\(\int \frac {e^{3+x \log (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)})} (-e+320 x+80 x \log (x)+5 x \log ^2(x)+(8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)) \log (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}))}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx\) [1644]

Optimal result

Integrand size = 157, antiderivative size = 22 \[ \int \frac {e^{3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx=e^{3+x \log \left (5 (x+\log (2))+\frac {e}{8+\log (x)}\right )} \] Output:

exp(ln(exp(1)/(ln(x)+8)+5*ln(2)+5*x)*x+3)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {e^{3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx=e^3 \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^x \] Input:

Integrate[(E^(3 + x*Log[(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/( 
8 + Log[x])])*(-E + 320*x + 80*x*Log[x] + 5*x*Log[x]^2 + (8*E + 320*x + 32 
0*Log[2] + (E + 80*x + 80*Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2)*Log[ 
(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])]))/(8*E + 32 
0*x + 320*Log[2] + (E + 80*x + 80*Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x] 
^2),x]
 

Output:

E^3*((E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^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.

\(\displaystyle \int \frac {\left (320 x+5 x \log ^2(x)+\left (320 x+(5 x+5 \log (2)) \log ^2(x)+(80 x+e+80 \log (2)) \log (x)+8 e+320 \log (2)\right ) \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+80 x \log (x)-e\right ) \exp \left (x \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+3\right )}{320 x+(5 x+5 \log (2)) \log ^2(x)+(80 x+e+80 \log (2)) \log (x)+8 e+320 \log (2)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (320 x+5 x \log ^2(x)+\left (320 x+(5 x+5 \log (2)) \log ^2(x)+(80 x+e+80 \log (2)) \log (x)+8 e+320 \log (2)\right ) \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+80 x \log (x)-e\right ) \exp \left (x \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+3\right )}{(\log (x)+8) \left (40 x+5 x \log (x)+5 \log (2) \log (x)+e \left (1+\frac {40 \log (2)}{e}\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {5 x \log ^2(x) \exp \left (x \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+3\right )}{(\log (x)+8) \left (40 x+5 x \log (x)+5 \log (2) \log (x)+e \left (1+\frac {40 \log (2)}{e}\right )\right )}+\frac {80 x \log (x) \exp \left (x \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+3\right )}{(\log (x)+8) \left (40 x+5 x \log (x)+5 \log (2) \log (x)+e \left (1+\frac {40 \log (2)}{e}\right )\right )}+\log \left (\frac {5 \log (x) (x+\log (2))+40 (x+\log (2))+e}{\log (x)+8}\right ) \exp \left (x \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+3\right )+\frac {\exp \left (x \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+4\right )}{(-\log (x)-8) \left (40 x+5 x \log (x)+5 \log (2) \log (x)+e \left (1+\frac {40 \log (2)}{e}\right )\right )}+\frac {320 x \exp \left (x \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+e+40 \log (2)}{\log (x)+8}\right )+3\right )}{(\log (x)+8) \left (40 x+5 x \log (x)+5 \log (2) \log (x)+e \left (1+\frac {40 \log (2)}{e}\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \exp \left (x \log \left (\frac {40 x+(5 x+5 \log (2)) \log (x)+40 \log (2)+e}{\log (x)+8}\right )+3\right ) \log \left (\frac {5 \log (x) (x+\log (2))+40 (x+\log (2))+e}{\log (x)+8}\right )dx+5 e^3 \int x \left (\frac {5 \log (x) (x+\log (2))+40 (x+\log (2))+e}{\log (x)+8}\right )^{x-1}dx-e^4 \int \frac {\left (\frac {5 \log (x) (x+\log (2))+40 (x+\log (2))+e}{\log (x)+8}\right )^{x-1}}{(\log (x)+8)^2}dx\)

Input:

Int[(E^(3 + x*Log[(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Lo 
g[x])])*(-E + 320*x + 80*x*Log[x] + 5*x*Log[x]^2 + (8*E + 320*x + 320*Log[ 
2] + (E + 80*x + 80*Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2)*Log[(E + 4 
0*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])]))/(8*E + 320*x + 
320*Log[2] + (E + 80*x + 80*Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 42.48 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59

