\(\int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x (-6 x^2-2 e^3 x^2-2 x^3)+e^{\frac {1-3 x+x \log (5)}{x}} (2 e^3-6 x^2+2 e^x x^2-2 x^3)}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x (-6 x^2-2 x^3)+e^{\frac {1-3 x+x \log (5)}{x}} (-6 x^2+2 e^x x^2-2 x^3)} \, dx\) [1431]

Optimal result

Integrand size = 210, antiderivative size = 26 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=x-\frac {2 e^3}{3-5 e^{-3+\frac {1}{x}}-e^x+x} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=x+\frac {2 e^6}{5 e^{\frac {1}{x}}+e^{3+x}-e^3 (3+x)} \] Input:

Integrate[(9*x^2 + 2*E^3*x^2 + E^(2*x)*x^2 + E^((2*(1 - 3*x + x*Log[5]))/x 
)*x^2 + 6*x^3 + x^4 + E^x*(-6*x^2 - 2*E^3*x^2 - 2*x^3) + E^((1 - 3*x + x*L 
og[5])/x)*(2*E^3 - 6*x^2 + 2*E^x*x^2 - 2*x^3))/(9*x^2 + E^(2*x)*x^2 + E^(( 
2*(1 - 3*x + x*Log[5]))/x)*x^2 + 6*x^3 + x^4 + E^x*(-6*x^2 - 2*x^3) + E^(( 
1 - 3*x + x*Log[5])/x)*(-6*x^2 + 2*E^x*x^2 - 2*x^3)),x]
 

Output:

x + (2*E^6)/(5*E^x^(-1) + E^(3 + x) - E^3*(3 + 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.

\(\displaystyle \int \frac {x^4+6 x^3+e^{2 x} x^2+2 e^3 x^2+9 x^2+x^2 e^{\frac {2 (-3 x+x \log (5)+1)}{x}}+e^x \left (-2 x^3-2 e^3 x^2-6 x^2\right )+\left (-2 x^3+2 e^x x^2-6 x^2+2 e^3\right ) e^{\frac {-3 x+x \log (5)+1}{x}}}{x^4+6 x^3+e^{2 x} x^2+9 x^2+x^2 e^{\frac {2 (-3 x+x \log (5)+1)}{x}}+e^x \left (-2 x^3-6 x^2\right )+\left (-2 x^3+2 e^x x^2-6 x^2\right ) e^{\frac {-3 x+x \log (5)+1}{x}}} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {x^4+6 x^3+e^{2 x} x^2+\left (9+2 e^3\right ) x^2+x^2 e^{\frac {2 (-3 x+x \log (5)+1)}{x}}+e^x \left (-2 x^3-2 e^3 x^2-6 x^2\right )+\left (-2 x^3+2 e^x x^2-6 x^2+2 e^3\right ) e^{\frac {-3 x+x \log (5)+1}{x}}}{x^4+6 x^3+e^{2 x} x^2+9 x^2+x^2 e^{\frac {2 (-3 x+x \log (5)+1)}{x}}+e^x \left (-2 x^3-6 x^2\right )+\left (-2 x^3+2 e^x x^2-6 x^2\right ) e^{\frac {-3 x+x \log (5)+1}{x}}}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^6 \left (x^4+6 x^3+e^{2 x} x^2+\left (9+2 e^3\right ) x^2+x^2 e^{\frac {2 (-3 x+x \log (5)+1)}{x}}+e^x \left (-2 x^3-2 e^3 x^2-6 x^2\right )+\left (-2 x^3+2 e^x x^2-6 x^2+2 e^3\right ) e^{\frac {-3 x+x \log (5)+1}{x}}\right )}{x^2 \left (e^3 x-e^{x+3}-5 e^{\frac {1}{x}}+3 e^3\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle e^6 \int \frac {x^4+6 x^3+25 e^{\frac {2 (1-3 x)}{x}} x^2+e^{2 x} x^2+\left (9+2 e^3\right ) x^2+10 e^{\frac {1-3 x}{x}} \left (-x^3+e^x x^2-3 x^2+e^3\right )-2 e^x \left (x^3+e^3 x^2+3 x^2\right )}{x^2 \left (e^3 x-e^{x+3}-5 e^{\frac {1}{x}}+3 e^3\right )^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle e^6 \left (10 \int \frac {e^{\frac {1}{x}}}{x^2 \left (-e^3 x+e^{x+3}+5 e^{\frac {1}{x}}-3 e^3\right )^2}dx-4 e^3 \int \frac {1}{\left (-e^3 x+e^{x+3}+5 e^{\frac {1}{x}}-3 e^3\right )^2}dx+10 \int \frac {e^{\frac {1}{x}}}{\left (-e^3 x+e^{x+3}+5 e^{\frac {1}{x}}-3 e^3\right )^2}dx-2 \int \frac {1}{-e^3 x+e^{x+3}+5 e^{\frac {1}{x}}-3 e^3}dx-2 e^3 \int \frac {x}{\left (e^3 x-e^{x+3}-5 e^{\frac {1}{x}}+3 e^3\right )^2}dx+\frac {x}{e^6}\right )\)

