Integrand size = 138, antiderivative size = 22 \[ \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)+e^{e^{-1+e^{e^x}}} \left (-16-16 x+e^{-1+e^{e^x}+e^x} \left (4 e^x x^2+4 e^x x \log (x)\right )\right )}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx=1+x+\frac {4 e^{e^{-1+e^{e^x}}}}{(x+\log (x))^4} \] Output:
1+x+4/(x+ln(x))^4*exp(exp(exp(exp(x))-1))
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)+e^{e^{-1+e^{e^x}}} \left (-16-16 x+e^{-1+e^{e^x}+e^x} \left (4 e^x x^2+4 e^x x \log (x)\right )\right )}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx=\frac {e x+\frac {4 e^{1+e^{-1+e^{e^x}}}}{(x+\log (x))^4}}{e} \] Input:
Integrate[(x^6 + 5*x^5*Log[x] + 10*x^4*Log[x]^2 + 10*x^3*Log[x]^3 + 5*x^2* Log[x]^4 + x*Log[x]^5 + E^E^(-1 + E^E^x)*(-16 - 16*x + E^(-1 + E^E^x + E^x )*(4*E^x*x^2 + 4*E^x*x*Log[x])))/(x^6 + 5*x^5*Log[x] + 10*x^4*Log[x]^2 + 1 0*x^3*Log[x]^3 + 5*x^2*Log[x]^4 + x*Log[x]^5),x]
Output:
(E*x + (4*E^(1 + E^(-1 + E^E^x)))/(x + Log[x])^4)/E
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.
| \(\displaystyle \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+e^{e^{e^{e^x}-1}} \left (e^{e^{e^x}+e^x-1} \left (4 e^x x^2+4 e^x x \log (x)\right )-16 x-16\right )+x \log ^5(x)}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+e^{e^{e^{e^x}-1}} \left (e^{e^{e^x}+e^x-1} \left (4 e^x x^2+4 e^x x \log (x)\right )-16 x-16\right )+x \log ^5(x)}{x (x+\log (x))^5}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x^5}{(x+\log (x))^5}+\frac {5 x^4 \log (x)}{(x+\log (x))^5}+\frac {10 x^3 \log ^2(x)}{(x+\log (x))^5}+\frac {10 x^2 \log ^3(x)}{(x+\log (x))^5}+\frac {\log ^5(x)}{(x+\log (x))^5}+\frac {5 x \log ^4(x)}{(x+\log (x))^5}+\frac {4 e^{x+e^{e^x}+e^{e^{e^x}-1}+e^x-1}}{(x+\log (x))^4}-\frac {16 e^{e^{e^{e^x}-1}}}{(x+\log (x))^5}-\frac {16 e^{e^{e^{e^x}-1}}}{x (x+\log (x))^5}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -16 \int \frac {e^{e^{-1+e^{e^x}}}}{(x+\log (x))^5}dx-16 \int \frac {e^{e^{-1+e^{e^x}}}}{x (x+\log (x))^5}dx+4 \int \frac {e^{x+e^{e^x}+e^{-1+e^{e^x}}+e^x-1}}{(x+\log (x))^4}dx+x\) |
Input:
Int[(x^6 + 5*x^5*Log[x] + 10*x^4*Log[x]^2 + 10*x^3*Log[x]^3 + 5*x^2*Log[x] ^4 + x*Log[x]^5 + E^E^(-1 + E^E^x)*(-16 - 16*x + E^(-1 + E^E^x + E^x)*(4*E ^x*x^2 + 4*E^x*x*Log[x])))/(x^6 + 5*x^5*Log[x] + 10*x^4*Log[x]^2 + 10*x^3* Log[x]^3 + 5*x^2*Log[x]^4 + x*Log[x]^5),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(44\) vs. \(2(18)=36\).
