\(\int \frac {-972-231 x+3 x^2+e^{3 e^3} (-3 x+3 x^2)+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} (-72 x+20 x^2+2 x^3)+(-288+152 x-12 x^2-2 x^3+e^{3 e^3} (8 x-2 x^2)) \log (x)+(16-8 x+x^2) \log ^2(x)} \, dx\) [444]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 143, antiderivative size = 33 \[ \int \frac {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx=2-\frac {3 x}{4-x-\frac {\left (5+e^{3 e^3}\right ) x}{9+x-\log (x)}} \] Output:

2-3*x/(4-x-x/(x+9-ln(x))*(5+exp(3*exp(3))))
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx=\frac {3 \left (36-\left (1+e^{3 e^3}\right ) x-4 \log (x)\right )}{-36+\left (10+e^{3 e^3}\right ) x+x^2-(-4+x) \log (x)} \] Input:

Integrate[(-972 - 231*x + 3*x^2 + E^(3*E^3)*(-3*x + 3*x^2) + (216 + 24*x)* 
Log[x] - 12*Log[x]^2)/(1296 - 720*x + 28*x^2 + E^(6*E^3)*x^2 + 20*x^3 + x^ 
4 + E^(3*E^3)*(-72*x + 20*x^2 + 2*x^3) + (-288 + 152*x - 12*x^2 - 2*x^3 + 
E^(3*E^3)*(8*x - 2*x^2))*Log[x] + (16 - 8*x + x^2)*Log[x]^2),x]
 

Output:

(3*(36 - (1 + E^(3*E^3))*x - 4*Log[x]))/(-36 + (10 + E^(3*E^3))*x + x^2 - 
(-4 + x)*Log[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.

\(\displaystyle \int \frac {3 x^2+e^{3 e^3} \left (3 x^2-3 x\right )-231 x-12 \log ^2(x)+(24 x+216) \log (x)-972}{x^4+20 x^3+e^{6 e^3} x^2+28 x^2+\left (x^2-8 x+16\right ) \log ^2(x)+e^{3 e^3} \left (2 x^3+20 x^2-72 x\right )+\left (-2 x^3-12 x^2+e^{3 e^3} \left (8 x-2 x^2\right )+152 x-288\right ) \log (x)-720 x+1296} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {3 x^2+e^{3 e^3} \left (3 x^2-3 x\right )-231 x-12 \log ^2(x)+(24 x+216) \log (x)-972}{x^4+20 x^3+\left (28+e^{6 e^3}\right ) x^2+\left (x^2-8 x+16\right ) \log ^2(x)+e^{3 e^3} \left (2 x^3+20 x^2-72 x\right )+\left (-2 x^3-12 x^2+e^{3 e^3} \left (8 x-2 x^2\right )+152 x-288\right ) \log (x)-720 x+1296}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {3 \left (\left (1+e^{3 e^3}\right ) x^2-\left (77+e^{3 e^3}\right ) x-4 \log ^2(x)+8 (x+9) \log (x)-324\right )}{\left (-x^2-\left (10+e^{3 e^3}\right ) x+(x-4) \log (x)+36\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 3 \int -\frac {-\left (\left (1+e^{3 e^3}\right ) x^2\right )+\left (77+e^{3 e^3}\right ) x+4 \log ^2(x)-8 (x+9) \log (x)+324}{\left (-x^2-\left (10+e^{3 e^3}\right ) x-(4-x) \log (x)+36\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -3 \int \frac {-\left (\left (1+e^{3 e^3}\right ) x^2\right )+\left (77+e^{3 e^3}\right ) x+4 \log ^2(x)-8 (x+9) \log (x)+324}{\left (-x^2-\left (10+e^{3 e^3}\right ) x-(4-x) \log (x)+36\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -3 \int \left (\frac {8 \left (5+e^{3 e^3}\right ) x}{(4-x)^2 \left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )}+\frac {\left (5+e^{3 e^3}\right ) \left (-x^3+9 x^2-4 \left (1-e^{3 e^3}\right ) x+16\right ) x}{(4-x)^2 \left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )^2}+\frac {4}{(x-4)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -3 \left (4 \left (5+e^{3 e^3}\right )^2 \int \frac {1}{\left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )^2}dx+64 \left (5+e^{3 e^3}\right )^2 \int \frac {1}{(4-x)^2 \left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )^2}dx-32 \left (5+e^{3 e^3}\right )^2 \int \frac {1}{(4-x) \left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )^2}dx+\left (5+e^{3 e^3}\right ) \int \frac {x}{\left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )^2}dx-\left (5+e^{3 e^3}\right ) \int \frac {x^2}{\left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )^2}dx+32 \left (5+e^{3 e^3}\right ) \int \frac {1}{(4-x)^2 \left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )}dx+8 \left (5+e^{3 e^3}\right ) \int \frac {1}{(x-4) \left (-x^2+\log (x) x-10 \left (1+\frac {e^{3 e^3}}{10}\right ) x-4 \log (x)+36\right )}dx+\frac {4}{4-x}\right )\)

