\(\int \frac {e^5 (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6)+(e^5 (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6)+e^5 (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5) \log (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2})) \log (x+\log (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}))}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+(-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7) \log (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2})} \, dx\) [2140]

Optimal result

Integrand size = 237, antiderivative size = 28 \[ \int \frac {e^5 \left (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6\right )+\left (e^5 \left (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6\right )+e^5 \left (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right ) \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+\left (-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )} \, dx=\frac {e^5 \left (x-\log \left (x+\log \left (5-4 \left (x+\frac {1}{-5+x^2}\right )\right )\right )\right )}{x} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^5 \left (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6\right )+\left (e^5 \left (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6\right )+e^5 \left (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right ) \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+\left (-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )} \, dx=-\frac {e^5 \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{x} \] Input:

Integrate[(E^5*(45*x - 92*x^2 - 14*x^3 + 40*x^4 + x^5 - 4*x^6) + (E^5*(-14 
5*x + 100*x^2 + 54*x^3 - 40*x^4 - 5*x^5 + 4*x^6) + E^5*(-145 + 100*x + 54* 
x^2 - 40*x^3 - 5*x^4 + 4*x^5)*Log[(-29 + 20*x + 5*x^2 - 4*x^3)/(-5 + x^2)] 
)*Log[x + Log[(-29 + 20*x + 5*x^2 - 4*x^3)/(-5 + x^2)]])/(-145*x^3 + 100*x 
^4 + 54*x^5 - 40*x^6 - 5*x^7 + 4*x^8 + (-145*x^2 + 100*x^3 + 54*x^4 - 40*x 
^5 - 5*x^6 + 4*x^7)*Log[(-29 + 20*x + 5*x^2 - 4*x^3)/(-5 + x^2)]),x]
 

Output:

-((E^5*Log[x + Log[(-29 + 20*x + 5*x^2 - 4*x^3)/(-5 + x^2)]])/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.

\(\displaystyle \int \frac {e^5 \left (-4 x^6+x^5+40 x^4-14 x^3-92 x^2+45 x\right )+\left (e^5 \left (4 x^5-5 x^4-40 x^3+54 x^2+100 x-145\right ) \log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )+e^5 \left (4 x^6-5 x^5-40 x^4+54 x^3+100 x^2-145 x\right )\right ) \log \left (\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )+x\right )}{4 x^8-5 x^7-40 x^6+54 x^5+100 x^4-145 x^3+\left (4 x^7-5 x^6-40 x^5+54 x^4+100 x^3-145 x^2\right ) \log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {e^5 \left (\log \left (\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )+x\right )+\frac {x \left (-4 x^5+x^4+40 x^3-14 x^2-92 x+45\right )}{\left (4 x^5-5 x^4-40 x^3+54 x^2+100 x-145\right ) \left (\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )+x\right )}\right )}{x^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle e^5 \int -\frac {\frac {x \left (-4 x^5+x^4+40 x^3-14 x^2-92 x+45\right )}{\left (-4 x^5+5 x^4+40 x^3-54 x^2-100 x+145\right ) \left (x+\log \left (\frac {4 x^3-5 x^2-20 x+29}{5-x^2}\right )\right )}-\log \left (x+\log \left (\frac {4 x^3-5 x^2-20 x+29}{5-x^2}\right )\right )}{x^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -e^5 \int \frac {\frac {x \left (-4 x^5+x^4+40 x^3-14 x^2-92 x+45\right )}{\left (-4 x^5+5 x^4+40 x^3-54 x^2-100 x+145\right ) \left (x+\log \left (\frac {4 x^3-5 x^2-20 x+29}{5-x^2}\right )\right )}-\log \left (x+\log \left (\frac {4 x^3-5 x^2-20 x+29}{5-x^2}\right )\right )}{x^2}dx\)

