\(\int \frac {e^{e^{\frac {9 x}{6+e^3+\log (\frac {5}{3})+3 \log (x)}}+\frac {9 x}{6+e^3+\log (\frac {5}{3})+3 \log (x)}} (27+9 e^3+9 \log (\frac {5}{3})+27 \log (x))}{36+12 e^3+e^6-(-12-2 e^3) \log (\frac {5}{3})+\log ^2(\frac {5}{3})+(36+6 e^3+6 \log (\frac {5}{3})) \log (x)+9 \log ^2(x)} \, dx\) [2305]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-2)]
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 112, antiderivative size = 25 \[ \int \frac {e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}+\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}} \left (27+9 e^3+9 \log \left (\frac {5}{3}\right )+27 \log (x)\right )}{36+12 e^3+e^6-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )+\log ^2\left (\frac {5}{3}\right )+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+9 \log ^2(x)} \, dx=e^{e^{\frac {3 x}{2+\frac {1}{3} \left (e^3+\log \left (\frac {5}{3}\right )\right )+\log (x)}}} \] Output:

exp(exp(3*x/(ln(x)+2-1/3*ln(3/5)+1/3*exp(3))))
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}+\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}} \left (27+9 e^3+9 \log \left (\frac {5}{3}\right )+27 \log (x)\right )}{36+12 e^3+e^6-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )+\log ^2\left (\frac {5}{3}\right )+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+9 \log ^2(x)} \, dx=e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}} \] Input:

Integrate[(E^(E^((9*x)/(6 + E^3 + Log[5/3] + 3*Log[x])) + (9*x)/(6 + E^3 + 
 Log[5/3] + 3*Log[x]))*(27 + 9*E^3 + 9*Log[5/3] + 27*Log[x]))/(36 + 12*E^3 
 + E^6 - (-12 - 2*E^3)*Log[5/3] + Log[5/3]^2 + (36 + 6*E^3 + 6*Log[5/3])*L 
og[x] + 9*Log[x]^2),x]
 

Output:

E^E^((9*x)/(6 + E^3 + Log[5/3] + 3*Log[x]))
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (27 \log (x)+9 e^3+27+9 \log \left (\frac {5}{3}\right )\right ) \exp \left (\frac {9 x}{3 \log (x)+e^3+6+\log \left (\frac {5}{3}\right )}+e^{\frac {9 x}{3 \log (x)+e^3+6+\log \left (\frac {5}{3}\right )}}\right )}{9 \log ^2(x)+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+e^6+12 e^3+36+\log ^2\left (\frac {5}{3}\right )-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {9 \left (3 \log (x)+3 \left (1+\frac {1}{3} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )\right ) \exp \left (\frac {9 x}{3 \log (x)+e^3+6+\log \left (\frac {5}{3}\right )}+e^{\frac {9 x}{3 \log (x)+e^3+6+\log \left (\frac {5}{3}\right )}}\right )}{\left (3 \log (x)+6 \left (1+\frac {1}{6} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 9 \int \frac {\exp \left (\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}+e^{\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}}\right ) \left (3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+3\right )}{\left (3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6\right )^2}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle 9 \int \frac {\exp \left (\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}+e^{\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}}\right ) \left (3 \log (x)+3 \left (1+\frac {1}{3} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )\right )}{\left (3 \log (x)+6 \left (1+\frac {1}{6} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 9 \int \left (\frac {\exp \left (\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}+e^{\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}}\right )}{3 \log (x)+6 \left (1+\frac {1}{6} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )}+\frac {\exp \left (\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}+e^{\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}}\right ) \left (-18 \left (1+\frac {1}{6} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )+9 \left (1+\frac {1}{3} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )\right )}{3 \left (3 \log (x)+6 \left (1+\frac {1}{6} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 9 \left (\int \frac {\exp \left (\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}+e^{\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}}\right )}{3 \log (x)+6 \left (1+\frac {1}{6} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )}dx-3 \int \frac {\exp \left (\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}+e^{\frac {9 x}{3 \log (x)+\log \left (\frac {5}{3}\right )+e^3+6}}\right )}{\left (3 \log (x)+6 \left (1+\frac {1}{6} \left (e^3+\log \left (\frac {5}{3}\right )\right )\right )\right )^2}dx\right )\)

Input:

Int[(E^(E^((9*x)/(6 + E^3 + Log[5/3] + 3*Log[x])) + (9*x)/(6 + E^3 + Log[5 
/3] + 3*Log[x]))*(27 + 9*E^3 + 9*Log[5/3] + 27*Log[x]))/(36 + 12*E^3 + E^6 
 - (-12 - 2*E^3)*Log[5/3] + Log[5/3]^2 + (36 + 6*E^3 + 6*Log[5/3])*Log[x] 
+ 9*Log[x]^2),x]
 

