\(\int \frac {5 e^{8+18 x} x^2+5 x^4+e^{.\frac {2}{5}/x} (-9 x+15 x^2)+e^{4+9 x} (10 x^3+e^{.\frac {2}{5}/x} (6+135 x^2))+(-6 e^{.\frac {2}{5}/x}-10 e^{4+9 x} x^2-10 x^3) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+(-120 e^{4+9 x} x^2-120 x^3) \log (x)+60 x^2 \log ^2(x)} \, dx\) [2966]

Optimal result

Integrand size = 172, antiderivative size = 37 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {1}{4} \left (\frac {x}{3}+\frac {e^{\left .\frac {2}{5}\right /x}}{-e^{4+9 x}-x+\log (x)}\right ) \] Output:

1/4*exp(2/5/x)/(ln(x)-x-exp(9*x+4))+1/12*x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {-3 e^{\left .\frac {2}{5}\right /x}+e^{4+9 x} x+x^2-x \log (x)}{12 \left (e^{4+9 x}+x-\log (x)\right )} \] Input:

Integrate[(5*E^(8 + 18*x)*x^2 + 5*x^4 + E^(2/(5*x))*(-9*x + 15*x^2) + E^(4 
 + 9*x)*(10*x^3 + E^(2/(5*x))*(6 + 135*x^2)) + (-6*E^(2/(5*x)) - 10*E^(4 + 
 9*x)*x^2 - 10*x^3)*Log[x] + 5*x^2*Log[x]^2)/(60*E^(8 + 18*x)*x^2 + 120*E^ 
(4 + 9*x)*x^3 + 60*x^4 + (-120*E^(4 + 9*x)*x^2 - 120*x^3)*Log[x] + 60*x^2* 
Log[x]^2),x]
 

Output:

(-3*E^(2/(5*x)) + E^(4 + 9*x)*x + x^2 - x*Log[x])/(12*(E^(4 + 9*x) + x - L 
og[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.

\(\displaystyle \int \frac {5 x^4+5 e^{18 x+8} x^2+e^{\left .\frac {2}{5}\right /x} \left (15 x^2-9 x\right )+5 x^2 \log ^2(x)+e^{9 x+4} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (135 x^2+6\right )\right )+\left (-10 x^3-10 e^{9 x+4} x^2-6 e^{\left .\frac {2}{5}\right /x}\right ) \log (x)}{60 x^4+120 e^{9 x+4} x^3+60 e^{18 x+8} x^2+60 x^2 \log ^2(x)+\left (-120 x^3-120 e^{9 x+4} x^2\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {5 x^4+5 e^{18 x+8} x^2+e^{\left .\frac {2}{5}\right /x} \left (15 x^2-9 x\right )+5 x^2 \log ^2(x)+e^{9 x+4} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (135 x^2+6\right )\right )+\left (-10 x^3-10 e^{9 x+4} x^2-6 e^{\left .\frac {2}{5}\right /x}\right ) \log (x)}{60 x^2 \left (x+e^{9 x+4}-\log (x)\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{60} \int \frac {5 x^4+5 e^{18 x+8} x^2+5 \log ^2(x) x^2-3 e^{\left .\frac {2}{5}\right /x} \left (3 x-5 x^2\right )+e^{9 x+4} \left (10 x^3+3 e^{\left .\frac {2}{5}\right /x} \left (45 x^2+2\right )\right )-2 \left (5 x^3+5 e^{9 x+4} x^2+3 e^{\left .\frac {2}{5}\right /x}\right ) \log (x)}{x^2 \left (x+e^{9 x+4}-\log (x)\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{60} \int \left (\frac {3 e^{\left .\frac {2}{5}\right /x} \left (45 x^2+2\right )}{x^2 \left (x+e^{9 x+4}-\log (x)\right )}-\frac {15 e^{\left .\frac {2}{5}\right /x} \left (9 x^2-9 \log (x) x-x+1\right )}{x \left (x+e^{9 x+4}-\log (x)\right )^2}+5\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{60} \left (6 \int \frac {e^{\left .\frac {2}{5}\right /x}}{x^2 \left (x+e^{9 x+4}-\log (x)\right )}dx+15 \int \frac {e^{\left .\frac {2}{5}\right /x}}{\left (x+e^{9 x+4}-\log (x)\right )^2}dx-15 \int \frac {e^{\left .\frac {2}{5}\right /x}}{x \left (x+e^{9 x+4}-\log (x)\right )^2}dx-135 \int \frac {e^{\left .\frac {2}{5}\right /x} x}{\left (x+e^{9 x+4}-\log (x)\right )^2}dx+135 \int \frac {e^{\left .\frac {2}{5}\right /x}}{x+e^{9 x+4}-\log (x)}dx+135 \int \frac {e^{\left .\frac {2}{5}\right /x} \log (x)}{\left (x+e^{9 x+4}-\log (x)\right )^2}dx+5 x\right )\)

