\(\int \frac {e^{\frac {-9+2 x \log (3+e^x+4 x^2)}{x \log (3+e^x+4 x^2)}} (9 e^x x+72 x^2+(27+9 e^x+36 x^2) \log (3+e^x+4 x^2))}{(3 x^2+e^x x^2+4 x^4) \log ^2(3+e^x+4 x^2)} \, dx\) [332]

Optimal result

Integrand size = 105, antiderivative size = 25 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=e^{\frac {2 x-\frac {9}{\log \left (3+e^x+4 x^2\right )}}{x}} \] Output:

exp((2*x-9/ln(exp(x)+4*x^2+3))/x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} \] Input:

Integrate[(E^((-9 + 2*x*Log[3 + E^x + 4*x^2])/(x*Log[3 + E^x + 4*x^2]))*(9 
*E^x*x + 72*x^2 + (27 + 9*E^x + 36*x^2)*Log[3 + E^x + 4*x^2]))/((3*x^2 + E 
^x*x^2 + 4*x^4)*Log[3 + E^x + 4*x^2]^2),x]
 

Output:

E^(2 - 9/(x*Log[3 + E^x + 4*x^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.

\(\displaystyle \int \frac {\left (72 x^2+\left (36 x^2+9 e^x+27\right ) \log \left (4 x^2+e^x+3\right )+9 e^x x\right ) \exp \left (\frac {2 x \log \left (4 x^2+e^x+3\right )-9}{x \log \left (4 x^2+e^x+3\right )}\right )}{\left (4 x^4+e^x x^2+3 x^2\right ) \log ^2\left (4 x^2+e^x+3\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {9 \left (\log \left (4 x^2+e^x+3\right )+x\right ) \exp \left (\frac {2 x \log \left (4 x^2+e^x+3\right )-9}{x \log \left (4 x^2+e^x+3\right )}\right )}{x^2 \log ^2\left (4 x^2+e^x+3\right )}-\frac {9 \left (4 x^2-8 x+3\right ) \exp \left (\frac {2 x \log \left (4 x^2+e^x+3\right )-9}{x \log \left (4 x^2+e^x+3\right )}\right )}{x \left (4 x^2+e^x+3\right ) \log ^2\left (4 x^2+e^x+3\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 9 \int \frac {\exp \left (\frac {2 x \log \left (4 x^2+e^x+3\right )-9}{x \log \left (4 x^2+e^x+3\right )}\right )}{x \log ^2\left (4 x^2+e^x+3\right )}dx+72 \int \frac {\exp \left (\frac {2 x \log \left (4 x^2+e^x+3\right )-9}{x \log \left (4 x^2+e^x+3\right )}\right )}{\left (4 x^2+e^x+3\right ) \log ^2\left (4 x^2+e^x+3\right )}dx-27 \int \frac {\exp \left (\frac {2 x \log \left (4 x^2+e^x+3\right )-9}{x \log \left (4 x^2+e^x+3\right )}\right )}{x \left (4 x^2+e^x+3\right ) \log ^2\left (4 x^2+e^x+3\right )}dx-36 \int \frac {\exp \left (\frac {2 x \log \left (4 x^2+e^x+3\right )-9}{x \log \left (4 x^2+e^x+3\right )}\right ) x}{\left (4 x^2+e^x+3\right ) \log ^2\left (4 x^2+e^x+3\right )}dx+9 \int \frac {\exp \left (\frac {2 x \log \left (4 x^2+e^x+3\right )-9}{x \log \left (4 x^2+e^x+3\right )}\right )}{x^2 \log \left (4 x^2+e^x+3\right )}dx\)

Input:

Int[(E^((-9 + 2*x*Log[3 + E^x + 4*x^2])/(x*Log[3 + E^x + 4*x^2]))*(9*E^x*x 
 + 72*x^2 + (27 + 9*E^x + 36*x^2)*Log[3 + E^x + 4*x^2]))/((3*x^2 + E^x*x^2 
 + 4*x^4)*Log[3 + E^x + 4*x^2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 27.52 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32

method	result	size
risch	\({\mathrm e}^{\frac {2 x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )-9}{x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )}}\)	\(33\)
parallelrisch	\({\mathrm e}^{\frac {2 x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )-9}{x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )}}\)	\(33\)

Input:

int(((9*exp(x)+36*x^2+27)*ln(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp((2*x*l 
n(exp(x)+4*x^2+3)-9)/x/ln(exp(x)+4*x^2+3))/(exp(x)*x^2+4*x^4+3*x^2)/ln(exp 
(x)+4*x^2+3)^2,x,method=_RETURNVERBOSE)
 