\(\Big \downarrow \) 2010

\(\displaystyle -e^5 \int \left (\frac {4 x^5-x^4-40 x^3+14 x^2+92 x-45}{x \left (x^2-5\right ) \left (4 x^3-5 x^2-20 x+29\right ) \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}-\frac {\log \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}{x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -e^5 \left (\frac {\int \frac {1}{\left (\sqrt {5}-x\right ) \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}dx}{\sqrt {5}}+\frac {9}{29} \int \frac {1}{x \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}dx+\frac {\int \frac {1}{\left (x+\sqrt {5}\right ) \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}dx}{\sqrt {5}}-\frac {690}{29} \int \frac {1}{\left (4 x^3-5 x^2-20 x+29\right ) \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}dx+\frac {248}{29} \int \frac {x}{\left (4 x^3-5 x^2-20 x+29\right ) \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}dx+\frac {80}{29} \int \frac {x^2}{\left (4 x^3-5 x^2-20 x+29\right ) \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}dx-\int \frac {\log \left (x+\log \left (\frac {-4 x^3+5 x^2+20 x-29}{x^2-5}\right )\right )}{x^2}dx\right )\)

Input:

Int[(E^5*(45*x - 92*x^2 - 14*x^3 + 40*x^4 + x^5 - 4*x^6) + (E^5*(-145*x + 
100*x^2 + 54*x^3 - 40*x^4 - 5*x^5 + 4*x^6) + E^5*(-145 + 100*x + 54*x^2 - 
40*x^3 - 5*x^4 + 4*x^5)*Log[(-29 + 20*x + 5*x^2 - 4*x^3)/(-5 + x^2)])*Log[ 
x + Log[(-29 + 20*x + 5*x^2 - 4*x^3)/(-5 + x^2)]])/(-145*x^3 + 100*x^4 + 5 
4*x^5 - 40*x^6 - 5*x^7 + 4*x^8 + (-145*x^2 + 100*x^3 + 54*x^4 - 40*x^5 - 5 
*x^6 + 4*x^7)*Log[(-29 + 20*x + 5*x^2 - 4*x^3)/(-5 + x^2)]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 36.66 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\(-\frac {{\mathrm e}^{5} \ln \left (\ln \left (-\frac {4 x^{3}-5 x^{2}-20 x +29}{x^{2}-5}\right )+x \right )}{x}\)	\(36\)

Input:

int((((4*x^5-5*x^4-40*x^3+54*x^2+100*x-145)*exp(5)*ln((-4*x^3+5*x^2+20*x-2 
9)/(x^2-5))+(4*x^6-5*x^5-40*x^4+54*x^3+100*x^2-145*x)*exp(5))*ln(ln((-4*x^ 
3+5*x^2+20*x-29)/(x^2-5))+x)+(-4*x^6+x^5+40*x^4-14*x^3-92*x^2+45*x)*exp(5) 
)/((4*x^7-5*x^6-40*x^5+54*x^4+100*x^3-145*x^2)*ln((-4*x^3+5*x^2+20*x-29)/( 
x^2-5))+4*x^8-5*x^7-40*x^6+54*x^5+100*x^4-145*x^3),x,method=_RETURNVERBOSE 
)
 

Output:

-exp(5)*ln(ln(-(4*x^3-5*x^2-20*x+29)/(x^2-5))+x)/x
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^5 \left (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6\right )+\left (e^5 \left (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6\right )+e^5 \left (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right ) \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+\left (-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )} \, dx=-\frac {e^{5} \log \left (x + \log \left (-\frac {4 \, x^{3} - 5 \, x^{2} - 20 \, x + 29}{x^{2} - 5}\right )\right )}{x} \] Input:

integrate((((4*x^5-5*x^4-40*x^3+54*x^2+100*x-145)*exp(5)*log((-4*x^3+5*x^2 
+20*x-29)/(x^2-5))+(4*x^6-5*x^5-40*x^4+54*x^3+100*x^2-145*x)*exp(5))*log(l 
og((-4*x^3+5*x^2+20*x-29)/(x^2-5))+x)+(-4*x^6+x^5+40*x^4-14*x^3-92*x^2+45* 
x)*exp(5))/((4*x^7-5*x^6-40*x^5+54*x^4+100*x^3-145*x^2)*log((-4*x^3+5*x^2+ 
20*x-29)/(x^2-5))+4*x^8-5*x^7-40*x^6+54*x^5+100*x^4-145*x^3),x, algorithm= 
"fricas")
 

