\(\int \frac {9-9 e^x+9 x+(36+58 x+6 x^2+e^x (-36-22 x-2 x^2)) \log (\frac {1}{5} (x^2-e^x x^2+x^3))+(9-9 e^x+9 x) \log ^2(\frac {1}{5} (x^2-e^x x^2+x^3))}{-x^2+e^x x^2-x^3+(-2 x^2+2 e^x x^2-2 x^3) \log ^2(\frac {1}{5} (x^2-e^x x^2+x^3))+(-x^2+e^x x^2-x^3) \log ^4(\frac {1}{5} (x^2-e^x x^2+x^3))} \, dx\) [2732]

Optimal result

Integrand size = 192, antiderivative size = 30 \[ \int \frac {9-9 e^x+9 x+\left (36+58 x+6 x^2+e^x \left (-36-22 x-2 x^2\right )\right ) \log \left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (9-9 e^x+9 x\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )}{-x^2+e^x x^2-x^3+\left (-2 x^2+2 e^x x^2-2 x^3\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (-x^2+e^x x^2-x^3\right ) \log ^4\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )} \, dx=\frac {1+\frac {9}{x}}{1+\log ^2\left (\frac {1}{5} x^2 \left (1-e^x+x\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {9-9 e^x+9 x+\left (36+58 x+6 x^2+e^x \left (-36-22 x-2 x^2\right )\right ) \log \left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (9-9 e^x+9 x\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )}{-x^2+e^x x^2-x^3+\left (-2 x^2+2 e^x x^2-2 x^3\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (-x^2+e^x x^2-x^3\right ) \log ^4\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )} \, dx=\frac {9+x}{x \left (1+\log ^2\left (\frac {1}{5} x^2 \left (1-e^x+x\right )\right )\right )} \] Input:

Integrate[(9 - 9*E^x + 9*x + (36 + 58*x + 6*x^2 + E^x*(-36 - 22*x - 2*x^2) 
)*Log[(x^2 - E^x*x^2 + x^3)/5] + (9 - 9*E^x + 9*x)*Log[(x^2 - E^x*x^2 + x^ 
3)/5]^2)/(-x^2 + E^x*x^2 - x^3 + (-2*x^2 + 2*E^x*x^2 - 2*x^3)*Log[(x^2 - E 
^x*x^2 + x^3)/5]^2 + (-x^2 + E^x*x^2 - x^3)*Log[(x^2 - E^x*x^2 + x^3)/5]^4 
),x]
 

Output:

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

\(\displaystyle \int \frac {\left (9 x-9 e^x+9\right ) \log ^2\left (\frac {1}{5} \left (x^3-e^x x^2+x^2\right )\right )+\left (6 x^2+e^x \left (-2 x^2-22 x-36\right )+58 x+36\right ) \log \left (\frac {1}{5} \left (x^3-e^x x^2+x^2\right )\right )-9 e^x+9 x+9}{-x^3+e^x x^2-x^2+\left (-x^3+e^x x^2-x^2\right ) \log ^4\left (\frac {1}{5} \left (x^3-e^x x^2+x^2\right )\right )+\left (-2 x^3+2 e^x x^2-2 x^2\right ) \log ^2\left (\frac {1}{5} \left (x^3-e^x x^2+x^2\right )\right )} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {-9 \log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )-2 x^2 \log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )-22 x \log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )-36 \log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )-9}{x^2 \left (\log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )+1\right )^2}-\frac {2 (x+9) \log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )}{\left (-x+e^x-1\right ) \left (\log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )+1\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \int \frac {\log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )}{\left (\log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )+1\right )^2}dx-18 \int \frac {\log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )}{\left (-x+e^x-1\right ) \left (\log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )+1\right )^2}dx-36 \int \frac {\log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )}{x^2 \left (\log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )+1\right )^2}dx-22 \int \frac {\log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )}{x \left (\log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )+1\right )^2}dx-2 \int \frac {x \log \left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )}{\left (-x+e^x-1\right ) \left (\log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )+1\right )^2}dx-9 \int \frac {1}{x^2 \left (\log ^2\left (\frac {1}{5} x^2 \left (x-e^x+1\right )\right )+1\right )}dx\)

Input:

