\(\int \frac {e^{\frac {3+e^{10} (-x^2-x^3)}{-e^{10} x^2+e^{10} x^2 \log (\frac {x}{5 \log (x)})}} (3+e^{10} (-x^2-x^3)+(3+e^{10} (x^2+2 x^3)) \log (x)+(-6-e^{10} x^3) \log (x) \log (\frac {x}{5 \log (x)}))}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log (\frac {x}{5 \log (x)})+e^{10} x^3 \log (x) \log ^2(\frac {x}{5 \log (x)})} \, dx\) [11]

Optimal result

Integrand size = 160, antiderivative size = 30 \[ \int \frac {e^{\frac {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}} \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx=e^{\frac {-1+\frac {3}{e^{10} x^2}-x}{-1+\log \left (\frac {x}{5 \log (x)}\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}} \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx=e^{\frac {3-e^{10} x^2 (1+x)}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}} \] Input:

Integrate[(E^((3 + E^10*(-x^2 - x^3))/(-(E^10*x^2) + E^10*x^2*Log[x/(5*Log 
[x])]))*(3 + E^10*(-x^2 - x^3) + (3 + E^10*(x^2 + 2*x^3))*Log[x] + (-6 - E 
^10*x^3)*Log[x]*Log[x/(5*Log[x])]))/(E^10*x^3*Log[x] - 2*E^10*x^3*Log[x]*L 
og[x/(5*Log[x])] + E^10*x^3*Log[x]*Log[x/(5*Log[x])]^2),x]
 

Output:

E^((3 - E^10*x^2*(1 + x))/(E^10*x^2*(-1 + Log[x/(5*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.

\(\displaystyle \int \frac {\left (\left (-e^{10} x^3-6\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} \left (-x^3-x^2\right )+\left (e^{10} \left (2 x^3+x^2\right )+3\right ) \log (x)+3\right ) \exp \left (\frac {e^{10} \left (-x^3-x^2\right )+3}{e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )-e^{10} x^2}\right )}{e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (\left (-e^{10} x^3-6\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} \left (-x^3-x^2\right )+\left (e^{10} \left (2 x^3+x^2\right )+3\right ) \log (x)+3\right ) \exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}-10\right )}{x^3 \log (x) \left (1-\log \left (\frac {x}{5 \log (x)}\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\left (-e^{10} x^3-6\right ) \exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}-10\right )}{x^3 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}+\frac {\left (e^{10} x^3+e^{10} x^2-3\right ) (\log (x)-1) \exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}-10\right )}{x^3 \log (x) \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {\exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}\right )}{\left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )^2}dx-3 \int \frac {\exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}-10\right )}{x^3 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )^2}dx+\int \frac {\exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}\right )}{x \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )^2}dx-\int \frac {\exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}\right )}{\log (x) \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )^2}dx+3 \int \frac {\exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}-10\right )}{x^3 \log (x) \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )^2}dx-\int \frac {\exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}\right )}{x \log (x) \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )^2}dx-\int \frac {\exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}\right )}{\log \left (\frac {x}{5 \log (x)}\right )-1}dx-6 \int \frac {\exp \left (\frac {-e^{10} x^3-e^{10} x^2+3}{e^{10} x^2 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}-10\right )}{x^3 \left (\log \left (\frac {x}{5 \log (x)}\right )-1\right )}dx\)

Input:

Int[(E^((3 + E^10*(-x^2 - x^3))/(-(E^10*x^2) + E^10*x^2*Log[x/(5*Log[x])]) 
)*(3 + E^10*(-x^2 - x^3) + (3 + E^10*(x^2 + 2*x^3))*Log[x] + (-6 - E^10*x^ 
3)*Log[x]*Log[x/(5*Log[x])]))/(E^10*x^3*Log[x] - 2*E^10*x^3*Log[x]*Log[x/( 
5*Log[x])] + E^10*x^3*Log[x]*Log[x/(5*Log[x])]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 147.50 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{10}+3\right ) {\mathrm e}^{-10}}{x^{2} \left (\ln \left (\frac {x}{5 \ln \left (x \right )}\right )-1\right )}}\)	\(40\)
risch	\({\mathrm e}^{\frac {2 \left (x^{3} {\mathrm e}^{10}+{\mathrm e}^{10} x^{2}-3\right ) {\mathrm e}^{-10}}{x^{2} \left (i \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{3}-i \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )+i \pi \,\operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )+2 \ln \left (5\right )+2 \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right )+2\right )}}\)	\(124\)

Input:

int(((-x^3*exp(5)^2-6)*ln(x)*ln(1/5*x/ln(x))+((2*x^3+x^2)*exp(5)^2+3)*ln(x 
)+(-x^3-x^2)*exp(5)^2+3)*exp(((-x^3-x^2)*exp(5)^2+3)/(x^2*exp(5)^2*ln(1/5* 
x/ln(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*ln(x)*ln(1/5*x/ln(x))^2-2*x^3*exp(5) 
^2*ln(x)*ln(1/5*x/ln(x))+x^3*exp(5)^2*ln(x)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^{\frac {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}} \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx=e^{\left (-\frac {{\left (x^{3} + x^{2}\right )} e^{10} - 3}{x^{2} e^{10} \log \left (\frac {x}{5 \, \log \left (x\right )}\right ) - x^{2} e^{10}}\right )} \] Input:

integrate(((-x^3*exp(5)^2-6)*log(x)*log(1/5*x/log(x))+((2*x^3+x^2)*exp(5)^ 
2+3)*log(x)+(-x^3-x^2)*exp(5)^2+3)*exp(((-x^3-x^2)*exp(5)^2+3)/(x^2*exp(5) 
^2*log(1/5*x/log(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*log(x)*log(1/5*x/log(x)) 
^2-2*x^3*exp(5)^2*log(x)*log(1/5*x/log(x))+x^3*exp(5)^2*log(x)),x, algorit 
hm="fricas")
 

Output:

e^(-((x^3 + x^2)*e^10 - 3)/(x^2*e^10*log(1/5*x/log(x)) - x^2*e^10))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}} \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((-x**3*exp(5)**2-6)*ln(x)*ln(1/5*x/ln(x))+((2*x**3+x**2)*exp(5) 
**2+3)*ln(x)+(-x**3-x**2)*exp(5)**2+3)*exp(((-x**3-x**2)*exp(5)**2+3)/(x** 
2*exp(5)**2*ln(1/5*x/ln(x))-x**2*exp(5)**2))/(x**3*exp(5)**2*ln(x)*ln(1/5* 
x/ln(x))**2-2*x**3*exp(5)**2*ln(x)*ln(1/5*x/ln(x))+x**3*exp(5)**2*ln(x)),x 
)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}} \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(((-x^3*exp(5)^2-6)*log(x)*log(1/5*x/log(x))+((2*x^3+x^2)*exp(5)^ 
2+3)*log(x)+(-x^3-x^2)*exp(5)^2+3)*exp(((-x^3-x^2)*exp(5)^2+3)/(x^2*exp(5) 
^2*log(1/5*x/log(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*log(x)*log(1/5*x/log(x)) 
^2-2*x^3*exp(5)^2*log(x)*log(1/5*x/log(x))+x^3*exp(5)^2*log(x)),x, algorit 
hm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [F]