Time = 95.42 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05
| method | result | size |
| risch | \(x +\frac {4 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}-1}}}{x^{4}+4 x^{3} \ln \left (x \right )+6 x^{2} \ln \left (x \right )^{2}+4 x \ln \left (x \right )^{3}+\ln \left (x \right )^{4}}\) | \(45\) |
| parallelrisch | \(\frac {4 x^{5}+16 x^{4} \ln \left (x \right )+24 x^{3} \ln \left (x \right )^{2}+16 x^{2} \ln \left (x \right )^{3}+4 x \ln \left (x \right )^{4}+16 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}-1}}}{4 x^{4}+16 x^{3} \ln \left (x \right )+24 x^{2} \ln \left (x \right )^{2}+16 x \ln \left (x \right )^{3}+4 \ln \left (x \right )^{4}}\) | \(83\) |
Input:
int((((4*x*exp(x)*ln(x)+4*exp(x)*x^2)*exp(exp(x))*exp(exp(exp(x))-1)-16*x- 16)*exp(exp(exp(exp(x))-1))+x*ln(x)^5+5*x^2*ln(x)^4+10*x^3*ln(x)^3+10*x^4* ln(x)^2+5*x^5*ln(x)+x^6)/(x*ln(x)^5+5*x^2*ln(x)^4+10*x^3*ln(x)^3+10*x^4*ln (x)^2+5*x^5*ln(x)+x^6),x,method=_RETURNVERBOSE)
Output:
x+4/(x^4+4*x^3*ln(x)+6*x^2*ln(x)^2+4*x*ln(x)^3+ln(x)^4)*exp(exp(exp(exp(x) )-1))
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (18) = 36\).
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.55 \[ \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)+e^{e^{-1+e^{e^x}}} \left (-16-16 x+e^{-1+e^{e^x}+e^x} \left (4 e^x x^2+4 e^x x \log (x)\right )\right )}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx=\frac {x^{5} + 4 \, x^{4} \log \left (x\right ) + 6 \, x^{3} \log \left (x\right )^{2} + 4 \, x^{2} \log \left (x\right )^{3} + x \log \left (x\right )^{4} + 4 \, e^{\left (e^{\left (e^{\left (e^{x}\right )} - 1\right )}\right )}}{x^{4} + 4 \, x^{3} \log \left (x\right ) + 6 \, x^{2} \log \left (x\right )^{2} + 4 \, x \log \left (x\right )^{3} + \log \left (x\right )^{4}} \] Input:
integrate((((4*x*exp(x)*log(x)+4*exp(x)*x^2)*exp(exp(x))*exp(exp(exp(x))-1 )-16*x-16)*exp(exp(exp(exp(x))-1))+x*log(x)^5+5*x^2*log(x)^4+10*x^3*log(x) ^3+10*x^4*log(x)^2+5*x^5*log(x)+x^6)/(x*log(x)^5+5*x^2*log(x)^4+10*x^3*log (x)^3+10*x^4*log(x)^2+5*x^5*log(x)+x^6),x, algorithm="fricas")
Output:
(x^5 + 4*x^4*log(x) + 6*x^3*log(x)^2 + 4*x^2*log(x)^3 + x*log(x)^4 + 4*e^( e^(e^(e^x) - 1)))/(x^4 + 4*x^3*log(x) + 6*x^2*log(x)^2 + 4*x*log(x)^3 + lo g(x)^4)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.69 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.18 \[ \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)+e^{e^{-1+e^{e^x}}} \left (-16-16 x+e^{-1+e^{e^x}+e^x} \left (4 e^x x^2+4 e^x x \log (x)\right )\right )}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx=x + \frac {4 e^{e^{e^{e^{x}} - 1}}}{x^{4} + 4 x^{3} \log {\left (x \right )} + 6 x^{2} \log {\left (x \right )}^{2} + 4 x \log {\left (x \right )}^{3} + \log {\left (x \right )}^{4}} \] Input:
integrate((((4*x*exp(x)*ln(x)+4*exp(x)*x**2)*exp(exp(x))*exp(exp(exp(x))-1 )-16*x-16)*exp(exp(exp(exp(x))-1))+x*ln(x)**5+5*x**2*ln(x)**4+10*x**3*ln(x )**3+10*x**4*ln(x)**2+5*x**5*ln(x)+x**6)/(x*ln(x)**5+5*x**2*ln(x)**4+10*x* *3*ln(x)**3+10*x**4*ln(x)**2+5*x**5*ln(x)+x**6),x)
Output:
x + 4*exp(exp(exp(exp(x)) - 1))/(x**4 + 4*x**3*log(x) + 6*x**2*log(x)**2 + 4*x*log(x)**3 + log(x)**4)
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (18) = 36\).
