3.17.59 \(\int \frac {e^{\frac {1}{-x-4 \log (3)+\log (x+e^x \log (x^2))}} (e^x (-16-4 x)-8 x+6 x^2+2 x^3+2 x^4+16 x^3 \log (3)+32 x^2 \log ^2(3)+e^x (2 x^3+16 x^2 \log (3)+32 x \log ^2(3)) \log (x^2)+(-4 x^3-16 x^2 \log (3)+e^x (-4 x^2-16 x \log (3)) \log (x^2)) \log (x+e^x \log (x^2))+(2 x^2+2 e^x x \log (x^2)) \log ^2(x+e^x \log (x^2)))}{x^4+8 x^3 \log (3)+16 x^2 \log ^2(3)+e^x (x^3+8 x^2 \log (3)+16 x \log ^2(3)) \log (x^2)+(-2 x^3-8 x^2 \log (3)+e^x (-2 x^2-8 x \log (3)) \log (x^2)) \log (x+e^x \log (x^2))+(x^2+e^x x \log (x^2)) \log ^2(x+e^x \log (x^2))} \, dx\)

Optimal. Leaf size=28 \[ 2 e^{\frac {1}{-x-4 \log (3)+\log \left (x+e^x \log \left (x^2\right )\right )}} (4+x) \]

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Rubi [F]  time = 180.00, 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 {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x - 4*Log[3] + Log[x + E^x*Log[x^2]])^(-1)*(E^x*(-16 - 4*x) - 8*x + 6*x^2 + 2*x^3 + 2*x^4 + 16*x^3*Lo
g[3] + 32*x^2*Log[3]^2 + E^x*(2*x^3 + 16*x^2*Log[3] + 32*x*Log[3]^2)*Log[x^2] + (-4*x^3 - 16*x^2*Log[3] + E^x*
(-4*x^2 - 16*x*Log[3])*Log[x^2])*Log[x + E^x*Log[x^2]] + (2*x^2 + 2*E^x*x*Log[x^2])*Log[x + E^x*Log[x^2]]^2))/
(x^4 + 8*x^3*Log[3] + 16*x^2*Log[3]^2 + E^x*(x^3 + 8*x^2*Log[3] + 16*x*Log[3]^2)*Log[x^2] + (-2*x^3 - 8*x^2*Lo
g[3] + E^x*(-2*x^2 - 8*x*Log[3])*Log[x^2])*Log[x + E^x*Log[x^2]] + (x^2 + E^x*x*Log[x^2])*Log[x + E^x*Log[x^2]
]^2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [F]  time = 0.50, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {1}{-x-4 \log (3)+\log \left (x+e^x \log \left (x^2\right )\right )}} \left (e^x (-16-4 x)-8 x+6 x^2+2 x^3+2 x^4+16 x^3 \log (3)+32 x^2 \log ^2(3)+e^x \left (2 x^3+16 x^2 \log (3)+32 x \log ^2(3)\right ) \log \left (x^2\right )+\left (-4 x^3-16 x^2 \log (3)+e^x \left (-4 x^2-16 x \log (3)\right ) \log \left (x^2\right )\right ) \log \left (x+e^x \log \left (x^2\right )\right )+\left (2 x^2+2 e^x x \log \left (x^2\right )\right ) \log ^2\left (x+e^x \log \left (x^2\right )\right )\right )}{x^4+8 x^3 \log (3)+16 x^2 \log ^2(3)+e^x \left (x^3+8 x^2 \log (3)+16 x \log ^2(3)\right ) \log \left (x^2\right )+\left (-2 x^3-8 x^2 \log (3)+e^x \left (-2 x^2-8 x \log (3)\right ) \log \left (x^2\right )\right ) \log \left (x+e^x \log \left (x^2\right )\right )+\left (x^2+e^x x \log \left (x^2\right )\right ) \log ^2\left (x+e^x \log \left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^(-x - 4*Log[3] + Log[x + E^x*Log[x^2]])^(-1)*(E^x*(-16 - 4*x) - 8*x + 6*x^2 + 2*x^3 + 2*x^4 + 16*
x^3*Log[3] + 32*x^2*Log[3]^2 + E^x*(2*x^3 + 16*x^2*Log[3] + 32*x*Log[3]^2)*Log[x^2] + (-4*x^3 - 16*x^2*Log[3]
+ E^x*(-4*x^2 - 16*x*Log[3])*Log[x^2])*Log[x + E^x*Log[x^2]] + (2*x^2 + 2*E^x*x*Log[x^2])*Log[x + E^x*Log[x^2]
]^2))/(x^4 + 8*x^3*Log[3] + 16*x^2*Log[3]^2 + E^x*(x^3 + 8*x^2*Log[3] + 16*x*Log[3]^2)*Log[x^2] + (-2*x^3 - 8*
x^2*Log[3] + E^x*(-2*x^2 - 8*x*Log[3])*Log[x^2])*Log[x + E^x*Log[x^2]] + (x^2 + E^x*x*Log[x^2])*Log[x + E^x*Lo
g[x^2]]^2),x]

