3e^4)+(12-3e^4)x^2-3x^2\log (2)+(-16x+6e^{4}x+2x\log (2))\log (3)+(6-3e^4){\log }^2(3)+e^{e^x}(-16x+6e^{4}x+2x\log (2)+e^x(-2x^2+x^2\log (2))+(12-6e^4)\log (3))}{e^{2e^x}{x}^4+x^6-2x^5\log (3)+x^4{\log }^2(3)+e^{e^x}(-2x^5+2x^4\log (3))}dx\\ & =\int \frac{e^{2e^x}(6-3e^4)+x^2(12-3e^4-3\log (2))+(-16x+6e^{4}x+2x\log (2))\log (3)+(6-3e^4){\log }^2(3)+e^{e^x}(-16x+6e^{4}x+2x\log (2)+e^x(-2x^2+x^2\log (2))+(12-6e^4)\log (3))}{e^{2e^x}{x}^4+x^6-2x^5\log (3)+x^4{\log }^2(3)+e^{e^x}(-2x^5+2x^4\log (3))}dx\\ & =\int \frac{6e^{2e^x}(1-\frac{e^4}{2})+e^{e^x+x}{x}^2(-2+\log (2))+6e^{4+e^x}(x-\log (3))-3e^4(x-\log (3){)}^2+6{\log }^2(3)+e^{e^x}(12\log (3)+x(-16+\log (4)))+x\log (3)(-16+\log (4))-x^2(-12+\log (8))}{x^4{(e^{e^x}-x+\log (3))}^2}dx\\ & =\int (-\frac{3e^4(x-\log (3){)}^2}{x^4{(-e^{e^x}+x-\log (3))}^2}+\frac{6{\log }^2(3)}{x^4{(-e^{e^x}+x-\log (3))}^2}-\frac{3e^{2e^x}(-2+e^4)}{x^4{(e^{e^x}-x+\log (3))}^2}+\frac{e^{e^x+x}(-2+\log (2))}{x^2{(e^{e^x}-x+\log (3))}^2}+\frac{6e^{4+e^x}(x-\log (3))}{x^4{(e^{e^x}-x+\log (3))}^2}+\frac{e^{e^x}(12\log (3)-x(16-\log (4)))}{x^4{(e^{e^x}-x+\log (3))}^2}+\frac{\log (3)(-16+\log (4))}{x^3{(-e^{e^x}+x-\log (3))}^2}-\frac{-12+\log (8)}{x^2{(-e^{e^x}+x-\log (3))}^2})dx\\ & =6\int \frac{e^{4+e^x}(x-\log (3))}{x^4{(e^{e^x}-x+\log (3))}^2}dx-\left(3e^4\right)\int \frac{(x-\log (3){)}^2}{x^4{(-e^{e^x}+x-\log (3))}^2}dx+\left(3(2-e^4)\right)\int \frac{e^{2e^x}}{x^4{(e^{e^x}-x+\log (3))}^2}dx+(-2+\log (2))\int \frac{e^{e^x+x}}{x^2{(e^{e^x}-x+\log (3))}^2}dx+(6{\log }^2(3))\int \frac{1}{x^4{(-e^{e^x}+x-\log (3))}^2}dx-(\log (3)(16-\log (4)))\int \frac{1}{x^3{(-e^{e^x}+x-\log (3))}^2}dx+(12-\log (8))\int \frac{1}{x^2{(-e^{e^x}+x-\log (3))}^2}dx+\int \frac{e^{e^x}(12\log (3)-x(16-\log (4)))}{x^4{(e^{e^x}-x+\log (3))}^2}dx\\ & =6\int (\frac{e^{4+e^x}}{x^3{(-e^{e^x}+x-\log (3))}^2}-\frac{e^{4+e^x}\log (3)}{x^4{(-e^{e^x}+x-\log (3))}^2})dx-\left(3e^4\right)\int (\frac{1}{x^2{(-e^{e^x}+x-\log (3))}^2}+\frac{{\log }^2(3)}{x^4{(-e^{e^x}+x-\log (3))}^2}-\frac{\log (9)}{x^3{(-e^{e^x}+x-\log (3))}^2})dx+\left(3(2-e^4)\right)\int \frac{e^{2e^x}}{x^4{(e^{e^x}-x+\log (3))}^2}dx+(-2+\log (2))\int \frac{e^{e^x+x}}{x^2{(e^{e^x}-x+\log (3))}^2}dx+(6{\log }^2(3))\int \frac{1}{x^4{(-e^{e^x}+x-\log (3))}^2}dx-(\log (3)(16-\log (4)))\int \frac{1}{x^3{(-e^{e^x}+x-\log (3))}^2}dx+(12-\log (8))\int \frac{1}{x^2{(-e^{e^x}+x-\log (3))}^2}dx+\int (\frac{12e^{e^x}\log (3)}{x^4{(-e^{e^x}+x-\log (3))}^2}+\frac{e^{e^x}(-16+\log (4))}{x^3{(-e^{e^x}+x-\log (3))}^2})dx\\ & =6\int \frac{e^{4+e^x}}{x^3{(-e^{e^x}+x-\log (3))}^2}dx-\left(3e^4\right)\int \frac{1}{x^2{(-e^{e^x}+x-\log (3))}^2}dx+\left(3(2-e^4)\right)\int \frac{e^{2e^x}}{x^4{(e^{e^x}-x+\log (3))}^2}dx+(-2+\log (2))\int \frac{e^{e^x+x}}{x^2{(e^{e^x}-x+\log (3))}^2}dx-(6\log (3))\int \frac{e^{4+e^x}}{x^4{(-e^{e^x}+x-\log (3))}^2}dx+(12\log (3))\int \frac{e^{e^x}}{x^4{(-e^{e^x}+x-\log (3))}^2}dx+(6{\log }^2(3))\int \frac{1}{x^4{(-e^{e^x}+x-\log (3))}^2}dx-(3e^4{\log }^2(3))\int \frac{1}{x^4{(-e^{e^x}+x-\log (3))}^2}dx-(\log (3)(16-\log (4)))\int \frac{1}{x^3{(-e^{e^x}+x-\log (3))}^2}dx+(-16+\log (4))\int \frac{e^{e^x}}{x^3{(-e^{e^x}+x-\log (3))}^2}dx+(12-\log (8))\int \frac{1}{x^2{(-e^{e^x}+x-\log (3))}^2}dx+(3e^4\log (9))\int \frac{1}{x^3{(-e^{e^x}+x-\log (3))}^2}dx\end{array}\end{array}$$