\[ \int \frac {e^{\frac {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}} \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx=\int { -\frac {{\left ({\left (x^{3} e^{10} + 6\right )} \log \left (x\right ) \log \left (\frac {x}{5 \, \log \left (x\right )}\right ) + {\left (x^{3} + x^{2}\right )} e^{10} - {\left ({\left (2 \, x^{3} + x^{2}\right )} e^{10} + 3\right )} \log \left (x\right ) - 3\right )} e^{\left (-\frac {{\left (x^{3} + x^{2}\right )} e^{10} - 3}{x^{2} e^{10} \log \left (\frac {x}{5 \, \log \left (x\right )}\right ) - x^{2} e^{10}}\right )}}{x^{3} e^{10} \log \left (x\right ) \log \left (\frac {x}{5 \, \log \left (x\right )}\right )^{2} - 2 \, x^{3} e^{10} \log \left (x\right ) \log \left (\frac {x}{5 \, \log \left (x\right )}\right ) + x^{3} e^{10} \log \left (x\right )} \,d x } \] Input:

integrate(((-x^3*exp(5)^2-6)*log(x)*log(1/5*x/log(x))+((2*x^3+x^2)*exp(5)^ 
2+3)*log(x)+(-x^3-x^2)*exp(5)^2+3)*exp(((-x^3-x^2)*exp(5)^2+3)/(x^2*exp(5) 
^2*log(1/5*x/log(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*log(x)*log(1/5*x/log(x)) 
^2-2*x^3*exp(5)^2*log(x)*log(1/5*x/log(x))+x^3*exp(5)^2*log(x)),x, algorit 
hm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 4.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.30 \[ \int \frac {e^{\frac {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}} \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx={\mathrm {e}}^{-\frac {3}{x^2\,{\mathrm {e}}^{10}+x^2\,{\mathrm {e}}^{10}\,\ln \left (5\right )-x^2\,{\mathrm {e}}^{10}\,\ln \left (\frac {x}{\ln \left (x\right )}\right )}}\,{\mathrm {e}}^{\frac {x}{\ln \left (5\right )-\ln \left (\frac {x}{\ln \left (x\right )}\right )+1}}\,{\mathrm {e}}^{\frac {1}{\ln \left (5\right )-\ln \left (\frac {x}{\ln \left (x\right )}\right )+1}} \] Input:

int(-(exp((exp(10)*(x^2 + x^3) - 3)/(x^2*exp(10) - x^2*exp(10)*log(x/(5*lo 
g(x)))))*(exp(10)*(x^2 + x^3) - log(x)*(exp(10)*(x^2 + 2*x^3) + 3) + log(x 
/(5*log(x)))*log(x)*(x^3*exp(10) + 6) - 3))/(x^3*exp(10)*log(x) - 2*x^3*ex 
p(10)*log(x/(5*log(x)))*log(x) + x^3*exp(10)*log(x/(5*log(x)))^2*log(x)),x 
)
 

Output:

exp(-3/(x^2*exp(10) + x^2*exp(10)*log(5) - x^2*exp(10)*log(x/log(x))))*exp 
(x/(log(5) - log(x/log(x)) + 1))*exp(1/(log(5) - log(x/log(x)) + 1))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}} \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx=\frac {e^{\frac {3}{\mathrm {log}\left (\frac {x}{5 \,\mathrm {log}\left (x \right )}\right ) e^{10} x^{2}-e^{10} x^{2}}}}{e^{\frac {x +1}{\mathrm {log}\left (\frac {x}{5 \,\mathrm {log}\left (x \right )}\right )-1}}} \] Input:

int(((-x^3*exp(5)^2-6)*log(x)*log(1/5*x/log(x))+((2*x^3+x^2)*exp(5)^2+3)*l 
og(x)+(-x^3-x^2)*exp(5)^2+3)*exp(((-x^3-x^2)*exp(5)^2+3)/(x^2*exp(5)^2*log 
(1/5*x/log(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*log(x)*log(1/5*x/log(x))^2-2*x 
^3*exp(5)^2*log(x)*log(1/5*x/log(x))+x^3*exp(5)^2*log(x)),x)
 

Output:

e**(3/(log(x/(5*log(x)))*e**10*x**2 - e**10*x**2))/e**((x + 1)/(log(x/(5*l 
og(x))) - 1))