method	result	size
parallelrisch	\({\mathrm e}^{x \ln \left (\frac {\left (5 \ln \left (2\right )+5 x \right ) \ln \left (x \right )+40 \ln \left (2\right )+{\mathrm e}+40 x}{\ln \left (x \right )+8}\right )+3}\)	\(35\)
risch	\(\left (\ln \left (x \right )+8\right )^{-x} \left ({\mathrm e}+\left (5 \ln \left (x \right )+40\right ) \ln \left (2\right )+\left (5 \ln \left (x \right )+40\right ) x \right )^{x} {\mathrm e}^{3+\frac {i x \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}+\left (5 \ln \left (x \right )+40\right ) \ln \left (2\right )+\left (5 \ln \left (x \right )+40\right ) x \right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}+\left (5 \ln \left (x \right )+40\right ) \ln \left (2\right )+\left (5 \ln \left (x \right )+40\right ) x \right )}{\ln \left (x \right )+8}\right )^{2}}{2}-\frac {i x \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}+\left (5 \ln \left (x \right )+40\right ) \ln \left (2\right )+\left (5 \ln \left (x \right )+40\right ) x \right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}+\left (5 \ln \left (x \right )+40\right ) \ln \left (2\right )+\left (5 \ln \left (x \right )+40\right ) x \right )}{\ln \left (x \right )+8}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )+8}\right )}{2}-\frac {i x \pi \operatorname {csgn}\left (\frac {i \left ({\mathrm e}+\left (5 \ln \left (x \right )+40\right ) \ln \left (2\right )+\left (5 \ln \left (x \right )+40\right ) x \right )}{\ln \left (x \right )+8}\right )^{3}}{2}+\frac {i x \pi \operatorname {csgn}\left (\frac {i \left ({\mathrm e}+\left (5 \ln \left (x \right )+40\right ) \ln \left (2\right )+\left (5 \ln \left (x \right )+40\right ) x \right )}{\ln \left (x \right )+8}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )+8}\right )}{2}}\)	\(249\)

Input:

int((((5*ln(2)+5*x)*ln(x)^2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1 
)+320*x)*ln(((5*ln(2)+5*x)*ln(x)+40*ln(2)+exp(1)+40*x)/(ln(x)+8))+5*x*ln(x 
)^2+80*x*ln(x)-exp(1)+320*x)*exp(x*ln(((5*ln(2)+5*x)*ln(x)+40*ln(2)+exp(1) 
+40*x)/(ln(x)+8))+3)/((5*ln(2)+5*x)*ln(x)^2+(80*ln(2)+exp(1)+80*x)*ln(x)+3 
20*ln(2)+8*exp(1)+320*x),x,method=_RETURNVERBOSE)
 

Output:

exp(x*ln(((5*ln(2)+5*x)*ln(x)+40*ln(2)+exp(1)+40*x)/(ln(x)+8))+3)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^{3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx=e^{\left (x \log \left (\frac {5 \, {\left (x + \log \left (2\right )\right )} \log \left (x\right ) + 40 \, x + e + 40 \, \log \left (2\right )}{\log \left (x\right ) + 8}\right ) + 3\right )} \] Input:

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log 
(2)+8*exp(1)+320*x)*log(((5*log(2)+5*x)*log(x)+40*log(2)+exp(1)+40*x)/(log 
(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)*l 
og(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*l 
og(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x),x, algorithm="fricas" 
)
 

Output:

e^(x*log((5*(x + log(2))*log(x) + 40*x + e + 40*log(2))/(log(x) + 8)) + 3)
 

Sympy [A] (verification not implemented)

Time = 3.42 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {e^{3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx=e^{x \log {\left (\frac {40 x + \left (5 x + 5 \log {\left (2 \right )}\right ) \log {\left (x \right )} + e + 40 \log {\left (2 \right )}}{\log {\left (x \right )} + 8} \right )} + 3} \] Input:

integrate((((5*ln(2)+5*x)*ln(x)**2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+ 
8*exp(1)+320*x)*ln(((5*ln(2)+5*x)*ln(x)+40*ln(2)+exp(1)+40*x)/(ln(x)+8))+5 
*x*ln(x)**2+80*x*ln(x)-exp(1)+320*x)*exp(x*ln(((5*ln(2)+5*x)*ln(x)+40*ln(2 
)+exp(1)+40*x)/(ln(x)+8))+3)/((5*ln(2)+5*x)*ln(x)**2+(80*ln(2)+exp(1)+80*x 
)*ln(x)+320*ln(2)+8*exp(1)+320*x),x)
 

Output:

exp(x*log((40*x + (5*x + 5*log(2))*log(x) + E + 40*log(2))/(log(x) + 8)) + 
 3)
 

Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {e^{3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx=e^{\left (x \log \left (5 \, {\left (x + \log \left (2\right )\right )} \log \left (x\right ) + 40 \, x + e + 40 \, \log \left (2\right )\right ) - x \log \left (\log \left (x\right ) + 8\right ) + 3\right )} \] Input:

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log 
(2)+8*exp(1)+320*x)*log(((5*log(2)+5*x)*log(x)+40*log(2)+exp(1)+40*x)/(log 
(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)*l 
og(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*l 
og(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x),x, algorithm="maxima" 
)
 

Output:

e^(x*log(5*(x + log(2))*log(x) + 40*x + e + 40*log(2)) - x*log(log(x) + 8) 
 + 3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (23) = 46\).