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Mathematica [B]  time = 0.22, size = 64, normalized size = 2.13 $$\begin{array}{c}\frac{-2+e^4-\frac{x(2-\log (2)+e^x(x(-2+\log (2))-\log (2)\log (3)+\log (9)))}{(-1+e^x(x-\log (3)))(e^{e^x}-x+\log (3))}}{x^3}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*E^x)*(6 - 3*E^4) + 12*x^2 - 3*E^4*x^2 - 3*x^2*Log[2] + (-16*x + 6*E^4*x + 2*x*Log[2])*Log[3] +
(6 - 3*E^4)*Log[3]^2 + E^E^x*(-16*x + 6*E^4*x + 2*x*Log[2] + E^x*(-2*x^2 + x^2*Log[2]) + (12 - 6*E^4)*Log[3])
)/(E^(2*E^x)*x^4 + x^6 - 2*x^5*Log[3] + x^4*Log[3]^2 + E^E^x*(-2*x^5 + 2*x^4*Log[3])),x]

[Out]

(-2 + E^4 - (x*(2 - Log[2] + E^x*(x*(-2 + Log[2]) - Log[2]*Log[3] + Log[9])))/((-1 + E^x*(x - Log[3]))*(E^E^x
- x + Log[3])))/x^3

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fricas [A]  time = 0.83, size = 51, normalized size = 1.70 $$\begin{array}{c}\frac{x{e}^4-(e^4-2)e^{\left(e^x\right)}-(e^4-2)\log (3)+x\log (2)-4x}{x^4-x^3{e}^{\left(e^x\right)}-x^3\log (3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp(4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-
16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3)^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)
/(x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log(3)+x^6),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.48, size = 56, normalized size = 1.87 $$\begin{array}{c}\frac{x{e}^4-e^4\log (3)+x\log (2)-4x-e^{(e^x+4)}+2e^{\left(e^x\right)}+2\log (3)}{x^4-x^3{e}^{\left(e^x\right)}-x^3\log (3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp(4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-
16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3)^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)
/(x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log(3)+x^6),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 33, normalized size = 1.10

