Integrand size = 89, antiderivative size = 23 \[ \int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx=x+\log \left (-\frac {1}{25} e^{-x} x^4+\log (x (x+\log (3)))\right ) \] Output:
ln(ln(x*(ln(3)+x))-1/25*x^4/exp(x))+x
\[ \int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx=\int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx \] Input:
Integrate[(-4*x^5 - 4*x^4*Log[3] + E^x*(50*x + 25*Log[3]) + E^x*(25*x^2 + 25*x*Log[3])*Log[x^2 + x*Log[3]])/(-x^6 - x^5*Log[3] + E^x*(25*x^2 + 25*x* Log[3])*Log[x^2 + x*Log[3]]),x]
Output:
Integrate[(-4*x^5 - 4*x^4*Log[3] + E^x*(50*x + 25*Log[3]) + E^x*(25*x^2 + 25*x*Log[3])*Log[x^2 + x*Log[3]])/(-x^6 - x^5*Log[3] + E^x*(25*x^2 + 25*x* Log[3])*Log[x^2 + x*Log[3]]), x]
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.
| \(\displaystyle \int \frac {-4 x^5-4 x^4 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )+e^x (50 x+25 \log (3))}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {4 x^5+4 x^4 \log (3)-e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )-e^x (50 x+25 \log (3))}{x (x+\log (3)) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x^2 \log (x (x+\log (3)))+2 x+x \log (3) \log (x (x+\log (3)))+\log (3)}{x (x+\log (3)) \log (x (x+\log (3)))}+\frac {x^3 \left (x^2 (-\log (x (x+\log (3))))-2 x+4 x \left (1-\frac {\log (3)}{4}\right ) \log (x (x+\log (3)))+4 \log (3) \log (x (x+\log (3)))-\log (3)\right )}{(x+\log (3)) \log (x (x+\log (3))) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \log ^5(3) \int \frac {1}{(x+\log (3)) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}dx-\log ^4(3) \int \frac {1}{x^4-25 e^x \log (x (x+\log (3)))}dx+(4-\log (3)) \log ^4(3) \int \frac {1}{(x+\log (3)) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}dx-4 \log ^4(3) \int \frac {1}{(x+\log (3)) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}dx-\log ^4(3) \int \frac {1}{(x+\log (3)) \log (x (x+\log (3))) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}dx-(4-\log (3)) \log ^3(3) \int \frac {1}{x^4-25 e^x \log (x (x+\log (3)))}dx+\log ^3(3) \int \frac {x}{x^4-25 e^x \log (x (x+\log (3)))}dx-\log ^3(3) \int \frac {1}{\log (x (x+\log (3))) \left (25 e^x \log (x (x+\log (3)))-x^4\right )}dx+\log ^2(3) \log (81) \int \frac {1}{x^4-25 e^x \log (x (x+\log (3)))}dx+(4-\log (3)) \log ^2(3) \int \frac {x}{x^4-25 e^x \log (x (x+\log (3)))}dx-4 \log ^2(3) \int \frac {x}{x^4-25 e^x \log (x (x+\log (3)))}dx-\log ^2(3) \int \frac {x}{\log (x (x+\log (3))) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}dx-\int \frac {x^4}{x^4-25 e^x \log (x (x+\log (3)))}dx+\log (3) \int \frac {x^3}{x^4-25 e^x \log (x (x+\log (3)))}dx+(4-\log (3)) \int \frac {x^3}{x^4-25 e^x \log (x (x+\log (3)))}dx-2 \int \frac {x^3}{\log (x (x+\log (3))) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}dx-\log ^2(3) \int \frac {x^2}{x^4-25 e^x \log (x (x+\log (3)))}dx+\log (81) \int \frac {x^2}{x^4-25 e^x \log (x (x+\log (3)))}dx-(4-\log (3)) \log (3) \int \frac {x^2}{x^4-25 e^x \log (x (x+\log (3)))}dx+\log (3) \int \frac {x^2}{\log (x (x+\log (3))) \left (x^4-25 e^x \log (x (x+\log (3)))\right )}dx+\int \frac {2 x+\log (3)}{x (x+\log (3)) \log (x (x+\log (3)))}dx+x\) |
Input:
Int[(-4*x^5 - 4*x^4*Log[3] + E^x*(50*x + 25*Log[3]) + E^x*(25*x^2 + 25*x*L og[3])*Log[x^2 + x*Log[3]])/(-x^6 - x^5*Log[3] + E^x*(25*x^2 + 25*x*Log[3] )*Log[x^2 + x*Log[3]]),x]
Output:
$Aborted
Time = 1.81 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74
| method | result | size |
| parallelrisch | \(\ln \left (x^{4}-25 \,{\mathrm e}^{x} \ln \left (x \left (\ln \left (3\right )+x \right )\right )\right )\) | \(17\) |
| risch | \(x +\ln \left (\ln \left (\ln \left (3\right )+x \right )+\frac {i \operatorname {csgn}\left (i x \left (\ln \left (3\right )+x \right )\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (3\right )+x \right )\right ) \pi }{2}-\frac {i \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (\ln \left (3\right )+x \right )\right ) \operatorname {csgn}\left (i \left (\ln \left (3\right )+x \right )\right ) \pi }{2}-\frac {i \operatorname {csgn}\left (i x \left (\ln \left (3\right )+x \right )\right )^{3} \pi }{2}+\frac {i \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (\ln \left (3\right )+x \right )\right )^{2} \pi }{2}-\frac {{\mathrm e}^{-x} x^{4}}{25}+\ln \left (x \right )\right )\) | \(105\) |
Input:
int(((25*x*ln(3)+25*x^2)*exp(x)*ln(x*ln(3)+x^2)+(25*ln(3)+50*x)*exp(x)-4*x ^4*ln(3)-4*x^5)/((25*x*ln(3)+25*x^2)*exp(x)*ln(x*ln(3)+x^2)-x^5*ln(3)-x^6) ,x,method=_RETURNVERBOSE)
Output:
ln(x^4-25*exp(x)*ln(x*(ln(3)+x)))
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx=x + \log \left (-{\left (x^{4} - 25 \, e^{x} \log \left (x^{2} + x \log \left (3\right )\right )\right )} e^{\left (-x\right )}\right ) \] Input:
integrate(((25*x*log(3)+25*x^2)*exp(x)*log(x*log(3)+x^2)+(25*log(3)+50*x)* exp(x)-4*x^4*log(3)-4*x^5)/((25*x*log(3)+25*x^2)*exp(x)*log(x*log(3)+x^2)- x^5*log(3)-x^6),x, algorithm="fricas")
Output:
x + log(-(x^4 - 25*e^x*log(x^2 + x*log(3)))*e^(-x))
Time = 0.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx=\log {\left (- \frac {x^{4}}{25 \log {\left (x^{2} + x \log {\left (3 \right )} \right )}} + e^{x} \right )} + \log {\left (\log {\left (x^{2} + x \log {\left (3 \right )} \right )} \right )} \] Input:
integrate(((25*x*ln(3)+25*x**2)*exp(x)*ln(x*ln(3)+x**2)+(25*ln(3)+50*x)*ex p(x)-4*x**4*ln(3)-4*x**5)/((25*x*ln(3)+25*x**2)*exp(x)*ln(x*ln(3)+x**2)-x* *5*ln(3)-x**6),x)
Output:
log(-x**4/(25*log(x**2 + x*log(3))) + exp(x)) + log(log(x**2 + x*log(3)))
Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx=x + \log \left (-\frac {1}{25} \, {\left (x^{4} - 25 \, e^{x} \log \left (x + \log \left (3\right )\right ) - 25 \, e^{x} \log \left (x\right )\right )} e^{\left (-x\right )}\right ) \] Input:
integrate(((25*x*log(3)+25*x^2)*exp(x)*log(x*log(3)+x^2)+(25*log(3)+50*x)* exp(x)-4*x^4*log(3)-4*x^5)/((25*x*log(3)+25*x^2)*exp(x)*log(x*log(3)+x^2)- x^5*log(3)-x^6),x, algorithm="maxima")
Output:
x + log(-1/25*(x^4 - 25*e^x*log(x + log(3)) - 25*e^x*log(x))*e^(-x))
Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx=\log \left (-x^{4} + 25 \, e^{x} \log \left (x^{2} + x \log \left (3\right )\right )\right ) \] Input:
integrate(((25*x*log(3)+25*x^2)*exp(x)*log(x*log(3)+x^2)+(25*log(3)+50*x)* exp(x)-4*x^4*log(3)-4*x^5)/((25*x*log(3)+25*x^2)*exp(x)*log(x*log(3)+x^2)- x^5*log(3)-x^6),x, algorithm="giac")
Output:
log(-x^4 + 25*e^x*log(x^2 + x*log(3)))
Time = 3.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx=x+\ln \left (\ln \left (x^2+\ln \left (3\right )\,x\right )-\frac {x^4\,{\mathrm {e}}^{-x}}{25}\right ) \] Input:
int(-(exp(x)*(50*x + 25*log(3)) - 4*x^4*log(3) - 4*x^5 + exp(x)*log(x*log( 3) + x^2)*(25*x*log(3) + 25*x^2))/(x^5*log(3) + x^6 - exp(x)*log(x*log(3) + x^2)*(25*x*log(3) + 25*x^2)),x)
Output:
x + log(log(x*log(3) + x^2) - (x^4*exp(-x))/25)
\[ \int \frac {-4 x^5-4 x^4 \log (3)+e^x (50 x+25 \log (3))+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )}{-x^6-x^5 \log (3)+e^x \left (25 x^2+25 x \log (3)\right ) \log \left (x^2+x \log (3)\right )} \, dx=25 \left (\int \frac {e^{x}}{25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) \mathrm {log}\left (3\right ) x +25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) x^{2}-\mathrm {log}\left (3\right ) x^{5}-x^{6}}d x \right ) \mathrm {log}\left (3\right )+50 \left (\int \frac {e^{x}}{25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) \mathrm {log}\left (3\right )+25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) x -\mathrm {log}\left (3\right ) x^{4}-x^{5}}d x \right )-4 \left (\int \frac {x^{4}}{25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) \mathrm {log}\left (3\right )+25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) x -\mathrm {log}\left (3\right ) x^{4}-x^{5}}d x \right )-4 \left (\int \frac {x^{3}}{25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) \mathrm {log}\left (3\right )+25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) x -\mathrm {log}\left (3\right ) x^{4}-x^{5}}d x \right ) \mathrm {log}\left (3\right )+25 \left (\int \frac {e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) x}{25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) \mathrm {log}\left (3\right )+25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) x -\mathrm {log}\left (3\right ) x^{4}-x^{5}}d x \right )+25 \left (\int \frac {e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right )}{25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) \mathrm {log}\left (3\right )+25 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right ) x +x^{2}\right ) x -\mathrm {log}\left (3\right ) x^{4}-x^{5}}d x \right ) \mathrm {log}\left (3\right ) \] Input:
int(((25*x*log(3)+25*x^2)*exp(x)*log(x*log(3)+x^2)+(25*log(3)+50*x)*exp(x) -4*x^4*log(3)-4*x^5)/((25*x*log(3)+25*x^2)*exp(x)*log(x*log(3)+x^2)-x^5*lo g(3)-x^6),x)
Output:
25*int(e**x/(25*e**x*log(log(3)*x + x**2)*log(3)*x + 25*e**x*log(log(3)*x + x**2)*x**2 - log(3)*x**5 - x**6),x)*log(3) + 50*int(e**x/(25*e**x*log(lo g(3)*x + x**2)*log(3) + 25*e**x*log(log(3)*x + x**2)*x - log(3)*x**4 - x** 5),x) - 4*int(x**4/(25*e**x*log(log(3)*x + x**2)*log(3) + 25*e**x*log(log( 3)*x + x**2)*x - log(3)*x**4 - x**5),x) - 4*int(x**3/(25*e**x*log(log(3)*x + x**2)*log(3) + 25*e**x*log(log(3)*x + x**2)*x - log(3)*x**4 - x**5),x)* log(3) + 25*int((e**x*log(log(3)*x + x**2)*x)/(25*e**x*log(log(3)*x + x**2 )*log(3) + 25*e**x*log(log(3)*x + x**2)*x - log(3)*x**4 - x**5),x) + 25*in t((e**x*log(log(3)*x + x**2))/(25*e**x*log(log(3)*x + x**2)*log(3) + 25*e* *x*log(log(3)*x + x**2)*x - log(3)*x**4 - x**5),x)*log(3)