Output:

-e^5*log(x + log(-(4*x^3 - 5*x^2 - 20*x + 29)/(x^2 - 5)))/x
 

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^5 \left (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6\right )+\left (e^5 \left (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6\right )+e^5 \left (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right ) \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+\left (-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )} \, dx=- \frac {e^{5} \log {\left (x + \log {\left (\frac {- 4 x^{3} + 5 x^{2} + 20 x - 29}{x^{2} - 5} \right )} \right )}}{x} \] Input:

integrate((((4*x**5-5*x**4-40*x**3+54*x**2+100*x-145)*exp(5)*ln((-4*x**3+5 
*x**2+20*x-29)/(x**2-5))+(4*x**6-5*x**5-40*x**4+54*x**3+100*x**2-145*x)*ex 
p(5))*ln(ln((-4*x**3+5*x**2+20*x-29)/(x**2-5))+x)+(-4*x**6+x**5+40*x**4-14 
*x**3-92*x**2+45*x)*exp(5))/((4*x**7-5*x**6-40*x**5+54*x**4+100*x**3-145*x 
**2)*ln((-4*x**3+5*x**2+20*x-29)/(x**2-5))+4*x**8-5*x**7-40*x**6+54*x**5+1 
00*x**4-145*x**3),x)
 

Output:

-exp(5)*log(x + log((-4*x**3 + 5*x**2 + 20*x - 29)/(x**2 - 5)))/x
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^5 \left (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6\right )+\left (e^5 \left (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6\right )+e^5 \left (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right ) \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+\left (-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )} \, dx=-\frac {e^{5} \log \left (x + \log \left (-4 \, x^{3} + 5 \, x^{2} + 20 \, x - 29\right ) - \log \left (x^{2} - 5\right )\right )}{x} \] Input:

integrate((((4*x^5-5*x^4-40*x^3+54*x^2+100*x-145)*exp(5)*log((-4*x^3+5*x^2 
+20*x-29)/(x^2-5))+(4*x^6-5*x^5-40*x^4+54*x^3+100*x^2-145*x)*exp(5))*log(l 
og((-4*x^3+5*x^2+20*x-29)/(x^2-5))+x)+(-4*x^6+x^5+40*x^4-14*x^3-92*x^2+45* 
x)*exp(5))/((4*x^7-5*x^6-40*x^5+54*x^4+100*x^3-145*x^2)*log((-4*x^3+5*x^2+ 
20*x-29)/(x^2-5))+4*x^8-5*x^7-40*x^6+54*x^5+100*x^4-145*x^3),x, algorithm= 
"maxima")
 

Output:

-e^5*log(x + log(-4*x^3 + 5*x^2 + 20*x - 29) - log(x^2 - 5))/x
 

Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^5 \left (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6\right )+\left (e^5 \left (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6\right )+e^5 \left (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right ) \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+\left (-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )} \, dx=-\frac {e^{5} \log \left (x + \log \left (-\frac {4 \, x^{3} - 5 \, x^{2} - 20 \, x + 29}{x^{2} - 5}\right )\right )}{x} \] Input:

integrate((((4*x^5-5*x^4-40*x^3+54*x^2+100*x-145)*exp(5)*log((-4*x^3+5*x^2 
+20*x-29)/(x^2-5))+(4*x^6-5*x^5-40*x^4+54*x^3+100*x^2-145*x)*exp(5))*log(l 
og((-4*x^3+5*x^2+20*x-29)/(x^2-5))+x)+(-4*x^6+x^5+40*x^4-14*x^3-92*x^2+45* 
x)*exp(5))/((4*x^7-5*x^6-40*x^5+54*x^4+100*x^3-145*x^2)*log((-4*x^3+5*x^2+ 
20*x-29)/(x^2-5))+4*x^8-5*x^7-40*x^6+54*x^5+100*x^4-145*x^3),x, algorithm= 
"giac")
 

