Integrand size = 103, antiderivative size = 24 \[ \int \frac {e^3 (-5-x)+e^3 (-5-x) \log (x)+5 e^3 \log (x) \log (x \log (x))+e^3 \left (4-10 x-x^2\right ) \log (x) \log ^2(x \log (x))}{x^2 \log (x)+\left (-8 x-2 x^3\right ) \log (x) \log (x \log (x))+\left (16+8 x^2+x^4\right ) \log (x) \log ^2(x \log (x))} \, dx=\frac {e^3 (5+x)}{4+x^2-\frac {x}{\log (x \log (x))}} \] Output:
exp(3)*(5+x)/(x^2+4-x/ln(x*ln(x)))
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^3 (-5-x)+e^3 (-5-x) \log (x)+5 e^3 \log (x) \log (x \log (x))+e^3 \left (4-10 x-x^2\right ) \log (x) \log ^2(x \log (x))}{x^2 \log (x)+\left (-8 x-2 x^3\right ) \log (x) \log (x \log (x))+\left (16+8 x^2+x^4\right ) \log (x) \log ^2(x \log (x))} \, dx=\frac {e^3 (5+x) \log (x \log (x))}{-x+\left (4+x^2\right ) \log (x \log (x))} \] Input:
Integrate[(E^3*(-5 - x) + E^3*(-5 - x)*Log[x] + 5*E^3*Log[x]*Log[x*Log[x]] + E^3*(4 - 10*x - x^2)*Log[x]*Log[x*Log[x]]^2)/(x^2*Log[x] + (-8*x - 2*x^ 3)*Log[x]*Log[x*Log[x]] + (16 + 8*x^2 + x^4)*Log[x]*Log[x*Log[x]]^2),x]
Output:
(E^3*(5 + x)*Log[x*Log[x]])/(-x + (4 + x^2)*Log[x*Log[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.
| \(\displaystyle \int \frac {e^3 \left (-x^2-10 x+4\right ) \log (x) \log ^2(x \log (x))+e^3 (-x-5)+5 e^3 \log (x) \log (x \log (x))+e^3 (-x-5) \log (x)}{\left (-2 x^3-8 x\right ) \log (x) \log (x \log (x))+x^2 \log (x)+\left (x^4+8 x^2+16\right ) \log (x) \log ^2(x \log (x))} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^3 \left (-\log (x) \left (\left (x^2+10 x-4\right ) \log ^2(x \log (x))+x-5 \log (x \log (x))+5\right )-x-5\right )}{\log (x) \left (x-\left (x^2+4\right ) \log (x \log (x))\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^3 \int -\frac {x+\log (x) \left (-\left (\left (-x^2-10 x+4\right ) \log ^2(x \log (x))\right )-5 \log (x \log (x))+x+5\right )+5}{\log (x) \left (x-\left (x^2+4\right ) \log (x \log (x))\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -e^3 \int \frac {x+\log (x) \left (-\left (\left (-x^2-10 x+4\right ) \log ^2(x \log (x))\right )-5 \log (x \log (x))+x+5\right )+5}{\log (x) \left (x-\left (x^2+4\right ) \log (x \log (x))\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -e^3 \int \left (\frac {x^2+10 x-4}{\left (x^2+4\right )^2}+\frac {2 x^3+15 x^2-8 x-20}{\left (x^2+4\right )^2 \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )}+\frac {(x+5) \left (\log (x) x^4+x^4+\log (x) x^3+8 \log (x) x^2+8 x^2-4 \log (x) x+16 \log (x)+16\right )}{\left (x^2+4\right )^2 \log (x) \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -e^3 \left (6 \int \frac {1}{\left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}dx-\left (\frac {5}{2}+3 i\right ) \int \frac {1}{(2 i-x) \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}dx+\int \frac {x}{\left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}dx+\left (\frac {5}{2}-3 i\right ) \int \frac {1}{(x+2 i) \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}dx+32 \int \frac {1}{\left (x^2+4\right )^2 \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}dx-40 \int \frac {x}{\left (x^2+4\right )^2 \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}dx+5 \int \frac {1}{\log (x) \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}dx+\int \frac {x}{\log (x) \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )^2}dx-\left (1-\frac {15 i}{4}\right ) \int \frac {1}{(2 i-x) \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )}dx+\left (1+\frac {15 i}{4}\right ) \int \frac {1}{(x+2 i) \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )}dx-80 \int \frac {1}{\left (x^2+4\right )^2 \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )}dx-16 \int \frac {x}{\left (x^2+4\right )^2 \left (\log (x \log (x)) x^2-x+4 \log (x \log (x))\right )}dx-\frac {x+5}{x^2+4}\right )\) |
Input:
Int[(E^3*(-5 - x) + E^3*(-5 - x)*Log[x] + 5*E^3*Log[x]*Log[x*Log[x]] + E^3 *(4 - 10*x - x^2)*Log[x]*Log[x*Log[x]]^2)/(x^2*Log[x] + (-8*x - 2*x^3)*Log [x]*Log[x*Log[x]] + (16 + 8*x^2 + x^4)*Log[x]*Log[x*Log[x]]^2),x]
Output:
$Aborted
Time = 23.79 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79
| method | result | size |
| parallelrisch | \(\frac {\ln \left (x \ln \left (x \right )\right ) x \,{\mathrm e}^{3}+5 \ln \left (x \ln \left (x \right )\right ) {\mathrm e}^{3}}{\ln \left (x \ln \left (x \right )\right ) x^{2}+4 \ln \left (x \ln \left (x \right )\right )-x}\) | \(43\) |
| risch | \(\frac {\left (5+x \right ) {\mathrm e}^{3}}{x^{2}+4}+\frac {2 i x \left (5+x \right ) {\mathrm e}^{3}}{\left (x^{2}+4\right ) \left (\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right ) x^{2}-\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2} x^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2} x^{2}+\pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3} x^{2}+4 \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )-4 \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-4 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}+4 \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}+2 i x^{2} \ln \left (\ln \left (x \right )\right )+2 i x^{2} \ln \left (x \right )-2 i x +8 i \ln \left (\ln \left (x \right )\right )+8 i \ln \left (x \right )\right )}\) | \(212\) |
Input:
int(((-x^2-10*x+4)*exp(3)*ln(x)*ln(x*ln(x))^2+5*exp(3)*ln(x)*ln(x*ln(x))+( -x-5)*exp(3)*ln(x)+(-x-5)*exp(3))/((x^4+8*x^2+16)*ln(x)*ln(x*ln(x))^2+(-2* x^3-8*x)*ln(x)*ln(x*ln(x))+x^2*ln(x)),x,method=_RETURNVERBOSE)
Output:
(ln(x*ln(x))*x*exp(3)+5*ln(x*ln(x))*exp(3))/(ln(x*ln(x))*x^2+4*ln(x*ln(x)) -x)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^3 (-5-x)+e^3 (-5-x) \log (x)+5 e^3 \log (x) \log (x \log (x))+e^3 \left (4-10 x-x^2\right ) \log (x) \log ^2(x \log (x))}{x^2 \log (x)+\left (-8 x-2 x^3\right ) \log (x) \log (x \log (x))+\left (16+8 x^2+x^4\right ) \log (x) \log ^2(x \log (x))} \, dx=\frac {{\left (x + 5\right )} e^{3} \log \left (x \log \left (x\right )\right )}{{\left (x^{2} + 4\right )} \log \left (x \log \left (x\right )\right ) - x} \] Input:
integrate(((-x^2-10*x+4)*exp(3)*log(x)*log(x*log(x))^2+5*exp(3)*log(x)*log (x*log(x))+(-x-5)*exp(3)*log(x)+(-x-5)*exp(3))/((x^4+8*x^2+16)*log(x)*log( x*log(x))^2+(-2*x^3-8*x)*log(x)*log(x*log(x))+x^2*log(x)),x, algorithm="fr icas")
Output:
(x + 5)*e^3*log(x*log(x))/((x^2 + 4)*log(x*log(x)) - x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (19) = 38\).
Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {e^3 (-5-x)+e^3 (-5-x) \log (x)+5 e^3 \log (x) \log (x \log (x))+e^3 \left (4-10 x-x^2\right ) \log (x) \log ^2(x \log (x))}{x^2 \log (x)+\left (-8 x-2 x^3\right ) \log (x) \log (x \log (x))+\left (16+8 x^2+x^4\right ) \log (x) \log ^2(x \log (x))} \, dx=\frac {x^{2} e^{3} + 5 x e^{3}}{- x^{3} - 4 x + \left (x^{4} + 8 x^{2} + 16\right ) \log {\left (x \log {\left (x \right )} \right )}} - \frac {- x e^{3} - 5 e^{3}}{x^{2} + 4} \] Input:
integrate(((-x**2-10*x+4)*exp(3)*ln(x)*ln(x*ln(x))**2+5*exp(3)*ln(x)*ln(x* ln(x))+(-x-5)*exp(3)*ln(x)+(-x-5)*exp(3))/((x**4+8*x**2+16)*ln(x)*ln(x*ln( x))**2+(-2*x**3-8*x)*ln(x)*ln(x*ln(x))+x**2*ln(x)),x)
Output:
(x**2*exp(3) + 5*x*exp(3))/(-x**3 - 4*x + (x**4 + 8*x**2 + 16)*log(x*log(x ))) - (-x*exp(3) - 5*exp(3))/(x**2 + 4)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (23) = 46\).
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {e^3 (-5-x)+e^3 (-5-x) \log (x)+5 e^3 \log (x) \log (x \log (x))+e^3 \left (4-10 x-x^2\right ) \log (x) \log ^2(x \log (x))}{x^2 \log (x)+\left (-8 x-2 x^3\right ) \log (x) \log (x \log (x))+\left (16+8 x^2+x^4\right ) \log (x) \log ^2(x \log (x))} \, dx=\frac {{\left (x e^{3} + 5 \, e^{3}\right )} \log \left (x\right ) + {\left (x e^{3} + 5 \, e^{3}\right )} \log \left (\log \left (x\right )\right )}{{\left (x^{2} + 4\right )} \log \left (x\right ) + {\left (x^{2} + 4\right )} \log \left (\log \left (x\right )\right ) - x} \] Input:
integrate(((-x^2-10*x+4)*exp(3)*log(x)*log(x*log(x))^2+5*exp(3)*log(x)*log (x*log(x))+(-x-5)*exp(3)*log(x)+(-x-5)*exp(3))/((x^4+8*x^2+16)*log(x)*log( x*log(x))^2+(-2*x^3-8*x)*log(x)*log(x*log(x))+x^2*log(x)),x, algorithm="ma xima")
Output:
((x*e^3 + 5*e^3)*log(x) + (x*e^3 + 5*e^3)*log(log(x)))/((x^2 + 4)*log(x) + (x^2 + 4)*log(log(x)) - x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (23) = 46\).