[Out]

Integrate[(E^(-x - 4*Log[3] + Log[x + E^x*Log[x^2]])^(-1)*(E^x*(-16 - 4*x) - 8*x + 6*x^2 + 2*x^3 + 2*x^4 + 16*
x^3*Log[3] + 32*x^2*Log[3]^2 + E^x*(2*x^3 + 16*x^2*Log[3] + 32*x*Log[3]^2)*Log[x^2] + (-4*x^3 - 16*x^2*Log[3]
+ E^x*(-4*x^2 - 16*x*Log[3])*Log[x^2])*Log[x + E^x*Log[x^2]] + (2*x^2 + 2*E^x*x*Log[x^2])*Log[x + E^x*Log[x^2]
]^2))/(x^4 + 8*x^3*Log[3] + 16*x^2*Log[3]^2 + E^x*(x^3 + 8*x^2*Log[3] + 16*x*Log[3]^2)*Log[x^2] + (-2*x^3 - 8*
x^2*Log[3] + E^x*(-2*x^2 - 8*x*Log[3])*Log[x^2])*Log[x + E^x*Log[x^2]] + (x^2 + E^x*x*Log[x^2])*Log[x + E^x*Lo
g[x^2]]^2), x]

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fricas [A]  time = 0.62, size = 28, normalized size = 1.00 \begin {gather*} 2 \, {\left (x + 4\right )} e^{\left (-\frac {1}{x + 4 \, \log \relax (3) - \log \left (e^{x} \log \left (x^{2}\right ) + x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)*log(x^2)+2*x^2)*log(exp(x)*log(x^2)+x)^2+((-16*x*log(3)-4*x^2)*exp(x)*log(x^2)-16*x^2*l
og(3)-4*x^3)*log(exp(x)*log(x^2)+x)+(32*x*log(3)^2+16*x^2*log(3)+2*x^3)*exp(x)*log(x^2)+(-16-4*x)*exp(x)+32*x^
2*log(3)^2+16*x^3*log(3)+2*x^4+2*x^3+6*x^2-8*x)*exp(1/(log(exp(x)*log(x^2)+x)-4*log(3)-x))/((x*exp(x)*log(x^2)
+x^2)*log(exp(x)*log(x^2)+x)^2+((-8*x*log(3)-2*x^2)*exp(x)*log(x^2)-8*x^2*log(3)-2*x^3)*log(exp(x)*log(x^2)+x)
+(16*x*log(3)^2+8*x^2*log(3)+x^3)*exp(x)*log(x^2)+16*x^2*log(3)^2+8*x^3*log(3)+x^4),x, algorithm="fricas")

[Out]

2*(x + 4)*e^(-1/(x + 4*log(3) - log(e^x*log(x^2) + x)))

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giac [B]  time = 4.20, size = 108, normalized size = 3.86 \begin {gather*} 2 \, x e^{\left (\frac {x - \log \left (e^{x} \log \left (x^{2}\right ) + x\right )}{4 \, {\left (x \log \relax (3) + 4 \, \log \relax (3)^{2} - \log \relax (3) \log \left (e^{x} \log \left (x^{2}\right ) + x\right )\right )}} - \frac {1}{4 \, \log \relax (3)}\right )} + 8 \, e^{\left (\frac {x - \log \left (e^{x} \log \left (x^{2}\right ) + x\right )}{4 \, {\left (x \log \relax (3) + 4 \, \log \relax (3)^{2} - \log \relax (3) \log \left (e^{x} \log \left (x^{2}\right ) + x\right )\right )}} - \frac {1}{4 \, \log \relax (3)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)*log(x^2)+2*x^2)*log(exp(x)*log(x^2)+x)^2+((-16*x*log(3)-4*x^2)*exp(x)*log(x^2)-16*x^2*l
og(3)-4*x^3)*log(exp(x)*log(x^2)+x)+(32*x*log(3)^2+16*x^2*log(3)+2*x^3)*exp(x)*log(x^2)+(-16-4*x)*exp(x)+32*x^
2*log(3)^2+16*x^3*log(3)+2*x^4+2*x^3+6*x^2-8*x)*exp(1/(log(exp(x)*log(x^2)+x)-4*log(3)-x))/((x*exp(x)*log(x^2)
+x^2)*log(exp(x)*log(x^2)+x)^2+((-8*x*log(3)-2*x^2)*exp(x)*log(x^2)-8*x^2*log(3)-2*x^3)*log(exp(x)*log(x^2)+x)
+(16*x*log(3)^2+8*x^2*log(3)+x^3)*exp(x)*log(x^2)+16*x^2*log(3)^2+8*x^3*log(3)+x^4),x, algorithm="giac")