Output:

exp((2*x*ln(exp(x)+4*x^2+3)-9)/x/ln(exp(x)+4*x^2+3))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=e^{\left (\frac {2 \, x \log \left (4 \, x^{2} + e^{x} + 3\right ) - 9}{x \log \left (4 \, x^{2} + e^{x} + 3\right )}\right )} \] Input:

integrate(((9*exp(x)+36*x^2+27)*log(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp 
((2*x*log(exp(x)+4*x^2+3)-9)/x/log(exp(x)+4*x^2+3))/(exp(x)*x^2+4*x^4+3*x^ 
2)/log(exp(x)+4*x^2+3)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(((9*exp(x)+36*x**2+27)*ln(exp(x)+4*x**2+3)+9*exp(x)*x+72*x**2)*e 
xp((2*x*ln(exp(x)+4*x**2+3)-9)/x/ln(exp(x)+4*x**2+3))/(exp(x)*x**2+4*x**4+ 
3*x**2)/ln(exp(x)+4*x**2+3)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(((9*exp(x)+36*x^2+27)*log(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp 
((2*x*log(exp(x)+4*x^2+3)-9)/x/log(exp(x)+4*x^2+3))/(exp(x)*x^2+4*x^4+3*x^ 
2)/log(exp(x)+4*x^2+3)^2,x, algorithm="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 [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=e^{\left (-\frac {9}{x \log \left (4 \, x^{2} + e^{x} + 3\right )} + 2\right )} \] Input:

integrate(((9*exp(x)+36*x^2+27)*log(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp 
((2*x*log(exp(x)+4*x^2+3)-9)/x/log(exp(x)+4*x^2+3))/(exp(x)*x^2+4*x^4+3*x^ 
2)/log(exp(x)+4*x^2+3)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.74 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx={\mathrm {e}}^2\,{\mathrm {e}}^{-\frac {9}{x\,\ln \left ({\mathrm {e}}^x+4\,x^2+3\right )}} \] Input:

int((exp((2*x*log(exp(x) + 4*x^2 + 3) - 9)/(x*log(exp(x) + 4*x^2 + 3)))*(l 
og(exp(x) + 4*x^2 + 3)*(9*exp(x) + 36*x^2 + 27) + 9*x*exp(x) + 72*x^2))/(l 
og(exp(x) + 4*x^2 + 3)^2*(x^2*exp(x) + 3*x^2 + 4*x^4)),x)
 

Output:

exp(2)*exp(-9/(x*log(exp(x) + 4*x^2 + 3)))
 

Reduce [F]

\[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=9 e^{2} \left (\int \frac {e^{x}}{e^{\frac {\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}+9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right )^{2} x +4 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right )^{2} x^{3}+3 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right )^{2} x}d x +\int \frac {e^{x}}{e^{\frac {\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}+9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}+4 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{4}+3 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}}d x +8 \left (\int \frac {1}{e^{\frac {\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}+9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right )^{2}+4 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right )^{2} x^{2}+3 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right )^{2}}d x \right )+3 \left (\int \frac {1}{e^{\frac {\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}+9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}+4 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{4}+3 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}}d x \right )+4 \left (\int \frac {1}{e^{\frac {\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}+9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right )+4 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x^{2}+3 e^{\frac {9}{\mathrm {log}\left (e^{x}+4 x^{2}+3\right ) x}} \mathrm {log}\left (e^{x}+4 x^{2}+3\right )}d x \right )\right ) \] Input:

int(((9*exp(x)+36*x^2+27)*log(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp((2*x* 
log(exp(x)+4*x^2+3)-9)/x/log(exp(x)+4*x^2+3))/(exp(x)*x^2+4*x^4+3*x^2)/log 
(exp(x)+4*x^2+3)^2,x)
 

Output:

9*e**2*(int(e**x/(e**((log(e**x + 4*x**2 + 3)*x**2 + 9)/(log(e**x + 4*x**2 
 + 3)*x))*log(e**x + 4*x**2 + 3)**2*x + 4*e**(9/(log(e**x + 4*x**2 + 3)*x) 
)*log(e**x + 4*x**2 + 3)**2*x**3 + 3*e**(9/(log(e**x + 4*x**2 + 3)*x))*log 
(e**x + 4*x**2 + 3)**2*x),x) + int(e**x/(e**((log(e**x + 4*x**2 + 3)*x**2 
+ 9)/(log(e**x + 4*x**2 + 3)*x))*log(e**x + 4*x**2 + 3)*x**2 + 4*e**(9/(lo 
g(e**x + 4*x**2 + 3)*x))*log(e**x + 4*x**2 + 3)*x**4 + 3*e**(9/(log(e**x + 
 4*x**2 + 3)*x))*log(e**x + 4*x**2 + 3)*x**2),x) + 8*int(1/(e**((log(e**x 
+ 4*x**2 + 3)*x**2 + 9)/(log(e**x + 4*x**2 + 3)*x))*log(e**x + 4*x**2 + 3) 
**2 + 4*e**(9/(log(e**x + 4*x**2 + 3)*x))*log(e**x + 4*x**2 + 3)**2*x**2 + 
 3*e**(9/(log(e**x + 4*x**2 + 3)*x))*log(e**x + 4*x**2 + 3)**2),x) + 3*int 
(1/(e**((log(e**x + 4*x**2 + 3)*x**2 + 9)/(log(e**x + 4*x**2 + 3)*x))*log( 
e**x + 4*x**2 + 3)*x**2 + 4*e**(9/(log(e**x + 4*x**2 + 3)*x))*log(e**x + 4 
*x**2 + 3)*x**4 + 3*e**(9/(log(e**x + 4*x**2 + 3)*x))*log(e**x + 4*x**2 + 
3)*x**2),x) + 4*int(1/(e**((log(e**x + 4*x**2 + 3)*x**2 + 9)/(log(e**x + 4 
*x**2 + 3)*x))*log(e**x + 4*x**2 + 3) + 4*e**(9/(log(e**x + 4*x**2 + 3)*x) 
)*log(e**x + 4*x**2 + 3)*x**2 + 3*e**(9/(log(e**x + 4*x**2 + 3)*x))*log(e* 
*x + 4*x**2 + 3)),x))