\(\int \frac {(18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} (90 x^2+36 x^3)) \log (\frac {1}{3} (e^x+x^2))+(-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} (-9+9 x^2)+e^{\frac {1}{x}+x} (-45-9 x^2+9 x^4)) \log ^2(\frac {1}{3} (e^x+x^2))}{e^x x^2+x^4} \, dx\) [2858]

Optimal result

Integrand size = 135, antiderivative size = 26 \[ \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx=9 e^{\frac {1}{x}} \left (5+e^x\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right ) \] Output:

exp(1/x)*(9*exp(x)+45)*ln(1/3*x^2+1/3*exp(x))^2
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx=9 e^{\frac {1}{x}} \left (5+e^x\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right ) \] Input:

Integrate[((18*E^(x^(-1) + 2*x)*x^2 + 180*E^x^(-1)*x^3 + E^(x^(-1) + x)*(9 
0*x^2 + 36*x^3))*Log[(E^x + x^2)/3] + (-45*E^x^(-1)*x^2 + E^(x^(-1) + 2*x) 
*(-9 + 9*x^2) + E^(x^(-1) + x)*(-45 - 9*x^2 + 9*x^4))*Log[(E^x + x^2)/3]^2 
)/(E^x*x^2 + x^4),x]
 

Output:

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

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.

\(\displaystyle \int \frac {\left (-45 e^{\frac {1}{x}} x^2+e^{2 x+\frac {1}{x}} \left (9 x^2-9\right )+e^{x+\frac {1}{x}} \left (9 x^4-9 x^2-45\right )\right ) \log ^2\left (\frac {1}{3} \left (x^2+e^x\right )\right )+\left (180 e^{\frac {1}{x}} x^3+18 e^{2 x+\frac {1}{x}} x^2+e^{x+\frac {1}{x}} \left (36 x^3+90 x^2\right )\right ) \log \left (\frac {1}{3} \left (x^2+e^x\right )\right )}{x^4+e^x x^2} \, dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 7299

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

Input:

Int[((18*E^(x^(-1) + 2*x)*x^2 + 180*E^x^(-1)*x^3 + E^(x^(-1) + x)*(90*x^2 
+ 36*x^3))*Log[(E^x + x^2)/3] + (-45*E^x^(-1)*x^2 + E^(x^(-1) + 2*x)*(-9 + 
 9*x^2) + E^(x^(-1) + x)*(-45 - 9*x^2 + 9*x^4))*Log[(E^x + x^2)/3]^2)/(E^x 
*x^2 + x^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 6.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\(9 \left ({\mathrm e}^{x}+5\right ) {\mathrm e}^{\frac {1}{x}} \ln \left (\frac {x^{2}}{3}+\frac {{\mathrm e}^{x}}{3}\right )^{2}\)	\(24\)
parallelrisch	\(9 \ln \left (\frac {x^{2}}{3}+\frac {{\mathrm e}^{x}}{3}\right )^{2} {\mathrm e}^{x} {\mathrm e}^{\frac {1}{x}}+45 \ln \left (\frac {x^{2}}{3}+\frac {{\mathrm e}^{x}}{3}\right )^{2} {\mathrm e}^{\frac {1}{x}}\)	\(42\)

Input:

int((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-45*x^2* 
exp(1/x))*ln(1/3*x^2+1/3*exp(x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^3+90*x^ 
2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*ln(1/3*x^2+1/3*exp(x)))/(exp(x)*x^2+x 
^4),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (23) = 46\).

Time = 0.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.62 \[ \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx=9 \, {\left (e^{\left (\frac {2 \, x^{2} + 1}{x} + \frac {x^{2} + 1}{x}\right )} + 5 \, e^{\left (\frac {2 \, {\left (x^{2} + 1\right )}}{x}\right )}\right )} e^{\left (-\frac {2 \, x^{2} + 1}{x}\right )} \log \left (\frac {1}{3} \, {\left (x^{2} e^{\left (\frac {x^{2} + 1}{x}\right )} + e^{\left (\frac {2 \, x^{2} + 1}{x}\right )}\right )} e^{\left (-\frac {x^{2} + 1}{x}\right )}\right )^{2} \] Input:

integrate((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-4 
5*x^2*exp(1/x))*log(1/3*x^2+1/3*exp(x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^ 
3+90*x^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*log(1/3*x^2+1/3*exp(x)))/(exp( 
x)*x^2+x^4),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx=\text {Timed out} \] Input:

integrate((((9*x**2-9)*exp(1/x)*exp(x)**2+(9*x**4-9*x**2-45)*exp(1/x)*exp( 
x)-45*x**2*exp(1/x))*ln(1/3*x**2+1/3*exp(x))**2+(18*x**2*exp(1/x)*exp(x)** 
2+(36*x**3+90*x**2)*exp(1/x)*exp(x)+180*x**3*exp(1/x))*ln(1/3*x**2+1/3*exp 
(x)))/(exp(x)*x**2+x**4),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (23) = 46\).

Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42 \[ \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx=9 \, {\left (e^{x} + 5\right )} e^{\frac {1}{x}} \log \left (x^{2} + e^{x}\right )^{2} - 18 \, {\left (e^{x} \log \left (3\right ) + 5 \, \log \left (3\right )\right )} e^{\frac {1}{x}} \log \left (x^{2} + e^{x}\right ) + 9 \, {\left (e^{x} \log \left (3\right )^{2} + 5 \, \log \left (3\right )^{2}\right )} e^{\frac {1}{x}} \] Input:

integrate((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-4 
5*x^2*exp(1/x))*log(1/3*x^2+1/3*exp(x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^ 
3+90*x^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*log(1/3*x^2+1/3*exp(x)))/(exp( 
x)*x^2+x^4),x, algorithm="maxima")
 

Output:

9*(e^x + 5)*e^(1/x)*log(x^2 + e^x)^2 - 18*(e^x*log(3) + 5*log(3))*e^(1/x)* 
log(x^2 + e^x) + 9*(e^x*log(3)^2 + 5*log(3)^2)*e^(1/x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-4 
5*x^2*exp(1/x))*log(1/3*x^2+1/3*exp(x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^ 
3+90*x^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*log(1/3*x^2+1/3*exp(x)))/(exp( 
x)*x^2+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%%%{-1,[0,14]%%%}+%%%{4,[0,13]%%%}+%%%{-3,[0,12]%%%}+%%%{-4,[0 
,11]%%%}+
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx=\int \frac {\ln \left (\frac {{\mathrm {e}}^x}{3}+\frac {x^2}{3}\right )\,\left ({\mathrm {e}}^{x+\frac {1}{x}}\,\left (36\,x^3+90\,x^2\right )+180\,x^3\,{\mathrm {e}}^{1/x}+18\,x^2\,{\mathrm {e}}^{2\,x+\frac {1}{x}}\right )-{\ln \left (\frac {{\mathrm {e}}^x}{3}+\frac {x^2}{3}\right )}^2\,\left ({\mathrm {e}}^{x+\frac {1}{x}}\,\left (-9\,x^4+9\,x^2+45\right )-{\mathrm {e}}^{2\,x+\frac {1}{x}}\,\left (9\,x^2-9\right )+45\,x^2\,{\mathrm {e}}^{1/x}\right )}{x^2\,{\mathrm {e}}^x+x^4} \,d x \] Input:

int((log(exp(x)/3 + x^2/3)*(180*x^3*exp(1/x) + exp(1/x)*exp(x)*(90*x^2 + 3 
6*x^3) + 18*x^2*exp(2*x)*exp(1/x)) - log(exp(x)/3 + x^2/3)^2*(45*x^2*exp(1 
/x) + exp(1/x)*exp(x)*(9*x^2 - 9*x^4 + 45) - exp(2*x)*exp(1/x)*(9*x^2 - 9) 
))/(x^2*exp(x) + x^4),x)
 

Output:

int((log(exp(x)/3 + x^2/3)*(exp(x + 1/x)*(90*x^2 + 36*x^3) + 180*x^3*exp(1 
/x) + 18*x^2*exp(2*x + 1/x)) - log(exp(x)/3 + x^2/3)^2*(exp(x + 1/x)*(9*x^ 
2 - 9*x^4 + 45) - exp(2*x + 1/x)*(9*x^2 - 9) + 45*x^2*exp(1/x)))/(x^2*exp( 
x) + x^4), x)
 

Reduce [F]

\[ \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx=\int \frac {\left (\left (9 x^{2}-9\right ) {\mathrm e}^{\frac {1}{x}} \left ({\mathrm e}^{x}\right )^{2}+\left (9 x^{4}-9 x^{2}-45\right ) {\mathrm e}^{\frac {1}{x}} {\mathrm e}^{x}-45 x^{2} {\mathrm e}^{\frac {1}{x}}\right ) \mathrm {log}\left (\frac {x^{2}}{3}+\frac {{\mathrm e}^{x}}{3}\right )^{2}+\left (18 x^{2} {\mathrm e}^{\frac {1}{x}} \left ({\mathrm e}^{x}\right )^{2}+\left (36 x^{3}+90 x^{2}\right ) {\mathrm e}^{\frac {1}{x}} {\mathrm e}^{x}+180 x^{3} {\mathrm e}^{\frac {1}{x}}\right ) \mathrm {log}\left (\frac {x^{2}}{3}+\frac {{\mathrm e}^{x}}{3}\right )}{{\mathrm e}^{x} x^{2}+x^{4}}d x \] Input:

int((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-45*x^2* 
exp(1/x))*log(1/3*x^2+1/3*exp(x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^3+90*x 
^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*log(1/3*x^2+1/3*exp(x)))/(exp(x)*x^2 
+x^4),x)
 

Output:

int((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-45*x^2* 
exp(1/x))*log(1/3*x^2+1/3*exp(x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^3+90*x 
^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*log(1/3*x^2+1/3*exp(x)))/(exp(x)*x^2 
+x^4),x)