[Out]

2*x*e^(1/4*(x - log(e^x*log(x^2) + x))/(x*log(3) + 4*log(3)^2 - log(3)*log(e^x*log(x^2) + x)) - 1/4/log(3)) +
8*e^(1/4*(x - log(e^x*log(x^2) + x))/(x*log(3) + 4*log(3)^2 - log(3)*log(e^x*log(x^2) + x)) - 1/4/log(3))

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maple [C]  time = 0.24, size = 59, normalized size = 2.11

method	result	size
risch	\(\left (2 x +8\right ) {\mathrm e}^{-\frac {1}{-\ln \left ({\mathrm e}^{x} \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )+x \right )+4 \ln \relax (3)+x}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(x)*ln(x^2)+2*x^2)*ln(exp(x)*ln(x^2)+x)^2+((-16*x*ln(3)-4*x^2)*exp(x)*ln(x^2)-16*x^2*ln(3)-4*x^3)
*ln(exp(x)*ln(x^2)+x)+(32*x*ln(3)^2+16*x^2*ln(3)+2*x^3)*exp(x)*ln(x^2)+(-16-4*x)*exp(x)+32*x^2*ln(3)^2+16*x^3*
ln(3)+2*x^4+2*x^3+6*x^2-8*x)*exp(1/(ln(exp(x)*ln(x^2)+x)-4*ln(3)-x))/((x*exp(x)*ln(x^2)+x^2)*ln(exp(x)*ln(x^2)
+x)^2+((-8*x*ln(3)-2*x^2)*exp(x)*ln(x^2)-8*x^2*ln(3)-2*x^3)*ln(exp(x)*ln(x^2)+x)+(16*x*ln(3)^2+8*x^2*ln(3)+x^3
)*exp(x)*ln(x^2)+16*x^2*ln(3)^2+8*x^3*ln(3)+x^4),x,method=_RETURNVERBOSE)

[Out]