Input:

Int[(9*x^2 + 2*E^3*x^2 + E^(2*x)*x^2 + E^((2*(1 - 3*x + x*Log[5]))/x)*x^2 
+ 6*x^3 + x^4 + E^x*(-6*x^2 - 2*E^3*x^2 - 2*x^3) + E^((1 - 3*x + x*Log[5]) 
/x)*(2*E^3 - 6*x^2 + 2*E^x*x^2 - 2*x^3))/(9*x^2 + E^(2*x)*x^2 + E^((2*(1 - 
 3*x + x*Log[5]))/x)*x^2 + 6*x^3 + x^4 + E^x*(-6*x^2 - 2*x^3) + E^((1 - 3* 
x + x*Log[5])/x)*(-6*x^2 + 2*E^x*x^2 - 2*x^3)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 5.97 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12

method	result	size
risch	\(x -\frac {2 \,{\mathrm e}^{3}}{x -{\mathrm e}^{x}-5 \,{\mathrm e}^{-\frac {-1+3 x}{x}}+3}\)	\(29\)
parallelrisch	\(-\frac {-x^{2}+{\mathrm e}^{x} x +x \,{\mathrm e}^{\frac {x \ln \left (5\right )-3 x +1}{x}}+2 \,{\mathrm e}^{3}-3 x}{x -{\mathrm e}^{x}-{\mathrm e}^{\frac {x \ln \left (5\right )-3 x +1}{x}}+3}\)	\(61\)

Input:

int((x^2*exp((x*ln(5)-3*x+1)/x)^2+(2*exp(x)*x^2+2*exp(3)-2*x^3-6*x^2)*exp( 
(x*ln(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^2*exp(3)-2*x^3-6*x^2)*exp(x)+2*x^2*e 
xp(3)+x^4+6*x^3+9*x^2)/(x^2*exp((x*ln(5)-3*x+1)/x)^2+(2*exp(x)*x^2-2*x^3-6 
*x^2)*exp((x*ln(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^3-6*x^2)*exp(x)+x^4+6*x^3+ 
9*x^2),x,method=_RETURNVERBOSE)
 

Output:

x-2*exp(3)/(x-exp(x)-5*exp(-(-1+3*x)/x)+3)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).

Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.27 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {x^{2} - x e^{x} - x e^{\left (\frac {x \log \left (5\right ) - 3 \, x + 1}{x}\right )} + 3 \, x - 2 \, e^{3}}{x - e^{x} - e^{\left (\frac {x \log \left (5\right ) - 3 \, x + 1}{x}\right )} + 3} \] Input:

integrate((x^2*exp((x*log(5)-3*x+1)/x)^2+(2*exp(x)*x^2+2*exp(3)-2*x^3-6*x^ 
2)*exp((x*log(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^2*exp(3)-2*x^3-6*x^2)*exp(x) 
+2*x^2*exp(3)+x^4+6*x^3+9*x^2)/(x^2*exp((x*log(5)-3*x+1)/x)^2+(2*exp(x)*x^ 
2-2*x^3-6*x^2)*exp((x*log(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^3-6*x^2)*exp(x)+ 
x^4+6*x^3+9*x^2),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=x + \frac {2 e^{3}}{- x + e^{x} + e^{\frac {- 3 x + x \log {\left (5 \right )} + 1}{x}} - 3} \] Input:

integrate((x**2*exp((x*ln(5)-3*x+1)/x)**2+(2*exp(x)*x**2+2*exp(3)-2*x**3-6 
*x**2)*exp((x*ln(5)-3*x+1)/x)+exp(x)**2*x**2+(-2*x**2*exp(3)-2*x**3-6*x**2 
)*exp(x)+2*x**2*exp(3)+x**4+6*x**3+9*x**2)/(x**2*exp((x*ln(5)-3*x+1)/x)**2 
+(2*exp(x)*x**2-2*x**3-6*x**2)*exp((x*ln(5)-3*x+1)/x)+exp(x)**2*x**2+(-2*x 
**3-6*x**2)*exp(x)+x**4+6*x**3+9*x**2),x)
 

Output:

x + 2*exp(3)/(-x + exp(x) + exp((-3*x + x*log(5) + 1)/x) - 3)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).

Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {x^{2} e^{3} + 3 \, x e^{3} - x e^{\left (x + 3\right )} - 5 \, x e^{\frac {1}{x}} - 2 \, e^{6}}{x e^{3} + 3 \, e^{3} - e^{\left (x + 3\right )} - 5 \, e^{\frac {1}{x}}} \] Input:

integrate((x^2*exp((x*log(5)-3*x+1)/x)^2+(2*exp(x)*x^2+2*exp(3)-2*x^3-6*x^ 
2)*exp((x*log(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^2*exp(3)-2*x^3-6*x^2)*exp(x) 
+2*x^2*exp(3)+x^4+6*x^3+9*x^2)/(x^2*exp((x*log(5)-3*x+1)/x)^2+(2*exp(x)*x^ 
2-2*x^3-6*x^2)*exp((x*log(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^3-6*x^2)*exp(x)+ 
x^4+6*x^3+9*x^2),x, algorithm="maxima")
 

Output:

(x^2*e^3 + 3*x*e^3 - x*e^(x + 3) - 5*x*e^(1/x) - 2*e^6)/(x*e^3 + 3*e^3 - e 
^(x + 3) - 5*e^(1/x))
 

Giac [F(-1)]

Timed out. \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=\text {Timed out} \] Input:

integrate((x^2*exp((x*log(5)-3*x+1)/x)^2+(2*exp(x)*x^2+2*exp(3)-2*x^3-6*x^ 
2)*exp((x*log(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^2*exp(3)-2*x^3-6*x^2)*exp(x) 
+2*x^2*exp(3)+x^4+6*x^3+9*x^2)/(x^2*exp((x*log(5)-3*x+1)/x)^2+(2*exp(x)*x^ 
2-2*x^3-6*x^2)*exp((x*log(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^3-6*x^2)*exp(x)+ 
x^4+6*x^3+9*x^2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 1.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=x-\frac {2\,{\mathrm {e}}^3}{x-{\mathrm {e}}^x-5\,{\mathrm {e}}^{1/x}\,{\mathrm {e}}^{-3}+3} \] Input:

int((exp((x*log(5) - 3*x + 1)/x)*(2*exp(3) + 2*x^2*exp(x) - 6*x^2 - 2*x^3) 
 + x^2*exp((2*(x*log(5) - 3*x + 1))/x) + x^2*exp(2*x) + 2*x^2*exp(3) - exp 
(x)*(2*x^2*exp(3) + 6*x^2 + 2*x^3) + 9*x^2 + 6*x^3 + x^4)/(x^2*exp((2*(x*l 
og(5) - 3*x + 1))/x) - exp(x)*(6*x^2 + 2*x^3) - exp((x*log(5) - 3*x + 1)/x 
)*(6*x^2 - 2*x^2*exp(x) + 2*x^3) + x^2*exp(2*x) + 9*x^2 + 6*x^3 + x^4),x)
 

Output:

x - (2*exp(3))/(x - exp(x) - 5*exp(1/x)*exp(-3) + 3)
 

Reduce [B] (verification not implemented)

Time = 2.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.50 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {5 e^{\frac {1}{x}} x +e^{x} e^{3} x +2 e^{6}-e^{3} x^{2}-3 e^{3} x}{5 e^{\frac {1}{x}}+e^{x} e^{3}-e^{3} x -3 e^{3}} \] Input:

int((x^2*exp((x*log(5)-3*x+1)/x)^2+(2*exp(x)*x^2+2*exp(3)-2*x^3-6*x^2)*exp 
((x*log(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^2*exp(3)-2*x^3-6*x^2)*exp(x)+2*x^2 
*exp(3)+x^4+6*x^3+9*x^2)/(x^2*exp((x*log(5)-3*x+1)/x)^2+(2*exp(x)*x^2-2*x^ 
3-6*x^2)*exp((x*log(5)-3*x+1)/x)+exp(x)^2*x^2+(-2*x^3-6*x^2)*exp(x)+x^4+6* 
x^3+9*x^2),x)
 

Output:

(5*e**(1/x)*x + e**x*e**3*x + 2*e**6 - e**3*x**2 - 3*e**3*x)/(5*e**(1/x) + 
 e**x*e**3 - e**3*x - 3*e**3)