\(\int \frac {-10 x+e^{3/4} x-x^4+(50-50 x^3-x^6+e^{3/4} (-5+4 x^3)) \log (x)+(-20 x+2 e^{3/4} x+x^4) \log (x) \log (\log (x))}{(25 x^2-10 x^5+x^8) \log (x)+(-10 x^3+2 x^6) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx\) [221]

Optimal result

Integrand size = 114, antiderivative size = 30 \[ \int \frac {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx=\frac {\frac {10-e^{3/4}}{x^2}+x}{-\frac {5}{x}+x^2+\log (\log (x))} \] Output:

(x+(10-exp(3/4))/x^2)/(ln(ln(x))+x^2-5/x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx=\frac {10-e^{3/4}+x^3}{x \left (-5+x^3+x \log (\log (x))\right )} \] Input:

Integrate[(-10*x + E^(3/4)*x - x^4 + (50 - 50*x^3 - x^6 + E^(3/4)*(-5 + 4* 
x^3))*Log[x] + (-20*x + 2*E^(3/4)*x + x^4)*Log[x]*Log[Log[x]])/((25*x^2 - 
10*x^5 + x^8)*Log[x] + (-10*x^3 + 2*x^6)*Log[x]*Log[Log[x]] + x^4*Log[x]*L 
og[Log[x]]^2),x]
 

Output:

(10 - E^(3/4) + x^3)/(x*(-5 + x^3 + x*Log[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.

\(\displaystyle \int \frac {-x^4+\left (x^4+2 e^{3/4} x-20 x\right ) \log (x) \log (\log (x))+\left (-x^6-50 x^3+e^{3/4} \left (4 x^3-5\right )+50\right ) \log (x)+e^{3/4} x-10 x}{x^4 \log (x) \log ^2(\log (x))+\left (2 x^6-10 x^3\right ) \log (x) \log (\log (x))+\left (x^8-10 x^5+25 x^2\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {-x^4+\left (x^4+2 e^{3/4} x-20 x\right ) \log (x) \log (\log (x))+\left (-x^6-50 x^3+e^{3/4} \left (4 x^3-5\right )+50\right ) \log (x)+\left (e^{3/4}-10\right ) x}{x^4 \log (x) \log ^2(\log (x))+\left (2 x^6-10 x^3\right ) \log (x) \log (\log (x))+\left (x^8-10 x^5+25 x^2\right ) \log (x)}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-x^4+\left (x^4+2 e^{3/4} x-20 x\right ) \log (x) \log (\log (x))+\left (-x^6-50 x^3+e^{3/4} \left (4 x^3-5\right )+50\right ) \log (x)+\left (e^{3/4}-10\right ) x}{x^2 \log (x) \left (-x^3-x \log (\log (x))+5\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^3+2 e^{3/4}-20}{x^2 \left (x^3+x \log (\log (x))-5\right )}-\frac {\left (x^3-e^{3/4}+10\right ) \left (2 x^3 \log (x)+x+5 \log (x)\right )}{x^2 \log (x) \left (x^3+x \log (\log (x))-5\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(-10*x + E^(3/4)*x - x^4 + (50 - 50*x^3 - x^6 + E^(3/4)*(-5 + 4*x^3))* 
Log[x] + (-20*x + 2*E^(3/4)*x + x^4)*Log[x]*Log[Log[x]])/((25*x^2 - 10*x^5 
 + x^8)*Log[x] + (-10*x^3 + 2*x^6)*Log[x]*Log[Log[x]] + x^4*Log[x]*Log[Log 
[x]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 4.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {-x^{3}+{\mathrm e}^{\frac {3}{4}}-10}{x \left (x^{3}+x \ln \left (\ln \left (x \right )\right )-5\right )}\)	\(27\)
parallelrisch	\(-\frac {-x^{3}+{\mathrm e}^{\frac {3}{4}}-10}{x \left (x^{3}+x \ln \left (\ln \left (x \right )\right )-5\right )}\)	\(27\)
default	\(-\frac {\ln \left (x \right ) \left (-2 x^{6}+2 \,{\mathrm e}^{\frac {3}{4}} x^{3}-25 x^{3}+5 \,{\mathrm e}^{\frac {3}{4}}-50\right )}{\left (2 x^{3} \ln \left (x \right )+5 \ln \left (x \right )+x \right ) x \left (x^{3}+x \ln \left (\ln \left (x \right )\right )-5\right )}-\frac {-x^{3}+{\mathrm e}^{\frac {3}{4}}-10}{\left (2 x^{3} \ln \left (x \right )+5 \ln \left (x \right )+x \right ) \left (x^{3}+x \ln \left (\ln \left (x \right )\right )-5\right )}\)	\(97\)
parts	\(-\frac {\ln \left (x \right ) \left (-2 x^{6}+2 \,{\mathrm e}^{\frac {3}{4}} x^{3}-25 x^{3}+5 \,{\mathrm e}^{\frac {3}{4}}-50\right )}{\left (2 x^{3} \ln \left (x \right )+5 \ln \left (x \right )+x \right ) x \left (x^{3}+x \ln \left (\ln \left (x \right )\right )-5\right )}-\frac {-x^{3}+{\mathrm e}^{\frac {3}{4}}-10}{\left (2 x^{3} \ln \left (x \right )+5 \ln \left (x \right )+x \right ) \left (x^{3}+x \ln \left (\ln \left (x \right )\right )-5\right )}\)	\(97\)