(2*x+8)*exp(-1/(-ln(exp(x)*(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)+x)+4*ln(3)+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {2 \, x^{4} e^{\left (-\frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )}}{x^{2} - x - 2 \, e^{x}} + \frac {16 \, x^{3} e^{\left (-\frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )} \log \relax (3)}{x^{2} - x - 2 \, e^{x}} + \frac {32 \, x^{2} e^{\left (-\frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )} \log \relax (3)^{2}}{x^{2} - x - 2 \, e^{x}} + \frac {4 \, x^{3} e^{\left (x - \frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )} \log \relax (x)}{x^{2} - x - 2 \, e^{x}} + \frac {32 \, x^{2} e^{\left (x - \frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )} \log \relax (3) \log \relax (x)}{x^{2} - x - 2 \, e^{x}} + \frac {64 \, x e^{\left (x - \frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )} \log \relax (3)^{2} \log \relax (x)}{x^{2} - x - 2 \, e^{x}} + \frac {2 \, x^{3} e^{\left (-\frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )}}{x^{2} - x - 2 \, e^{x}} + \frac {6 \, x^{2} e^{\left (-\frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )}}{x^{2} - x - 2 \, e^{x}} - \frac {4 \, x e^{\left (x - \frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )}}{x^{2} - x - 2 \, e^{x}} - \frac {8 \, x e^{\left (-\frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )}}{x^{2} - x - 2 \, e^{x}} - \frac {16 \, e^{\left (x - \frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )}}{x^{2} - x - 2 \, e^{x}} + 2 \, \int -\frac {{\left (2 \, {\left (x + 4 \, \log \relax (3)\right )} \log \left (2 \, e^{x} \log \relax (x) + x\right ) - \log \left (2 \, e^{x} \log \relax (x) + x\right )^{2}\right )} e^{\left (-\frac {1}{x + 4 \, \log \relax (3) - \log \left (2 \, e^{x} \log \relax (x) + x\right )}\right )}}{x^{2} + 8 \, x \log \relax (3) + 16 \, \log \relax (3)^{2} - 2 \, {\left (x + 4 \, \log \relax (3)\right )} \log \left (2 \, e^{x} \log \relax (x) + x\right ) + \log \left (2 \, e^{x} \log \relax (x) + x\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)*log(x^2)+2*x^2)*log(exp(x)*log(x^2)+x)^2+((-16*x*log(3)-4*x^2)*exp(x)*log(x^2)-16*x^2*l
og(3)-4*x^3)*log(exp(x)*log(x^2)+x)+(32*x*log(3)^2+16*x^2*log(3)+2*x^3)*exp(x)*log(x^2)+(-16-4*x)*exp(x)+32*x^
2*log(3)^2+16*x^3*log(3)+2*x^4+2*x^3+6*x^2-8*x)*exp(1/(log(exp(x)*log(x^2)+x)-4*log(3)-x))/((x*exp(x)*log(x^2)
+x^2)*log(exp(x)*log(x^2)+x)^2+((-8*x*log(3)-2*x^2)*exp(x)*log(x^2)-8*x^2*log(3)-2*x^3)*log(exp(x)*log(x^2)+x)
+(16*x*log(3)^2+8*x^2*log(3)+x^3)*exp(x)*log(x^2)+16*x^2*log(3)^2+8*x^3*log(3)+x^4),x, algorithm="maxima")

[Out]

2*x^4*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 - x - 2*e^x) + 16*x^3*e^(-1/(x + 4*log(3) - log(2*e^x
*log(x) + x)))*log(3)/(x^2 - x - 2*e^x) + 32*x^2*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))*log(3)^2/(x^2 -
x - 2*e^x) + 4*x^3*e^(x - 1/(x + 4*log(3) - log(2*e^x*log(x) + x)))*log(x)/(x^2 - x - 2*e^x) + 32*x^2*e^(x -
1/(x + 4*log(3) - log(2*e^x*log(x) + x)))*log(3)*log(x)/(x^2 - x - 2*e^x) + 64*x*e^(x - 1/(x + 4*log(3) - log(
2*e^x*log(x) + x)))*log(3)^2*log(x)/(x^2 - x - 2*e^x) + 2*x^3*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x
^2 - x - 2*e^x) + 6*x^2*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 - x - 2*e^x) - 4*x*e^(x - 1/(x + 4*
log(3) - log(2*e^x*log(x) + x)))/(x^2 - x - 2*e^x) - 8*x*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 -
x - 2*e^x) - 16*e^(x - 1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 - x - 2*e^x) + 2*integrate(-(2*(x + 4*lo
g(3))*log(2*e^x*log(x) + x) - log(2*e^x*log(x) + x)^2)*e^(-1/(x + 4*log(3) - log(2*e^x*log(x) + x)))/(x^2 + 8*
x*log(3) + 16*log(3)^2 - 2*(x + 4*log(3))*log(2*e^x*log(x) + x) + log(2*e^x*log(x) + x)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {1}{x+4\,\ln \relax (3)-\ln \left (x+\ln \left (x^2\right )\,{\mathrm {e}}^x\right )}}\,\left (32\,x^2\,{\ln \relax (3)}^2-8\,x+{\ln \left (x+\ln \left (x^2\right )\,{\mathrm {e}}^x\right )}^2\,\left (2\,x^2+2\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^x\right )-{\mathrm {e}}^x\,\left (4\,x+16\right )+16\,x^3\,\ln \relax (3)-\ln \left (x+\ln \left (x^2\right )\,{\mathrm {e}}^x\right )\,\left (16\,x^2\,\ln \relax (3)+4\,x^3+\ln \left (x^2\right )\,{\mathrm {e}}^x\,\left (4\,x^2+16\,\ln \relax (3)\,x\right )\right )+6\,x^2+2\,x^3+2\,x^4+\ln \left (x^2\right )\,{\mathrm {e}}^x\,\left (2\,x^3+16\,\ln \relax (3)\,x^2+32\,{\ln \relax (3)}^2\,x\right )\right )}{16\,x^2\,{\ln \relax (3)}^2+8\,x^3\,\ln \relax (3)+{\ln \left (x+\ln \left (x^2\right )\,{\mathrm {e}}^x\right )}^2\,\left (x^2+x\,\ln \left (x^2\right )\,{\mathrm {e}}^x\right )-\ln \left (x+\ln \left (x^2\right )\,{\mathrm {e}}^x\right )\,\left (8\,x^2\,\ln \relax (3)+2\,x^3+\ln \left (x^2\right )\,{\mathrm {e}}^x\,\left (2\,x^2+8\,\ln \relax (3)\,x\right )\right )+x^4+\ln \left (x^2\right )\,{\mathrm {e}}^x\,\left (x^3+8\,\ln \relax (3)\,x^2+16\,{\ln \relax (3)}^2\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-1/(x + 4*log(3) - log(x + log(x^2)*exp(x))))*(32*x^2*log(3)^2 - 8*x + log(x + log(x^2)*exp(x))^2*(2*
x^2 + 2*x*log(x^2)*exp(x)) - exp(x)*(4*x + 16) + 16*x^3*log(3) - log(x + log(x^2)*exp(x))*(16*x^2*log(3) + 4*x
^3 + log(x^2)*exp(x)*(16*x*log(3) + 4*x^2)) + 6*x^2 + 2*x^3 + 2*x^4 + log(x^2)*exp(x)*(32*x*log(3)^2 + 16*x^2*
log(3) + 2*x^3)))/(16*x^2*log(3)^2 + 8*x^3*log(3) + log(x + log(x^2)*exp(x))^2*(x^2 + x*log(x^2)*exp(x)) - log
(x + log(x^2)*exp(x))*(8*x^2*log(3) + 2*x^3 + log(x^2)*exp(x)*(8*x*log(3) + 2*x^2)) + x^4 + log(x^2)*exp(x)*(1
6*x*log(3)^2 + 8*x^2*log(3) + x^3)),x)

