\(\int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} (-1-x+3 x^2+2 x^3)+(1+2 x+x^2) \log (x))}{x^2+2 x^3+x^4} \, dx\) [114]

Optimal result

Integrand size = 101, antiderivative size = 30 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {5+e^{5+\frac {x+2 x^2}{1+x}}+x-\log (x)}{x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {5+e^{4+2 x+\frac {1}{1+x}}+x}{x}} x^{-1/x} \] Input:

Integrate[(E^((5 + E^((5 + 6*x + 2*x^2)/(1 + x)) + x - Log[x])/x)*(-6 - 12 
*x - 6*x^2 + E^((5 + 6*x + 2*x^2)/(1 + x))*(-1 - x + 3*x^2 + 2*x^3) + (1 + 
 2*x + x^2)*Log[x]))/(x^2 + 2*x^3 + x^4),x]
 

Output:

E^((5 + E^(4 + 2*x + (1 + x)^(-1)) + x)/x)/x^x^(-1)
 

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.

\(\displaystyle \int \frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}+x-\log (x)+5}{x}} \left (-6 x^2+\left (x^2+2 x+1\right ) \log (x)+e^{\frac {2 x^2+6 x+5}{x+1}} \left (2 x^3+3 x^2-x-1\right )-12 x-6\right )}{x^4+2 x^3+x^2} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}+x-\log (x)+5}{x}} \left (-6 x^2+\left (x^2+2 x+1\right ) \log (x)+e^{\frac {2 x^2+6 x+5}{x+1}} \left (2 x^3+3 x^2-x-1\right )-12 x-6\right )}{x^2 \left (x^2+2 x+1\right )}dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}+x-\log (x)+5}{x}} \left (-6 x^2+\left (x^2+2 x+1\right ) \log (x)+e^{\frac {2 x^2+6 x+5}{x+1}} \left (2 x^3+3 x^2-x-1\right )-12 x-6\right )}{x^2 (x+1)^2}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {e^{\frac {2 x^2+6 x+5}{x+1}}}{x}+\frac {5}{x}-\frac {\log (x)}{x}+1} \left (-6 x^2+\left (x^2+2 x+1\right ) \log (x)+e^{\frac {2 x^2+6 x+5}{x+1}} \left (2 x^3+3 x^2-x-1\right )-12 x-6\right )}{x^2 (x+1)^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(E^((5 + E^((5 + 6*x + 2*x^2)/(1 + x)) + x - Log[x])/x)*(-6 - 12*x - 6 
*x^2 + E^((5 + 6*x + 2*x^2)/(1 + x))*(-1 - x + 3*x^2 + 2*x^3) + (1 + 2*x + 
 x^2)*Log[x]))/(x^2 + 2*x^3 + x^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.82 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10

method	result	size
parallelrisch	\({\mathrm e}^{-\frac {\ln \left (x \right )-{\mathrm e}^{\frac {2 x^{2}+6 x +5}{1+x}}-5-x}{x}}\)	\(33\)
risch	\(x^{-\frac {1}{x}} {\mathrm e}^{\frac {{\mathrm e}^{\frac {2 x^{2}+6 x +5}{1+x}}+5+x}{x}}\)	\(34\)

Input:

int(((x^2+2*x+1)*ln(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6*x^2-12 
*x-6)*exp((-ln(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2),x,metho 
d=_RETURNVERBOSE)
 

Output:

exp(-(ln(x)-exp((2*x^2+6*x+5)/(1+x))-5-x)/x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (\frac {x + e^{\left (\frac {2 \, x^{2} + 6 \, x + 5}{x + 1}\right )} - \log \left (x\right ) + 5}{x}\right )} \] Input:

integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6 
*x^2-12*x-6)*exp((-log(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2) 
,x, algorithm="fricas")
 

Output:

e^((x + e^((2*x^2 + 6*x + 5)/(x + 1)) - log(x) + 5)/x)
 

Sympy [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {x + e^{\frac {2 x^{2} + 6 x + 5}{x + 1}} - \log {\left (x \right )} + 5}{x}} \] Input:

integrate(((x**2+2*x+1)*ln(x)+(2*x**3+3*x**2-x-1)*exp((2*x**2+6*x+5)/(1+x) 
)-6*x**2-12*x-6)*exp((-ln(x)+exp((2*x**2+6*x+5)/(1+x))+5+x)/x)/(x**4+2*x** 
3+x**2),x)
 

Output:

exp((x + exp((2*x**2 + 6*x + 5)/(x + 1)) - log(x) + 5)/x)
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (\frac {e^{\left (2 \, x + \frac {1}{x + 1} + 4\right )}}{x} - \frac {\log \left (x\right )}{x} + \frac {5}{x} + 1\right )} \] Input:

integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6 
*x^2-12*x-6)*exp((-log(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2) 
,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (\frac {e^{\left (\frac {2 \, x^{2}}{x + 1} + \frac {6 \, x}{x + 1} + \frac {5}{x + 1}\right )}}{x} - \frac {\log \left (x\right )}{x} + \frac {5}{x} + 1\right )} \] Input:

integrate(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6 
*x^2-12*x-6)*exp((-log(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2) 
,x, algorithm="giac")
 

Output:

e^(e^(2*x^2/(x + 1) + 6*x/(x + 1) + 5/(x + 1))/x - log(x)/x + 5/x + 1)
 

Mupad [B] (verification not implemented)

Time = 3.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=\frac {\mathrm {e}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {6\,x}{x+1}}\,{\mathrm {e}}^{\frac {2\,x^2}{x+1}}\,{\mathrm {e}}^{\frac {5}{x+1}}}{x}}}{x^{1/x}} \] Input:

int(-(exp((x + exp((6*x + 2*x^2 + 5)/(x + 1)) - log(x) + 5)/x)*(12*x + exp 
((6*x + 2*x^2 + 5)/(x + 1))*(x - 3*x^2 - 2*x^3 + 1) - log(x)*(2*x + x^2 + 
1) + 6*x^2 + 6))/(x^2 + 2*x^3 + x^4),x)
 

Output:

(exp(1)*exp(5/x)*exp((exp((6*x)/(x + 1))*exp((2*x^2)/(x + 1))*exp(5/(x + 1 
)))/x))/x^(1/x)
 

Reduce [F]

\[ \int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+x^4} \, dx=\int \frac {\left (\left (x^{2}+2 x +1\right ) \mathrm {log}\left (x \right )+\left (2 x^{3}+3 x^{2}-x -1\right ) {\mathrm e}^{\frac {2 x^{2}+6 x +5}{x +1}}-6 x^{2}-12 x -6\right ) {\mathrm e}^{\frac {-\mathrm {log}\left (x \right )+{\mathrm e}^{\frac {2 x^{2}+6 x +5}{x +1}}+5+x}{x}}}{x^{4}+2 x^{3}+x^{2}}d x \] Input:

int(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6*x^2-1 
2*x-6)*exp((-log(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2),x)
 

Output:

int(((x^2+2*x+1)*log(x)+(2*x^3+3*x^2-x-1)*exp((2*x^2+6*x+5)/(1+x))-6*x^2-1 
2*x-6)*exp((-log(x)+exp((2*x^2+6*x+5)/(1+x))+5+x)/x)/(x^4+2*x^3+x^2),x)