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {e^3 (-5-x)+e^3 (-5-x) \log (x)+5 e^3 \log (x) \log (x \log (x))+e^3 \left (4-10 x-x^2\right ) \log (x) \log ^2(x \log (x))}{x^2 \log (x)+\left (-8 x-2 x^3\right ) \log (x) \log (x \log (x))+\left (16+8 x^2+x^4\right ) \log (x) \log ^2(x \log (x))} \, dx=\frac {x e^{3} \log \left (x\right ) + x e^{3} \log \left (\log \left (x\right )\right ) + 5 \, e^{3} \log \left (x\right ) + 5 \, e^{3} \log \left (\log \left (x\right )\right )}{x^{2} \log \left (x\right ) + x^{2} \log \left (\log \left (x\right )\right ) - x + 4 \, \log \left (x\right ) + 4 \, \log \left (\log \left (x\right )\right )} \] Input:
integrate(((-x^2-10*x+4)*exp(3)*log(x)*log(x*log(x))^2+5*exp(3)*log(x)*log (x*log(x))+(-x-5)*exp(3)*log(x)+(-x-5)*exp(3))/((x^4+8*x^2+16)*log(x)*log( x*log(x))^2+(-2*x^3-8*x)*log(x)*log(x*log(x))+x^2*log(x)),x, algorithm="gi ac")
Output:
(x*e^3*log(x) + x*e^3*log(log(x)) + 5*e^3*log(x) + 5*e^3*log(log(x)))/(x^2 *log(x) + x^2*log(log(x)) - x + 4*log(x) + 4*log(log(x)))
Time = 4.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^3 (-5-x)+e^3 (-5-x) \log (x)+5 e^3 \log (x) \log (x \log (x))+e^3 \left (4-10 x-x^2\right ) \log (x) \log ^2(x \log (x))}{x^2 \log (x)+\left (-8 x-2 x^3\right ) \log (x) \log (x \log (x))+\left (16+8 x^2+x^4\right ) \log (x) \log ^2(x \log (x))} \, dx=\frac {\ln \left (x\,\ln \left (x\right )\right )\,{\mathrm {e}}^3\,\left (x+5\right )}{4\,\ln \left (x\,\ln \left (x\right )\right )-x+x^2\,\ln \left (x\,\ln \left (x\right )\right )} \] Input:
int(-(exp(3)*(x + 5) + exp(3)*log(x)*(x + 5) - 5*log(x*log(x))*exp(3)*log( x) + log(x*log(x))^2*exp(3)*log(x)*(10*x + x^2 - 4))/(x^2*log(x) + log(x*l og(x))^2*log(x)*(8*x^2 + x^4 + 16) - log(x*log(x))*log(x)*(8*x + 2*x^3)),x )
Output:
(log(x*log(x))*exp(3)*(x + 5))/(4*log(x*log(x)) - x + x^2*log(x*log(x)))
Time = 0.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.46 \[ \int \frac {e^3 (-5-x)+e^3 (-5-x) \log (x)+5 e^3 \log (x) \log (x \log (x))+e^3 \left (4-10 x-x^2\right ) \log (x) \log ^2(x \log (x))}{x^2 \log (x)+\left (-8 x-2 x^3\right ) \log (x) \log (x \log (x))+\left (16+8 x^2+x^4\right ) \log (x) \log ^2(x \log (x))} \, dx=\frac {e^{3} \left (-\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (\mathrm {log}\left (x \right ) x \right ) x^{2}-4 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (\mathrm {log}\left (x \right ) x \right )+\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x +\mathrm {log}\left (\mathrm {log}\left (x \right ) x \right )^{2} x^{2}+4 \mathrm {log}\left (\mathrm {log}\left (x \right ) x \right )^{2}-\mathrm {log}\left (\mathrm {log}\left (x \right ) x \right ) \mathrm {log}\left (x \right ) x^{2}-4 \,\mathrm {log}\left (\mathrm {log}\left (x \right ) x \right ) \mathrm {log}\left (x \right )+5 \,\mathrm {log}\left (\mathrm {log}\left (x \right ) x \right )+\mathrm {log}\left (x \right ) x \right )}{\mathrm {log}\left (\mathrm {log}\left (x \right ) x \right ) x^{2}+4 \,\mathrm {log}\left (\mathrm {log}\left (x \right ) x \right )-x} \] Input:
int(((-x^2-10*x+4)*exp(3)*log(x)*log(x*log(x))^2+5*exp(3)*log(x)*log(x*log (x))+(-x-5)*exp(3)*log(x)+(-x-5)*exp(3))/((x^4+8*x^2+16)*log(x)*log(x*log( x))^2+(-2*x^3-8*x)*log(x)*log(x*log(x))+x^2*log(x)),x)
Output:
(e**3*( - log(log(x))*log(log(x)*x)*x**2 - 4*log(log(x))*log(log(x)*x) + l og(log(x))*x + log(log(x)*x)**2*x**2 + 4*log(log(x)*x)**2 - log(log(x)*x)* log(x)*x**2 - 4*log(log(x)*x)*log(x) + 5*log(log(x)*x) + log(x)*x))/(log(l og(x)*x)*x**2 + 4*log(log(x)*x) - x)