Input:

int(((2*x*exp(3/4)+x^4-20*x)*ln(x)*ln(ln(x))+((4*x^3-5)*exp(3/4)-x^6-50*x^ 
3+50)*ln(x)+x*exp(3/4)-x^4-10*x)/(x^4*ln(x)*ln(ln(x))^2+(2*x^6-10*x^3)*ln( 
x)*ln(ln(x))+(x^8-10*x^5+25*x^2)*ln(x)),x,method=_RETURNVERBOSE)
 

Output:

-(-x^3+exp(3/4)-10)/x/(x^3+x*ln(ln(x))-5)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx=\frac {x^{3} - e^{\frac {3}{4}} + 10}{x^{4} + x^{2} \log \left (\log \left (x\right )\right ) - 5 \, x} \] Input:

integrate(((2*x*exp(3/4)+x^4-20*x)*log(x)*log(log(x))+((4*x^3-5)*exp(3/4)- 
x^6-50*x^3+50)*log(x)+x*exp(3/4)-x^4-10*x)/(x^4*log(x)*log(log(x))^2+(2*x^ 
6-10*x^3)*log(x)*log(log(x))+(x^8-10*x^5+25*x^2)*log(x)),x, algorithm="fri 
cas")
 

Output:

(x^3 - e^(3/4) + 10)/(x^4 + x^2*log(log(x)) - 5*x)
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx=\frac {x^{3} - e^{\frac {3}{4}} + 10}{x^{4} + x^{2} \log {\left (\log {\left (x \right )} \right )} - 5 x} \] Input:

integrate(((2*x*exp(3/4)+x**4-20*x)*ln(x)*ln(ln(x))+((4*x**3-5)*exp(3/4)-x 
**6-50*x**3+50)*ln(x)+x*exp(3/4)-x**4-10*x)/(x**4*ln(x)*ln(ln(x))**2+(2*x* 
*6-10*x**3)*ln(x)*ln(ln(x))+(x**8-10*x**5+25*x**2)*ln(x)),x)
 

Output:

(x**3 - exp(3/4) + 10)/(x**4 + x**2*log(log(x)) - 5*x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(((2*x*exp(3/4)+x^4-20*x)*log(x)*log(log(x))+((4*x^3-5)*exp(3/4)- 
x^6-50*x^3+50)*log(x)+x*exp(3/4)-x^4-10*x)/(x^4*log(x)*log(log(x))^2+(2*x^ 
6-10*x^3)*log(x)*log(log(x))+(x^8-10*x^5+25*x^2)*log(x)),x, algorithm="max 
ima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx=\frac {x^{3} - e^{\frac {3}{4}} + 10}{x^{4} + x^{2} \log \left (\log \left (x\right )\right ) - 5 \, x} \] Input:

integrate(((2*x*exp(3/4)+x^4-20*x)*log(x)*log(log(x))+((4*x^3-5)*exp(3/4)- 
x^6-50*x^3+50)*log(x)+x*exp(3/4)-x^4-10*x)/(x^4*log(x)*log(log(x))^2+(2*x^ 
6-10*x^3)*log(x)*log(log(x))+(x^8-10*x^5+25*x^2)*log(x)),x, algorithm="gia 
c")
 

Output:

(x^3 - e^(3/4) + 10)/(x^4 + x^2*log(log(x)) - 5*x)
 

Mupad [B] (verification not implemented)

Time = 2.86 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.37 \[ \int \frac {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx=\frac {x^4\,\ln \left (x\right )-5\,{\mathrm {e}}^{3/4}\,{\ln \left (x\right )}^2+x\,\left (10\,\ln \left (x\right )-{\mathrm {e}}^{3/4}\,\ln \left (x\right )\right )-x^3\,\left (2\,{\mathrm {e}}^{3/4}\,{\ln \left (x\right )}^2-25\,{\ln \left (x\right )}^2\right )+50\,{\ln \left (x\right )}^2+2\,x^6\,{\ln \left (x\right )}^2}{x\,\left (2\,x^3\,{\ln \left (x\right )}^2+x\,\ln \left (x\right )+5\,{\ln \left (x\right )}^2\right )\,\left (x\,\ln \left (\ln \left (x\right )\right )+x^3-5\right )} \] Input:

int((x*exp(3/4) - 10*x + log(x)*(exp(3/4)*(4*x^3 - 5) - 50*x^3 - x^6 + 50) 
 - x^4 + log(log(x))*log(x)*(2*x*exp(3/4) - 20*x + x^4))/(log(x)*(25*x^2 - 
 10*x^5 + x^8) - log(log(x))*log(x)*(10*x^3 - 2*x^6) + x^4*log(log(x))^2*l 
og(x)),x)
 

Output:

(x^4*log(x) - 5*exp(3/4)*log(x)^2 + x*(10*log(x) - exp(3/4)*log(x)) - x^3* 
(2*exp(3/4)*log(x)^2 - 25*log(x)^2) + 50*log(x)^2 + 2*x^6*log(x)^2)/(x*(5* 
log(x)^2 + 2*x^3*log(x)^2 + x*log(x))*(x*log(log(x)) + x^3 - 5))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx=\frac {-e^{\frac {3}{4}}+x^{3}+10}{x \left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x +x^{3}-5\right )} \] Input:

int(((2*x*exp(3/4)+x^4-20*x)*log(x)*log(log(x))+((4*x^3-5)*exp(3/4)-x^6-50 
*x^3+50)*log(x)+x*exp(3/4)-x^4-10*x)/(x^4*log(x)*log(log(x))^2+(2*x^6-10*x 
^3)*log(x)*log(log(x))+(x^8-10*x^5+25*x^2)*log(x)),x)
 

Output:

( - e**(3/4) + x**3 + 10)/(x*(log(log(x))*x + x**3 - 5))