Output:

$Aborted
 
Maple [A] (verified)

Time = 4.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80

method result size
parallelrisch \({\mathrm e}^{{\mathrm e}^{-\frac {9 x}{-3 \ln \left (x \right )+\ln \left (\frac {3}{5}\right )-{\mathrm e}^{3}-6}}}\) \(20\)
risch \({\mathrm e}^{{\mathrm e}^{\frac {9 x}{3 \ln \left (x \right )-\ln \left (3\right )+\ln \left (5\right )+{\mathrm e}^{3}+6}}}\) \(22\)
norman \(\frac {\left (-6+\ln \left (3\right )-\ln \left (5\right )-{\mathrm e}^{3}\right ) {\mathrm e}^{{\mathrm e}^{\frac {9 x}{3 \ln \left (x \right )-\ln \left (\frac {3}{5}\right )+{\mathrm e}^{3}+6}}}-3 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{\frac {9 x}{3 \ln \left (x \right )-\ln \left (\frac {3}{5}\right )+{\mathrm e}^{3}+6}}}}{-3 \ln \left (x \right )+\ln \left (\frac {3}{5}\right )-{\mathrm e}^{3}-6}\) \(72\)

Input:

int((27*ln(x)-9*ln(3/5)+9*exp(3)+27)*exp(9*x/(3*ln(x)-ln(3/5)+exp(3)+6))*e 
xp(exp(9*x/(3*ln(x)-ln(3/5)+exp(3)+6)))/(9*ln(x)^2+(-6*ln(3/5)+6*exp(3)+36 
)*ln(x)+ln(3/5)^2+(-2*exp(3)-12)*ln(3/5)+exp(3)^2+12*exp(3)+36),x,method=_ 
RETURNVERBOSE)
 

Output:

exp(exp(-9*x/(-3*ln(x)+ln(3/5)-exp(3)-6)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (19) = 38\).

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.76 \[ \int \frac {e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}+\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}} \left (27+9 e^3+9 \log \left (\frac {5}{3}\right )+27 \log (x)\right )}{36+12 e^3+e^6-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )+\log ^2\left (\frac {5}{3}\right )+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+9 \log ^2(x)} \, dx=e^{\left (\frac {{\left (e^{3} - \log \left (\frac {3}{5}\right ) + 3 \, \log \left (x\right ) + 6\right )} e^{\left (\frac {9 \, x}{e^{3} - \log \left (\frac {3}{5}\right ) + 3 \, \log \left (x\right ) + 6}\right )} + 9 \, x}{e^{3} - \log \left (\frac {3}{5}\right ) + 3 \, \log \left (x\right ) + 6} - \frac {9 \, x}{e^{3} - \log \left (\frac {3}{5}\right ) + 3 \, \log \left (x\right ) + 6}\right )} \] Input:

integrate((27*log(x)-9*log(3/5)+9*exp(3)+27)*exp(9*x/(3*log(x)-log(3/5)+ex 
p(3)+6))*exp(exp(9*x/(3*log(x)-log(3/5)+exp(3)+6)))/(9*log(x)^2+(-6*log(3/ 
5)+6*exp(3)+36)*log(x)+log(3/5)^2+(-2*exp(3)-12)*log(3/5)+exp(3)^2+12*exp( 
3)+36),x, algorithm="fricas")
 

Output:

e^(((e^3 - log(3/5) + 3*log(x) + 6)*e^(9*x/(e^3 - log(3/5) + 3*log(x) + 6) 
) + 9*x)/(e^3 - log(3/5) + 3*log(x) + 6) - 9*x/(e^3 - log(3/5) + 3*log(x) 
+ 6))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}+\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}} \left (27+9 e^3+9 \log \left (\frac {5}{3}\right )+27 \log (x)\right )}{36+12 e^3+e^6-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )+\log ^2\left (\frac {5}{3}\right )+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+9 \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((27*ln(x)-9*ln(3/5)+9*exp(3)+27)*exp(9*x/(3*ln(x)-ln(3/5)+exp(3) 
+6))*exp(exp(9*x/(3*ln(x)-ln(3/5)+exp(3)+6)))/(9*ln(x)**2+(-6*ln(3/5)+6*ex 
p(3)+36)*ln(x)+ln(3/5)**2+(-2*exp(3)-12)*ln(3/5)+exp(3)**2+12*exp(3)+36),x 
)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}+\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}} \left (27+9 e^3+9 \log \left (\frac {5}{3}\right )+27 \log (x)\right )}{36+12 e^3+e^6-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )+\log ^2\left (\frac {5}{3}\right )+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+9 \log ^2(x)} \, dx=e^{\left (e^{\left (\frac {9 \, x}{e^{3} + \log \left (5\right ) - \log \left (3\right ) + 3 \, \log \left (x\right ) + 6}\right )}\right )} \] Input:

integrate((27*log(x)-9*log(3/5)+9*exp(3)+27)*exp(9*x/(3*log(x)-log(3/5)+ex 
p(3)+6))*exp(exp(9*x/(3*log(x)-log(3/5)+exp(3)+6)))/(9*log(x)^2+(-6*log(3/ 
5)+6*exp(3)+36)*log(x)+log(3/5)^2+(-2*exp(3)-12)*log(3/5)+exp(3)^2+12*exp( 
3)+36),x, algorithm="maxima")
 

Output:

e^(e^(9*x/(e^3 + log(5) - log(3) + 3*log(x) + 6)))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (19) = 38\).

Time = 1.41 (sec) , antiderivative size = 157, normalized size of antiderivative = 6.28 \[ \int \frac {e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}+\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}} \left (27+9 e^3+9 \log \left (\frac {5}{3}\right )+27 \log (x)\right )}{36+12 e^3+e^6-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )+\log ^2\left (\frac {5}{3}\right )+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+9 \log ^2(x)} \, dx=e^{\left (\frac {e^{\left (\frac {9 \, x}{e^{3} + \log \left (5\right ) - \log \left (3\right ) + 3 \, \log \left (x\right ) + 6}\right )} \log \left (5\right ) - e^{\left (\frac {9 \, x}{e^{3} + \log \left (5\right ) - \log \left (3\right ) + 3 \, \log \left (x\right ) + 6}\right )} \log \left (3\right ) + 3 \, e^{\left (\frac {9 \, x}{e^{3} + \log \left (5\right ) - \log \left (3\right ) + 3 \, \log \left (x\right ) + 6}\right )} \log \left (x\right ) + 9 \, x + 6 \, e^{\left (\frac {9 \, x}{e^{3} + \log \left (5\right ) - \log \left (3\right ) + 3 \, \log \left (x\right ) + 6}\right )} + e^{\left (\frac {9 \, x}{e^{3} + \log \left (5\right ) - \log \left (3\right ) + 3 \, \log \left (x\right ) + 6} + 3\right )}}{e^{3} + \log \left (5\right ) - \log \left (3\right ) + 3 \, \log \left (x\right ) + 6} - \frac {9 \, x}{e^{3} + \log \left (5\right ) - \log \left (3\right ) + 3 \, \log \left (x\right ) + 6}\right )} \] Input:

integrate((27*log(x)-9*log(3/5)+9*exp(3)+27)*exp(9*x/(3*log(x)-log(3/5)+ex 
p(3)+6))*exp(exp(9*x/(3*log(x)-log(3/5)+exp(3)+6)))/(9*log(x)^2+(-6*log(3/ 
5)+6*exp(3)+36)*log(x)+log(3/5)^2+(-2*exp(3)-12)*log(3/5)+exp(3)^2+12*exp( 
3)+36),x, algorithm="giac")
 

Output:

e^((e^(9*x/(e^3 + log(5) - log(3) + 3*log(x) + 6))*log(5) - e^(9*x/(e^3 + 
log(5) - log(3) + 3*log(x) + 6))*log(3) + 3*e^(9*x/(e^3 + log(5) - log(3) 
+ 3*log(x) + 6))*log(x) + 9*x + 6*e^(9*x/(e^3 + log(5) - log(3) + 3*log(x) 
 + 6)) + e^(9*x/(e^3 + log(5) - log(3) + 3*log(x) + 6) + 3))/(e^3 + log(5) 
 - log(3) + 3*log(x) + 6) - 9*x/(e^3 + log(5) - log(3) + 3*log(x) + 6))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}+\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}} \left (27+9 e^3+9 \log \left (\frac {5}{3}\right )+27 \log (x)\right )}{36+12 e^3+e^6-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )+\log ^2\left (\frac {5}{3}\right )+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+9 \log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^{\frac {9\,x}{{\mathrm {e}}^3-\ln \left (\frac {3}{5}\right )+3\,\ln \left (x\right )+6}}\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {9\,x}{{\mathrm {e}}^3-\ln \left (\frac {3}{5}\right )+3\,\ln \left (x\right )+6}}}\,\left (9\,{\mathrm {e}}^3-9\,\ln \left (\frac {3}{5}\right )+27\,\ln \left (x\right )+27\right )}{9\,{\ln \left (x\right )}^2+\left (6\,{\mathrm {e}}^3-6\,\ln \left (\frac {3}{5}\right )+36\right )\,\ln \left (x\right )+12\,{\mathrm {e}}^3+{\mathrm {e}}^6+{\ln \left (\frac {3}{5}\right )}^2-\ln \left (\frac {3}{5}\right )\,\left (2\,{\mathrm {e}}^3+12\right )+36} \,d x \] Input:

int((exp((9*x)/(exp(3) - log(3/5) + 3*log(x) + 6))*exp(exp((9*x)/(exp(3) - 
 log(3/5) + 3*log(x) + 6)))*(9*exp(3) - 9*log(3/5) + 27*log(x) + 27))/(12* 
exp(3) + exp(6) + log(x)*(6*exp(3) - 6*log(3/5) + 36) + 9*log(x)^2 + log(3 
/5)^2 - log(3/5)*(2*exp(3) + 12) + 36),x)
 

Output:

int((exp((9*x)/(exp(3) - log(3/5) + 3*log(x) + 6))*exp(exp((9*x)/(exp(3) - 
 log(3/5) + 3*log(x) + 6)))*(9*exp(3) - 9*log(3/5) + 27*log(x) + 27))/(12* 
exp(3) + exp(6) + log(x)*(6*exp(3) - 6*log(3/5) + 36) + 9*log(x)^2 + log(3 
/5)^2 - log(3/5)*(2*exp(3) + 12) + 36), x)
 

Reduce [F]

\[ \int \frac {e^{e^{\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}}+\frac {9 x}{6+e^3+\log \left (\frac {5}{3}\right )+3 \log (x)}} \left (27+9 e^3+9 \log \left (\frac {5}{3}\right )+27 \log (x)\right )}{36+12 e^3+e^6-\left (-12-2 e^3\right ) \log \left (\frac {5}{3}\right )+\log ^2\left (\frac {5}{3}\right )+\left (36+6 e^3+6 \log \left (\frac {5}{3}\right )\right ) \log (x)+9 \log ^2(x)} \, dx =\text {Too large to display} \] Input:

int((27*log(x)-9*log(3/5)+9*exp(3)+27)*exp(9*x/(3*log(x)-log(3/5)+exp(3)+6 
))*exp(exp(9*x/(3*log(x)-log(3/5)+exp(3)+6)))/(9*log(x)^2+(-6*log(3/5)+6*e 
xp(3)+36)*log(x)+log(3/5)^2+(-2*exp(3)-12)*log(3/5)+exp(3)^2+12*exp(3)+36) 
,x)
 

Output:

9*( - int(e**(1/e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6)))/(e**((9*x)/(l 
og(3/5) - 3*log(x) - e**3 - 6))*log(3/5)**2 - 6*e**((9*x)/(log(3/5) - 3*lo 
g(x) - e**3 - 6))*log(3/5)*log(x) - 2*e**((9*x)/(log(3/5) - 3*log(x) - e** 
3 - 6))*log(3/5)*e**3 - 12*e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*log 
(3/5) + 9*e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*log(x)**2 + 6*e**((9 
*x)/(log(3/5) - 3*log(x) - e**3 - 6))*log(x)*e**3 + 36*e**((9*x)/(log(3/5) 
 - 3*log(x) - e**3 - 6))*log(x) + e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 
6))*e**6 + 12*e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*e**3 + 36*e**((9 
*x)/(log(3/5) - 3*log(x) - e**3 - 6))),x)*log(3/5) + int(e**(1/e**((9*x)/( 
log(3/5) - 3*log(x) - e**3 - 6)))/(e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 
 6))*log(3/5)**2 - 6*e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*log(3/5)* 
log(x) - 2*e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*log(3/5)*e**3 - 12* 
e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*log(3/5) + 9*e**((9*x)/(log(3/ 
5) - 3*log(x) - e**3 - 6))*log(x)**2 + 6*e**((9*x)/(log(3/5) - 3*log(x) - 
e**3 - 6))*log(x)*e**3 + 36*e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*lo 
g(x) + e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*e**6 + 12*e**((9*x)/(lo 
g(3/5) - 3*log(x) - e**3 - 6))*e**3 + 36*e**((9*x)/(log(3/5) - 3*log(x) - 
e**3 - 6))),x)*e**3 + 3*int(e**(1/e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 
6)))/(e**((9*x)/(log(3/5) - 3*log(x) - e**3 - 6))*log(3/5)**2 - 6*e**((9*x 
)/(log(3/5) - 3*log(x) - e**3 - 6))*log(3/5)*log(x) - 2*e**((9*x)/(log(...