Input:

Int[(5*E^(8 + 18*x)*x^2 + 5*x^4 + E^(2/(5*x))*(-9*x + 15*x^2) + E^(4 + 9*x 
)*(10*x^3 + E^(2/(5*x))*(6 + 135*x^2)) + (-6*E^(2/(5*x)) - 10*E^(4 + 9*x)* 
x^2 - 10*x^3)*Log[x] + 5*x^2*Log[x]^2)/(60*E^(8 + 18*x)*x^2 + 120*E^(4 + 9 
*x)*x^3 + 60*x^4 + (-120*E^(4 + 9*x)*x^2 - 120*x^3)*Log[x] + 60*x^2*Log[x] 
^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 23.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {x}{12}-\frac {{\mathrm e}^{\frac {2}{5 x}}}{4 \left (x +{\mathrm e}^{9 x +4}-\ln \left (x \right )\right )}\)	\(27\)
parallelrisch	\(\frac {5 x^{2}-5 x \ln \left (x \right )+5 x \,{\mathrm e}^{9 x +4}-15 \,{\mathrm e}^{\frac {2}{5 x}}}{60 x +60 \,{\mathrm e}^{9 x +4}-60 \ln \left (x \right )}\)	\(45\)

Input:

int((5*x^2*ln(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*ln(x)+5*x^2*ex 
p(9*x+4)^2+((135*x^2+6)*exp(2/5/x)+10*x^3)*exp(9*x+4)+(15*x^2-9*x)*exp(2/5 
/x)+5*x^4)/(60*x^2*ln(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*ln(x)+60*x^2*exp( 
9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x,method=_RETURNVERBOSE)
 

Output:

1/12*x-1/4*exp(2/5/x)/(x+exp(9*x+4)-ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {x^{2} + x e^{\left (9 \, x + 4\right )} - x \log \left (x\right ) - 3 \, e^{\left (\frac {2}{5 \, x}\right )}}{12 \, {\left (x + e^{\left (9 \, x + 4\right )} - \log \left (x\right )\right )}} \] Input:

integrate((5*x^2*log(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*log(x)+ 
5*x^2*exp(9*x+4)^2+((135*x^2+6)*exp(2/5/x)+10*x^3)*exp(9*x+4)+(15*x^2-9*x) 
*exp(2/5/x)+5*x^4)/(60*x^2*log(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*log(x)+6 
0*x^2*exp(9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x, algorithm="fricas")
 

Output:

1/12*(x^2 + x*e^(9*x + 4) - x*log(x) - 3*e^(2/5/x))/(x + e^(9*x + 4) - log 
(x))
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {x}{12} - \frac {e^{\frac {2}{5 x}}}{4 x + 4 e^{9 x + 4} - 4 \log {\left (x \right )}} \] Input:

integrate((5*x**2*ln(x)**2+(-10*x**2*exp(9*x+4)-6*exp(2/5/x)-10*x**3)*ln(x 
)+5*x**2*exp(9*x+4)**2+((135*x**2+6)*exp(2/5/x)+10*x**3)*exp(9*x+4)+(15*x* 
*2-9*x)*exp(2/5/x)+5*x**4)/(60*x**2*ln(x)**2+(-120*x**2*exp(9*x+4)-120*x** 
3)*ln(x)+60*x**2*exp(9*x+4)**2+120*x**3*exp(9*x+4)+60*x**4),x)
 