method	result	size
risch	$\frac{{\mathrm{e}}^4}{x^3}-\frac{2}{x^3}-\frac{\ln (2)-2}{x^2(\ln (3)+{\mathrm{e}}^{{\mathrm{e}}^x}-x)}$	$33$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*ln(2)-2*x^2)*exp(x)+(-6*exp(4)+12)*ln(3)+2*x*ln(2)+6*x*exp(4)-16*x)*exp
(exp(x))+(-3*exp(4)+6)*ln(3)^2+(2*x*ln(2)+6*x*exp(4)-16*x)*ln(3)-3*x^2*ln(2)-3*x^2*exp(4)+12*x^2)/(x^4*exp(exp
(x))^2+(2*x^4*ln(3)-2*x^5)*exp(exp(x))+x^4*ln(3)^2-2*x^5*ln(3)+x^6),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.91, size = 50, normalized size = 1.67 $$\begin{array}{c}\frac{x(e^4+\log (2)-4)-(e^4-2)e^{\left(e^x\right)}-e^4\log (3)+2\log (3)}{x^4-x^3{e}^{\left(e^x\right)}-x^3\log (3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp(4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-
16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3)^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)
/(x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log(3)+x^6),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}-\int \frac{{\mathrm{e}}^{2{\mathrm{e}}^x}(3{\mathrm{e}}^4-6)+{\ln (3)}^2(3{\mathrm{e}}^4-6)-\ln (3)(6x{\mathrm{e}}^4-16x+2x\ln (2))+3x^2{\mathrm{e}}^4+3x^2\ln (2)-{\mathrm{e}}^{{\mathrm{e}}^x}(6x{\mathrm{e}}^4-16x+2x\ln (2)+{\mathrm{e}}^x(x^2\ln (2)-2x^2)-\ln (3)(6{\mathrm{e}}^4-12))-12x^2}{x^4{\ln (3)}^2-2x^5\ln (3)+x^6+x^4{\mathrm{e}}^{2{\mathrm{e}}^x}+{\mathrm{e}}^{{\mathrm{e}}^x}(2x^4\ln (3)-2x^5)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*exp(x))*(3*exp(4) - 6) + log(3)^2*(3*exp(4) - 6) - log(3)*(6*x*exp(4) - 16*x + 2*x*log(2)) + 3*x^2
*exp(4) + 3*x^2*log(2) - exp(exp(x))*(6*x*exp(4) - 16*x + 2*x*log(2) + exp(x)*(x^2*log(2) - 2*x^2) - log(3)*(6
*exp(4) - 12)) - 12*x^2)/(x^4*log(3)^2 - 2*x^5*log(3) + x^6 + x^4*exp(2*exp(x)) + exp(exp(x))*(2*x^4*log(3) -
2*x^5)),x)

[Out]

-int((exp(2*exp(x))*(3*exp(4) - 6) + log(3)^2*(3*exp(4) - 6) - log(3)*(6*x*exp(4) - 16*x + 2*x*log(2)) + 3*x^2
*exp(4) + 3*x^2*log(2) - exp(exp(x))*(6*x*exp(4) - 16*x + 2*x*log(2) + exp(x)*(x^2*log(2) - 2*x^2) - log(3)*(6
*exp(4) - 12)) - 12*x^2)/(x^4*log(3)^2 - 2*x^5*log(3) + x^6 + x^4*exp(2*exp(x)) + exp(exp(x))*(2*x^4*log(3) -
2*x^5)), x)

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sympy [A]  time = 0.21, size = 34, normalized size = 1.13 $$\begin{array}{c}\frac{2-\log (2)}{-x^3+x^2{e}^{e^x}+x^2\log (3)}-\frac{6-3e^4}{3x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(4)+6)*exp(exp(x))**2+((x**2*ln(2)-2*x**2)*exp(x)+(-6*exp(4)+12)*ln(3)+2*x*ln(2)+6*x*exp(4)-
16*x)*exp(exp(x))+(-3*exp(4)+6)*ln(3)**2+(2*x*ln(2)+6*x*exp(4)-16*x)*ln(3)-3*x**2*ln(2)-3*x**2*exp(4)+12*x**2)
/(x**4*exp(exp(x))**2+(2*x**4*ln(3)-2*x**5)*exp(exp(x))+x**4*ln(3)**2-2*x**5*ln(3)+x**6),x)

[Out]

(2 - log(2))/(-x**3 + x**2*exp(exp(x)) + x**2*log(3)) - (6 - 3*exp(4))/(3*x**3)

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