[Out]

int((exp(-1/(x + 4*log(3) - log(x + log(x^2)*exp(x))))*(32*x^2*log(3)^2 - 8*x + log(x + log(x^2)*exp(x))^2*(2*
x^2 + 2*x*log(x^2)*exp(x)) - exp(x)*(4*x + 16) + 16*x^3*log(3) - log(x + log(x^2)*exp(x))*(16*x^2*log(3) + 4*x
^3 + log(x^2)*exp(x)*(16*x*log(3) + 4*x^2)) + 6*x^2 + 2*x^3 + 2*x^4 + log(x^2)*exp(x)*(32*x*log(3)^2 + 16*x^2*
log(3) + 2*x^3)))/(16*x^2*log(3)^2 + 8*x^3*log(3) + log(x + log(x^2)*exp(x))^2*(x^2 + x*log(x^2)*exp(x)) - log
(x + log(x^2)*exp(x))*(8*x^2*log(3) + 2*x^3 + log(x^2)*exp(x)*(8*x*log(3) + 2*x^2)) + x^4 + log(x^2)*exp(x)*(1
6*x*log(3)^2 + 8*x^2*log(3) + x^3)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x)*ln(x**2)+2*x**2)*ln(exp(x)*ln(x**2)+x)**2+((-16*x*ln(3)-4*x**2)*exp(x)*ln(x**2)-16*x**2
*ln(3)-4*x**3)*ln(exp(x)*ln(x**2)+x)+(32*x*ln(3)**2+16*x**2*ln(3)+2*x**3)*exp(x)*ln(x**2)+(-16-4*x)*exp(x)+32*
x**2*ln(3)**2+16*x**3*ln(3)+2*x**4+2*x**3+6*x**2-8*x)*exp(1/(ln(exp(x)*ln(x**2)+x)-4*ln(3)-x))/((x*exp(x)*ln(x
**2)+x**2)*ln(exp(x)*ln(x**2)+x)**2+((-8*x*ln(3)-2*x**2)*exp(x)*ln(x**2)-8*x**2*ln(3)-2*x**3)*ln(exp(x)*ln(x**
2)+x)+(16*x*ln(3)**2+8*x**2*ln(3)+x**3)*exp(x)*ln(x**2)+16*x**2*ln(3)**2+8*x**3*ln(3)+x**4),x)

[Out]

Timed out

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