Time = 0.33 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.64 \[ \int \frac {e^{3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx=e^{\left (x \log \left (\frac {5 \, x \log \left (x\right )}{\log \left (x\right ) + 8} + \frac {5 \, \log \left (2\right ) \log \left (x\right )}{\log \left (x\right ) + 8} + \frac {40 \, x}{\log \left (x\right ) + 8} + \frac {e}{\log \left (x\right ) + 8} + \frac {40 \, \log \left (2\right )}{\log \left (x\right ) + 8}\right ) + 3\right )} \] Input:

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log 
(2)+8*exp(1)+320*x)*log(((5*log(2)+5*x)*log(x)+40*log(2)+exp(1)+40*x)/(log 
(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)*l 
og(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*l 
og(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x),x, algorithm="giac")
 

Output:

e^(x*log(5*x*log(x)/(log(x) + 8) + 5*log(2)*log(x)/(log(x) + 8) + 40*x/(lo 
g(x) + 8) + e/(log(x) + 8) + 40*log(2)/(log(x) + 8)) + 3)
 

Mupad [B] (verification not implemented)

Time = 2.59 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {e^{3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx={\mathrm {e}}^3\,{\left (\frac {40\,x+\mathrm {e}+40\,\ln \left (2\right )+5\,\ln \left (2\right )\,\ln \left (x\right )+5\,x\,\ln \left (x\right )}{\ln \left (x\right )+8}\right )}^x \] Input:

int((exp(x*log((40*x + exp(1) + 40*log(2) + log(x)*(5*x + 5*log(2)))/(log( 
x) + 8)) + 3)*(320*x - exp(1) + 5*x*log(x)^2 + log((40*x + exp(1) + 40*log 
(2) + log(x)*(5*x + 5*log(2)))/(log(x) + 8))*(320*x + 8*exp(1) + 320*log(2 
) + log(x)^2*(5*x + 5*log(2)) + log(x)*(80*x + exp(1) + 80*log(2))) + 80*x 
*log(x)))/(320*x + 8*exp(1) + 320*log(2) + log(x)^2*(5*x + 5*log(2)) + log 
(x)*(80*x + exp(1) + 80*log(2))),x)
 

Output:

exp(3)*((40*x + exp(1) + 40*log(2) + 5*log(2)*log(x) + 5*x*log(x))/(log(x) 
 + 8))^x
 

Reduce [F]

\[ \int \frac {e^{3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx=\text {too large to display} \] Input:

int((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(2)+8* 
exp(1)+320*x)*log(((5*log(2)+5*x)*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8) 
)+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)*log(x)+ 
40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*log(2)+ 
exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x),x)
 

Output:

e**3*( - int((5*log(x)*log(2) + 5*log(x)*x + 40*log(2) + e + 40*x)**x/(5*( 
log(x) + 8)**x*log(x)**2*log(2) + 5*(log(x) + 8)**x*log(x)**2*x + 80*(log( 
x) + 8)**x*log(x)*log(2) + (log(x) + 8)**x*log(x)*e + 80*(log(x) + 8)**x*l 
og(x)*x + 320*(log(x) + 8)**x*log(2) + 8*(log(x) + 8)**x*e + 320*(log(x) + 
 8)**x*x),x)*e + 5*int(((5*log(x)*log(2) + 5*log(x)*x + 40*log(2) + e + 40 
*x)**x*log(x)**2*x)/(5*(log(x) + 8)**x*log(x)**2*log(2) + 5*(log(x) + 8)** 
x*log(x)**2*x + 80*(log(x) + 8)**x*log(x)*log(2) + (log(x) + 8)**x*log(x)* 
e + 80*(log(x) + 8)**x*log(x)*x + 320*(log(x) + 8)**x*log(2) + 8*(log(x) + 
 8)**x*e + 320*(log(x) + 8)**x*x),x) + 5*int(((5*log(x)*log(2) + 5*log(x)* 
x + 40*log(2) + e + 40*x)**x*log((5*log(x)*log(2) + 5*log(x)*x + 40*log(2) 
 + e + 40*x)/(log(x) + 8))*log(x)**2*x)/(5*(log(x) + 8)**x*log(x)**2*log(2 
) + 5*(log(x) + 8)**x*log(x)**2*x + 80*(log(x) + 8)**x*log(x)*log(2) + (lo 
g(x) + 8)**x*log(x)*e + 80*(log(x) + 8)**x*log(x)*x + 320*(log(x) + 8)**x* 
log(2) + 8*(log(x) + 8)**x*e + 320*(log(x) + 8)**x*x),x) + 5*int(((5*log(x 
)*log(2) + 5*log(x)*x + 40*log(2) + e + 40*x)**x*log((5*log(x)*log(2) + 5* 
log(x)*x + 40*log(2) + e + 40*x)/(log(x) + 8))*log(x)**2)/(5*(log(x) + 8)* 
*x*log(x)**2*log(2) + 5*(log(x) + 8)**x*log(x)**2*x + 80*(log(x) + 8)**x*l 
og(x)*log(2) + (log(x) + 8)**x*log(x)*e + 80*(log(x) + 8)**x*log(x)*x + 32 
0*(log(x) + 8)**x*log(2) + 8*(log(x) + 8)**x*e + 320*(log(x) + 8)**x*x),x) 
*log(2) + 80*int(((5*log(x)*log(2) + 5*log(x)*x + 40*log(2) + e + 40*x)...