Input:

Int[(-972 - 231*x + 3*x^2 + E^(3*E^3)*(-3*x + 3*x^2) + (216 + 24*x)*Log[x] 
 - 12*Log[x]^2)/(1296 - 720*x + 28*x^2 + E^(6*E^3)*x^2 + 20*x^3 + x^4 + E^ 
(3*E^3)*(-72*x + 20*x^2 + 2*x^3) + (-288 + 152*x - 12*x^2 - 2*x^3 + E^(3*E 
^3)*(8*x - 2*x^2))*Log[x] + (16 - 8*x + x^2)*Log[x]^2),x]
 

Output:

$Aborted
 
Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36

method result size
norman \(\frac {-12 \ln \left (x \right )+\left (-3 \,{\mathrm e}^{3 \,{\mathrm e}^{3}}-3\right ) x +108}{{\mathrm e}^{3 \,{\mathrm e}^{3}} x -x \ln \left (x \right )+x^{2}+4 \ln \left (x \right )+10 x -36}\) \(45\)
parallelrisch \(\frac {108-3 x -3 \,{\mathrm e}^{3 \,{\mathrm e}^{3}} x -12 \ln \left (x \right )}{{\mathrm e}^{3 \,{\mathrm e}^{3}} x -x \ln \left (x \right )+x^{2}+4 \ln \left (x \right )+10 x -36}\) \(45\)
default \(\frac {-12 \ln \left (x \right )+3 \left (-{\mathrm e}^{3 \,{\mathrm e}^{3}}-1\right ) x +108}{{\mathrm e}^{3 \,{\mathrm e}^{3}} x -x \ln \left (x \right )+x^{2}+4 \ln \left (x \right )+10 x -36}\) \(46\)
risch \(\frac {12}{x -4}-\frac {3 \left (5+{\mathrm e}^{3 \,{\mathrm e}^{3}}\right ) x^{2}}{\left (x -4\right ) \left ({\mathrm e}^{3 \,{\mathrm e}^{3}} x -x \ln \left (x \right )+x^{2}+4 \ln \left (x \right )+10 x -36\right )}\) \(52\)

Input:

int((-12*ln(x)^2+(24*x+216)*ln(x)+(3*x^2-3*x)*exp(3*exp(3))+3*x^2-231*x-97 
2)/((x^2-8*x+16)*ln(x)^2+((-2*x^2+8*x)*exp(3*exp(3))-2*x^3-12*x^2+152*x-28 
8)*ln(x)+x^2*exp(3*exp(3))^2+(2*x^3+20*x^2-72*x)*exp(3*exp(3))+x^4+20*x^3+ 
28*x^2-720*x+1296),x,method=_RETURNVERBOSE)
 

Output:

(-12*ln(x)+(-3*exp(exp(3))^3-3)*x+108)/(exp(3*exp(3))*x-x*ln(x)+x^2+4*ln(x 
)+10*x-36)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx=-\frac {3 \, {\left (x e^{\left (3 \, e^{3}\right )} + x + 4 \, \log \left (x\right ) - 36\right )}}{x^{2} + x e^{\left (3 \, e^{3}\right )} - {\left (x - 4\right )} \log \left (x\right ) + 10 \, x - 36} \] Input:

integrate((-12*log(x)^2+(24*x+216)*log(x)+(3*x^2-3*x)*exp(3*exp(3))+3*x^2- 
231*x-972)/((x^2-8*x+16)*log(x)^2+((-2*x^2+8*x)*exp(3*exp(3))-2*x^3-12*x^2 
+152*x-288)*log(x)+x^2*exp(3*exp(3))^2+(2*x^3+20*x^2-72*x)*exp(3*exp(3))+x 
^4+20*x^3+28*x^2-720*x+1296),x, algorithm="fricas")
 

Output:

-3*(x*e^(3*e^3) + x + 4*log(x) - 36)/(x^2 + x*e^(3*e^3) - (x - 4)*log(x) + 
 10*x - 36)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (24) = 48\).

Time = 0.18 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int \frac {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx=\frac {15 x^{2} + 3 x^{2} e^{3 e^{3}}}{- x^{3} - x^{2} e^{3 e^{3}} - 6 x^{2} + 76 x + 4 x e^{3 e^{3}} + \left (x^{2} - 8 x + 16\right ) \log {\left (x \right )} - 144} + \frac {12}{x - 4} \] Input:

integrate((-12*ln(x)**2+(24*x+216)*ln(x)+(3*x**2-3*x)*exp(3*exp(3))+3*x**2 
-231*x-972)/((x**2-8*x+16)*ln(x)**2+((-2*x**2+8*x)*exp(3*exp(3))-2*x**3-12 
*x**2+152*x-288)*ln(x)+x**2*exp(3*exp(3))**2+(2*x**3+20*x**2-72*x)*exp(3*e 
xp(3))+x**4+20*x**3+28*x**2-720*x+1296),x)
 

Output:

(15*x**2 + 3*x**2*exp(3*exp(3)))/(-x**3 - x**2*exp(3*exp(3)) - 6*x**2 + 76 
*x + 4*x*exp(3*exp(3)) + (x**2 - 8*x + 16)*log(x) - 144) + 12/(x - 4)
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx=-\frac {3 \, {\left (x {\left (e^{\left (3 \, e^{3}\right )} + 1\right )} + 4 \, \log \left (x\right ) - 36\right )}}{x^{2} + x {\left (e^{\left (3 \, e^{3}\right )} + 10\right )} - {\left (x - 4\right )} \log \left (x\right ) - 36} \] Input:

integrate((-12*log(x)^2+(24*x+216)*log(x)+(3*x^2-3*x)*exp(3*exp(3))+3*x^2- 
231*x-972)/((x^2-8*x+16)*log(x)^2+((-2*x^2+8*x)*exp(3*exp(3))-2*x^3-12*x^2 
+152*x-288)*log(x)+x^2*exp(3*exp(3))^2+(2*x^3+20*x^2-72*x)*exp(3*exp(3))+x 
^4+20*x^3+28*x^2-720*x+1296),x, algorithm="maxima")
 

Output:

-3*(x*(e^(3*e^3) + 1) + 4*log(x) - 36)/(x^2 + x*(e^(3*e^3) + 10) - (x - 4) 
*log(x) - 36)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx=-\frac {3 \, {\left (x e^{\left (3 \, e^{3}\right )} + x + 4 \, \log \left (x\right ) - 36\right )}}{x^{2} + x e^{\left (3 \, e^{3}\right )} - x \log \left (x\right ) + 10 \, x + 4 \, \log \left (x\right ) - 36} \] Input:

integrate((-12*log(x)^2+(24*x+216)*log(x)+(3*x^2-3*x)*exp(3*exp(3))+3*x^2- 
231*x-972)/((x^2-8*x+16)*log(x)^2+((-2*x^2+8*x)*exp(3*exp(3))-2*x^3-12*x^2 
+152*x-288)*log(x)+x^2*exp(3*exp(3))^2+(2*x^3+20*x^2-72*x)*exp(3*exp(3))+x 
^4+20*x^3+28*x^2-720*x+1296),x, algorithm="giac")
 

Output:

-3*(x*e^(3*e^3) + x + 4*log(x) - 36)/(x^2 + x*e^(3*e^3) - x*log(x) + 10*x 
+ 4*log(x) - 36)
 

Mupad [B] (verification not implemented)

Time = 4.52 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx=-\frac {12\,\ln \left (x\right )+x\,\left (3\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}+3\right )-108}{10\,x+4\,\ln \left (x\right )+x\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}-x\,\ln \left (x\right )+x^2-36} \] Input:

int(-(231*x + 12*log(x)^2 - log(x)*(24*x + 216) - 3*x^2 + exp(3*exp(3))*(3 
*x - 3*x^2) + 972)/(exp(3*exp(3))*(20*x^2 - 72*x + 2*x^3) - 720*x + x^2*ex 
p(6*exp(3)) + log(x)^2*(x^2 - 8*x + 16) + 28*x^2 + 20*x^3 + x^4 - log(x)*( 
12*x^2 - 152*x + 2*x^3 - exp(3*exp(3))*(8*x - 2*x^2) + 288) + 1296),x)
 

Output:

-(12*log(x) + x*(3*exp(3*exp(3)) + 3) - 108)/(10*x + 4*log(x) + x*exp(3*ex 
p(3)) - x*log(x) + x^2 - 36)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {-972-231 x+3 x^2+e^{3 e^3} \left (-3 x+3 x^2\right )+(216+24 x) \log (x)-12 \log ^2(x)}{1296-720 x+28 x^2+e^{6 e^3} x^2+20 x^3+x^4+e^{3 e^3} \left (-72 x+20 x^2+2 x^3\right )+\left (-288+152 x-12 x^2-2 x^3+e^{3 e^3} \left (8 x-2 x^2\right )\right ) \log (x)+\left (16-8 x+x^2\right ) \log ^2(x)} \, dx=\frac {3 x \left (-\mathrm {log}\left (x \right )+x +9\right )}{e^{3 e^{3}} x -\mathrm {log}\left (x \right ) x +4 \,\mathrm {log}\left (x \right )+x^{2}+10 x -36} \] Input:

int((-12*log(x)^2+(24*x+216)*log(x)+(3*x^2-3*x)*exp(3*exp(3))+3*x^2-231*x- 
972)/((x^2-8*x+16)*log(x)^2+((-2*x^2+8*x)*exp(3*exp(3))-2*x^3-12*x^2+152*x 
-288)*log(x)+x^2*exp(3*exp(3))^2+(2*x^3+20*x^2-72*x)*exp(3*exp(3))+x^4+20* 
x^3+28*x^2-720*x+1296),x)
 

Output:

(3*x*( - log(x) + x + 9))/(e**(3*e**3)*x - log(x)*x + 4*log(x) + x**2 + 10 
*x - 36)