Output:

-e^5*log(x + log(-(4*x^3 - 5*x^2 - 20*x + 29)/(x^2 - 5)))/x
 

Mupad [B] (verification not implemented)

Time = 3.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^5 \left (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6\right )+\left (e^5 \left (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6\right )+e^5 \left (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right ) \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+\left (-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )} \, dx=-\frac {\ln \left (x+\ln \left (\frac {-4\,x^3+5\,x^2+20\,x-29}{x^2-5}\right )\right )\,{\mathrm {e}}^5}{x} \] Input:

int(-(exp(5)*(45*x - 92*x^2 - 14*x^3 + 40*x^4 + x^5 - 4*x^6) - log(x + log 
((20*x + 5*x^2 - 4*x^3 - 29)/(x^2 - 5)))*(exp(5)*(145*x - 100*x^2 - 54*x^3 
 + 40*x^4 + 5*x^5 - 4*x^6) - log((20*x + 5*x^2 - 4*x^3 - 29)/(x^2 - 5))*ex 
p(5)*(100*x + 54*x^2 - 40*x^3 - 5*x^4 + 4*x^5 - 145)))/(log((20*x + 5*x^2 
- 4*x^3 - 29)/(x^2 - 5))*(145*x^2 - 100*x^3 - 54*x^4 + 40*x^5 + 5*x^6 - 4* 
x^7) + 145*x^3 - 100*x^4 - 54*x^5 + 40*x^6 + 5*x^7 - 4*x^8),x)
 

Output:

-(log(x + log((20*x + 5*x^2 - 4*x^3 - 29)/(x^2 - 5)))*exp(5))/x
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^5 \left (45 x-92 x^2-14 x^3+40 x^4+x^5-4 x^6\right )+\left (e^5 \left (-145 x+100 x^2+54 x^3-40 x^4-5 x^5+4 x^6\right )+e^5 \left (-145+100 x+54 x^2-40 x^3-5 x^4+4 x^5\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right ) \log \left (x+\log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )\right )}{-145 x^3+100 x^4+54 x^5-40 x^6-5 x^7+4 x^8+\left (-145 x^2+100 x^3+54 x^4-40 x^5-5 x^6+4 x^7\right ) \log \left (\frac {-29+20 x+5 x^2-4 x^3}{-5+x^2}\right )} \, dx=-\frac {\mathrm {log}\left (\mathrm {log}\left (\frac {-4 x^{3}+5 x^{2}+20 x -29}{x^{2}-5}\right )+x \right ) e^{5}}{x} \] Input:

int((((4*x^5-5*x^4-40*x^3+54*x^2+100*x-145)*exp(5)*log((-4*x^3+5*x^2+20*x- 
29)/(x^2-5))+(4*x^6-5*x^5-40*x^4+54*x^3+100*x^2-145*x)*exp(5))*log(log((-4 
*x^3+5*x^2+20*x-29)/(x^2-5))+x)+(-4*x^6+x^5+40*x^4-14*x^3-92*x^2+45*x)*exp 
(5))/((4*x^7-5*x^6-40*x^5+54*x^4+100*x^3-145*x^2)*log((-4*x^3+5*x^2+20*x-2 
9)/(x^2-5))+4*x^8-5*x^7-40*x^6+54*x^5+100*x^4-145*x^3),x)
 

Output:

( - log(log(( - 4*x**3 + 5*x**2 + 20*x - 29)/(x**2 - 5)) + x)*e**5)/x