Int[(9 - 9*E^x + 9*x + (36 + 58*x + 6*x^2 + E^x*(-36 - 22*x - 2*x^2))*Log[ 
(x^2 - E^x*x^2 + x^3)/5] + (9 - 9*E^x + 9*x)*Log[(x^2 - E^x*x^2 + x^3)/5]^ 
2)/(-x^2 + E^x*x^2 - x^3 + (-2*x^2 + 2*E^x*x^2 - 2*x^3)*Log[(x^2 - E^x*x^2 
 + x^3)/5]^2 + (-x^2 + E^x*x^2 - x^3)*Log[(x^2 - E^x*x^2 + x^3)/5]^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 10.73 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(\frac {x +9}{x \left (\ln \left (-\frac {x^{2} \left ({\mathrm e}^{x}-x -1\right )}{5}\right )^{2}+1\right )}\)	\(27\)
risch	\(\text {Expression too large to display}\)	\(1503\)

Input:

int(((-9*exp(x)+9*x+9)*ln(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+((-2*x^2-22*x 
-36)*exp(x)+6*x^2+58*x+36)*ln(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)-9*exp(x)+9* 
x+9)/((exp(x)*x^2-x^3-x^2)*ln(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^4+(2*exp(x) 
*x^2-2*x^3-2*x^2)*ln(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+exp(x)*x^2-x^3-x^2 
),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {9-9 e^x+9 x+\left (36+58 x+6 x^2+e^x \left (-36-22 x-2 x^2\right )\right ) \log \left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (9-9 e^x+9 x\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )}{-x^2+e^x x^2-x^3+\left (-2 x^2+2 e^x x^2-2 x^3\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (-x^2+e^x x^2-x^3\right ) \log ^4\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )} \, dx=\frac {x + 9}{x \log \left (\frac {1}{5} \, x^{3} - \frac {1}{5} \, x^{2} e^{x} + \frac {1}{5} \, x^{2}\right )^{2} + x} \] Input:

integrate(((-9*exp(x)+9*x+9)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+((-2*x 
^2-22*x-36)*exp(x)+6*x^2+58*x+36)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)-9*e 
xp(x)+9*x+9)/((exp(x)*x^2-x^3-x^2)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^4+ 
(2*exp(x)*x^2-2*x^3-2*x^2)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+exp(x)*x 
^2-x^3-x^2),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {9-9 e^x+9 x+\left (36+58 x+6 x^2+e^x \left (-36-22 x-2 x^2\right )\right ) \log \left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (9-9 e^x+9 x\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )}{-x^2+e^x x^2-x^3+\left (-2 x^2+2 e^x x^2-2 x^3\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (-x^2+e^x x^2-x^3\right ) \log ^4\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )} \, dx=\frac {x + 9}{x \log {\left (\frac {x^{3}}{5} - \frac {x^{2} e^{x}}{5} + \frac {x^{2}}{5} \right )}^{2} + x} \] Input:

integrate(((-9*exp(x)+9*x+9)*ln(-1/5*exp(x)*x**2+1/5*x**3+1/5*x**2)**2+((- 
2*x**2-22*x-36)*exp(x)+6*x**2+58*x+36)*ln(-1/5*exp(x)*x**2+1/5*x**3+1/5*x* 
*2)-9*exp(x)+9*x+9)/((exp(x)*x**2-x**3-x**2)*ln(-1/5*exp(x)*x**2+1/5*x**3+ 
1/5*x**2)**4+(2*exp(x)*x**2-2*x**3-2*x**2)*ln(-1/5*exp(x)*x**2+1/5*x**3+1/ 
5*x**2)**2+exp(x)*x**2-x**3-x**2),x)
 

Output:

(x + 9)/(x*log(x**3/5 - x**2*exp(x)/5 + x**2/5)**2 + x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (27) = 54\).

Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.03 \[ \int \frac {9-9 e^x+9 x+\left (36+58 x+6 x^2+e^x \left (-36-22 x-2 x^2\right )\right ) \log \left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (9-9 e^x+9 x\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )}{-x^2+e^x x^2-x^3+\left (-2 x^2+2 e^x x^2-2 x^3\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (-x^2+e^x x^2-x^3\right ) \log ^4\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )} \, dx=\frac {x + 9}{x \log \left (x - e^{x} + 1\right )^{2} - 4 \, x \log \left (5\right ) \log \left (x\right ) + 4 \, x \log \left (x\right )^{2} + {\left (\log \left (5\right )^{2} + 1\right )} x - 2 \, {\left (x \log \left (5\right ) - 2 \, x \log \left (x\right )\right )} \log \left (x - e^{x} + 1\right )} \] Input:

integrate(((-9*exp(x)+9*x+9)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+((-2*x 
^2-22*x-36)*exp(x)+6*x^2+58*x+36)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)-9*e 
xp(x)+9*x+9)/((exp(x)*x^2-x^3-x^2)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^4+ 
(2*exp(x)*x^2-2*x^3-2*x^2)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+exp(x)*x 
^2-x^3-x^2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 22.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {9-9 e^x+9 x+\left (36+58 x+6 x^2+e^x \left (-36-22 x-2 x^2\right )\right ) \log \left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (9-9 e^x+9 x\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )}{-x^2+e^x x^2-x^3+\left (-2 x^2+2 e^x x^2-2 x^3\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (-x^2+e^x x^2-x^3\right ) \log ^4\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )} \, dx=\frac {x + 9}{x \log \left (\frac {1}{5} \, x^{3} - \frac {1}{5} \, x^{2} e^{x} + \frac {1}{5} \, x^{2}\right )^{2} + x} \] Input:

integrate(((-9*exp(x)+9*x+9)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+((-2*x 
^2-22*x-36)*exp(x)+6*x^2+58*x+36)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)-9*e 
xp(x)+9*x+9)/((exp(x)*x^2-x^3-x^2)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^4+ 
(2*exp(x)*x^2-2*x^3-2*x^2)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+exp(x)*x 
^2-x^3-x^2),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.86 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {9-9 e^x+9 x+\left (36+58 x+6 x^2+e^x \left (-36-22 x-2 x^2\right )\right ) \log \left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (9-9 e^x+9 x\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )}{-x^2+e^x x^2-x^3+\left (-2 x^2+2 e^x x^2-2 x^3\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (-x^2+e^x x^2-x^3\right ) \log ^4\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )} \, dx=\frac {\left (x+9\right )\,\left (x-x\,{\mathrm {e}}^x+x^2\right )}{x^2\,\left ({\ln \left (\frac {x^2}{5}-\frac {x^2\,{\mathrm {e}}^x}{5}+\frac {x^3}{5}\right )}^2+1\right )\,\left (x-{\mathrm {e}}^x+1\right )} \] Input:

int(-(9*x - 9*exp(x) + log(x^2/5 - (x^2*exp(x))/5 + x^3/5)^2*(9*x - 9*exp( 
x) + 9) + log(x^2/5 - (x^2*exp(x))/5 + x^3/5)*(58*x - exp(x)*(22*x + 2*x^2 
 + 36) + 6*x^2 + 36) + 9)/(log(x^2/5 - (x^2*exp(x))/5 + x^3/5)^4*(x^2 - x^ 
2*exp(x) + x^3) - x^2*exp(x) + x^2 + x^3 + log(x^2/5 - (x^2*exp(x))/5 + x^ 
3/5)^2*(2*x^2 - 2*x^2*exp(x) + 2*x^3)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int \frac {9-9 e^x+9 x+\left (36+58 x+6 x^2+e^x \left (-36-22 x-2 x^2\right )\right ) \log \left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (9-9 e^x+9 x\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )}{-x^2+e^x x^2-x^3+\left (-2 x^2+2 e^x x^2-2 x^3\right ) \log ^2\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )+\left (-x^2+e^x x^2-x^3\right ) \log ^4\left (\frac {1}{5} \left (x^2-e^x x^2+x^3\right )\right )} \, dx=\frac {-\mathrm {log}\left (-\frac {e^{x} x^{2}}{5}+\frac {x^{3}}{5}+\frac {x^{2}}{5}\right )^{2} x +9}{x \left (\mathrm {log}\left (-\frac {e^{x} x^{2}}{5}+\frac {x^{3}}{5}+\frac {x^{2}}{5}\right )^{2}+1\right )} \] Input:

int(((-9*exp(x)+9*x+9)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+((-2*x^2-22* 
x-36)*exp(x)+6*x^2+58*x+36)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)-9*exp(x)+ 
9*x+9)/((exp(x)*x^2-x^3-x^2)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^4+(2*exp 
(x)*x^2-2*x^3-2*x^2)*log(-1/5*exp(x)*x^2+1/5*x^3+1/5*x^2)^2+exp(x)*x^2-x^3 
-x^2),x)
 

Output:

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