\(\int \frac {e^{\frac {-x+(5+x^3) \log (e^{e^{6 x}} x)}{\log (e^{e^{6 x}} x)}} (1+6 e^{6 x} x-\log (e^{e^{6 x}} x)+3 x^2 \log ^2(e^{e^{6 x}} x))}{\log ^2(e^{e^{6 x}} x)} \, dx\) [95]

Optimal result

Integrand size = 87, antiderivative size = 22 \[ \int \frac {e^{\frac {-x+\left (5+x^3\right ) \log \left (e^{e^{6 x}} x\right )}{\log \left (e^{e^{6 x}} x\right )}} \left (1+6 e^{6 x} x-\log \left (e^{e^{6 x}} x\right )+3 x^2 \log ^2\left (e^{e^{6 x}} x\right )\right )}{\log ^2\left (e^{e^{6 x}} x\right )} \, dx=e^{5+x^3-\frac {x}{\log \left (e^{e^{6 x}} x\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-x+\left (5+x^3\right ) \log \left (e^{e^{6 x}} x\right )}{\log \left (e^{e^{6 x}} x\right )}} \left (1+6 e^{6 x} x-\log \left (e^{e^{6 x}} x\right )+3 x^2 \log ^2\left (e^{e^{6 x}} x\right )\right )}{\log ^2\left (e^{e^{6 x}} x\right )} \, dx=e^{5+x^3-\frac {x}{\log \left (e^{e^{6 x}} x\right )}} \] Input:

Integrate[(E^((-x + (5 + x^3)*Log[E^E^(6*x)*x])/Log[E^E^(6*x)*x])*(1 + 6*E 
^(6*x)*x - Log[E^E^(6*x)*x] + 3*x^2*Log[E^E^(6*x)*x]^2))/Log[E^E^(6*x)*x]^ 
2,x]
 

Output:

E^(5 + x^3 - x/Log[E^E^(6*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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(E^((-x + (5 + x^3)*Log[E^E^(6*x)*x])/Log[E^E^(6*x)*x])*(1 + 6*E^(6*x) 
*x - Log[E^E^(6*x)*x] + 3*x^2*Log[E^E^(6*x)*x]^2))/Log[E^E^(6*x)*x]^2,x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.47 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59

method	result	size
derivativedivides	\({\mathrm e}^{\frac {\left (x^{3}+5\right ) \ln \left (x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )-x}{\ln \left (x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )}}\)	\(35\)
default	\({\mathrm e}^{\frac {\left (x^{3}+5\right ) \ln \left (x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )-x}{\ln \left (x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )}}\)	\(35\)
parallelrisch	\({\mathrm e}^{\frac {\left (x^{3}+5\right ) \ln \left (x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )-x}{\ln \left (x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )}}\)	\(35\)
risch	\({\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{2} x^{3}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \right ) x^{3}-i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{3} x^{3}+i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{2} \operatorname {csgn}\left (i x \right ) x^{3}+5 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{2}-5 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \right )-5 i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{3}+5 i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{2} \operatorname {csgn}\left (i x \right )+2 x^{3} \ln \left (x \right )+2 \ln \left ({\mathrm e}^{{\mathrm e}^{6 x}}\right ) x^{3}+10 \ln \left (x \right )+10 \ln \left ({\mathrm e}^{{\mathrm e}^{6 x}}\right )-2 x}{i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right ) \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{3}+i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{6 x}}\right )^{2} \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left ({\mathrm e}^{{\mathrm e}^{6 x}}\right )}}\)	\(334\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {e^{\frac {-x+\left (5+x^3\right ) \log \left (e^{e^{6 x}} x\right )}{\log \left (e^{e^{6 x}} x\right )}} \left (1+6 e^{6 x} x-\log \left (e^{e^{6 x}} x\right )+3 x^2 \log ^2\left (e^{e^{6 x}} x\right )\right )}{\log ^2\left (e^{e^{6 x}} x\right )} \, dx=e^{\left (\frac {{\left (x^{3} + 5\right )} \log \left (x e^{\left (e^{\left (6 \, x\right )}\right )}\right ) - x}{\log \left (x e^{\left (e^{\left (6 \, x\right )}\right )}\right )}\right )} \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {-x+\left (5+x^3\right ) \log \left (e^{e^{6 x}} x\right )}{\log \left (e^{e^{6 x}} x\right )}} \left (1+6 e^{6 x} x-\log \left (e^{e^{6 x}} x\right )+3 x^2 \log ^2\left (e^{e^{6 x}} x\right )\right )}{\log ^2\left (e^{e^{6 x}} x\right )} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {-x+\left (5+x^3\right ) \log \left (e^{e^{6 x}} x\right )}{\log \left (e^{e^{6 x}} x\right )}} \left (1+6 e^{6 x} x-\log \left (e^{e^{6 x}} x\right )+3 x^2 \log ^2\left (e^{e^{6 x}} x\right )\right )}{\log ^2\left (e^{e^{6 x}} x\right )} \, dx=e^{\left (x^{3} - \frac {x}{e^{\left (6 \, x\right )} + \log \left (x\right )} + 5\right )} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {-x+\left (5+x^3\right ) \log \left (e^{e^{6 x}} x\right )}{\log \left (e^{e^{6 x}} x\right )}} \left (1+6 e^{6 x} x-\log \left (e^{e^{6 x}} x\right )+3 x^2 \log ^2\left (e^{e^{6 x}} x\right )\right )}{\log ^2\left (e^{e^{6 x}} x\right )} \, dx=e^{\left (x^{3} - \frac {x}{\log \left (x e^{\left (e^{\left (6 \, x\right )}\right )}\right )} + 5\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.95 \[ \int \frac {e^{\frac {-x+\left (5+x^3\right ) \log \left (e^{e^{6 x}} x\right )}{\log \left (e^{e^{6 x}} x\right )}} \left (1+6 e^{6 x} x-\log \left (e^{e^{6 x}} x\right )+3 x^2 \log ^2\left (e^{e^{6 x}} x\right )\right )}{\log ^2\left (e^{e^{6 x}} x\right )} \, dx=x^{\frac {x^3+5}{{\mathrm {e}}^{6\,x}+\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {x}{{\mathrm {e}}^{6\,x}+\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^{6\,x}}{{\mathrm {e}}^{6\,x}+\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^{6\,x}}{{\mathrm {e}}^{6\,x}+\ln \left (x\right )}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\frac {-x+\left (5+x^3\right ) \log \left (e^{e^{6 x}} x\right )}{\log \left (e^{e^{6 x}} x\right )}} \left (1+6 e^{6 x} x-\log \left (e^{e^{6 x}} x\right )+3 x^2 \log ^2\left (e^{e^{6 x}} x\right )\right )}{\log ^2\left (e^{e^{6 x}} x\right )} \, dx=\frac {e^{x^{3}} e^{5}}{e^{\frac {x}{\mathrm {log}\left (e^{e^{6 x}} x \right )}}} \] Input:

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

Output:

(e**(x**3)*e**5)/e**(x/log(e**(e**(6*x))*x))