Output:

x/12 - exp(2/(5*x))/(4*x + 4*exp(9*x + 4) - 4*log(x))
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {x^{2} + x e^{\left (9 \, x + 4\right )} - x \log \left (x\right ) - 3 \, e^{\left (\frac {2}{5 \, x}\right )}}{12 \, {\left (x + e^{\left (9 \, x + 4\right )} - \log \left (x\right )\right )}} \] Input:

integrate((5*x^2*log(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*log(x)+ 
5*x^2*exp(9*x+4)^2+((135*x^2+6)*exp(2/5/x)+10*x^3)*exp(9*x+4)+(15*x^2-9*x) 
*exp(2/5/x)+5*x^4)/(60*x^2*log(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*log(x)+6 
0*x^2*exp(9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x, algorithm="maxima")
 

Output:

1/12*(x^2 + x*e^(9*x + 4) - x*log(x) - 3*e^(2/5/x))/(x + e^(9*x + 4) - log 
(x))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((5*x^2*log(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*log(x)+ 
5*x^2*exp(9*x+4)^2+((135*x^2+6)*exp(2/5/x)+10*x^3)*exp(9*x+4)+(15*x^2-9*x) 
*exp(2/5/x)+5*x^4)/(60*x^2*log(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*log(x)+6 
0*x^2*exp(9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-109350,[0,3,6,0]%%%}+%%%{-4860,[0,3,4,0]%%%}+%%%{273375,[ 
0,2,8,0]%
 

Mupad [B] (verification not implemented)

Time = 4.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {x}{12}-\frac {{\mathrm {e}}^{\frac {2}{5\,x}}}{4\,\left (x+{\mathrm {e}}^{9\,x+4}-\ln \left (x\right )\right )} \] Input:

int((exp(9*x + 4)*(exp(2/(5*x))*(135*x^2 + 6) + 10*x^3) - exp(2/(5*x))*(9* 
x - 15*x^2) + 5*x^2*log(x)^2 - log(x)*(6*exp(2/(5*x)) + 10*x^2*exp(9*x + 4 
) + 10*x^3) + 5*x^2*exp(18*x + 8) + 5*x^4)/(60*x^2*log(x)^2 - log(x)*(120* 
x^2*exp(9*x + 4) + 120*x^3) + 120*x^3*exp(9*x + 4) + 60*x^2*exp(18*x + 8) 
+ 60*x^4),x)
 

Output:

x/12 - exp(2/(5*x))/(4*(x + exp(9*x + 4) - log(x)))
 

Reduce [B] (verification not implemented)

Time = 1.78 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.32 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {-3 e^{\frac {2}{5 x}}+e^{9 x} e^{4} x -\mathrm {log}\left (x \right ) x +x^{2}}{12 e^{9 x} e^{4}-12 \,\mathrm {log}\left (x \right )+12 x} \] Input:

int((5*x^2*log(x)^2+(-10*x^2*exp(9*x+4)-6*exp(2/5/x)-10*x^3)*log(x)+5*x^2* 
exp(9*x+4)^2+((135*x^2+6)*exp(2/5/x)+10*x^3)*exp(9*x+4)+(15*x^2-9*x)*exp(2 
/5/x)+5*x^4)/(60*x^2*log(x)^2+(-120*x^2*exp(9*x+4)-120*x^3)*log(x)+60*x^2* 
exp(9*x+4)^2+120*x^3*exp(9*x+4)+60*x^4),x)
 

Output:

( - 3*e**(2/(5*x)) + e**(9*x)*e**4*x - log(x)*x + x**2)/(12*(e**(9*x)*e**4 
 - log(x) + x))