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.55 \[ \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)+e^{e^{-1+e^{e^x}}} \left (-16-16 x+e^{-1+e^{e^x}+e^x} \left (4 e^x x^2+4 e^x x \log (x)\right )\right )}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx=\frac {x^{5} + 4 \, x^{4} \log \left (x\right ) + 6 \, x^{3} \log \left (x\right )^{2} + 4 \, x^{2} \log \left (x\right )^{3} + x \log \left (x\right )^{4} + 4 \, e^{\left (e^{\left (e^{\left (e^{x}\right )} - 1\right )}\right )}}{x^{4} + 4 \, x^{3} \log \left (x\right ) + 6 \, x^{2} \log \left (x\right )^{2} + 4 \, x \log \left (x\right )^{3} + \log \left (x\right )^{4}} \] Input:
integrate((((4*x*exp(x)*log(x)+4*exp(x)*x^2)*exp(exp(x))*exp(exp(exp(x))-1 )-16*x-16)*exp(exp(exp(exp(x))-1))+x*log(x)^5+5*x^2*log(x)^4+10*x^3*log(x) ^3+10*x^4*log(x)^2+5*x^5*log(x)+x^6)/(x*log(x)^5+5*x^2*log(x)^4+10*x^3*log (x)^3+10*x^4*log(x)^2+5*x^5*log(x)+x^6),x, algorithm="maxima")
Output:
(x^5 + 4*x^4*log(x) + 6*x^3*log(x)^2 + 4*x^2*log(x)^3 + x*log(x)^4 + 4*e^( e^(e^(e^x) - 1)))/(x^4 + 4*x^3*log(x) + 6*x^2*log(x)^2 + 4*x*log(x)^3 + lo g(x)^4)
\[ \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)+e^{e^{-1+e^{e^x}}} \left (-16-16 x+e^{-1+e^{e^x}+e^x} \left (4 e^x x^2+4 e^x x \log (x)\right )\right )}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx=\int { \frac {x^{6} + 5 \, x^{5} \log \left (x\right ) + 10 \, x^{4} \log \left (x\right )^{2} + 10 \, x^{3} \log \left (x\right )^{3} + 5 \, x^{2} \log \left (x\right )^{4} + x \log \left (x\right )^{5} + 4 \, {\left ({\left (x^{2} e^{x} + x e^{x} \log \left (x\right )\right )} e^{\left (e^{x} + e^{\left (e^{x}\right )} - 1\right )} - 4 \, x - 4\right )} e^{\left (e^{\left (e^{\left (e^{x}\right )} - 1\right )}\right )}}{x^{6} + 5 \, x^{5} \log \left (x\right ) + 10 \, x^{4} \log \left (x\right )^{2} + 10 \, x^{3} \log \left (x\right )^{3} + 5 \, x^{2} \log \left (x\right )^{4} + x \log \left (x\right )^{5}} \,d x } \] Input:
integrate((((4*x*exp(x)*log(x)+4*exp(x)*x^2)*exp(exp(x))*exp(exp(exp(x))-1 )-16*x-16)*exp(exp(exp(exp(x))-1))+x*log(x)^5+5*x^2*log(x)^4+10*x^3*log(x) ^3+10*x^4*log(x)^2+5*x^5*log(x)+x^6)/(x*log(x)^5+5*x^2*log(x)^4+10*x^3*log (x)^3+10*x^4*log(x)^2+5*x^5*log(x)+x^6),x, algorithm="giac")
Output:
integrate((x^6 + 5*x^5*log(x) + 10*x^4*log(x)^2 + 10*x^3*log(x)^3 + 5*x^2* log(x)^4 + x*log(x)^5 + 4*((x^2*e^x + x*e^x*log(x))*e^(e^x + e^(e^x) - 1) - 4*x - 4)*e^(e^(e^(e^x) - 1)))/(x^6 + 5*x^5*log(x) + 10*x^4*log(x)^2 + 10 *x^3*log(x)^3 + 5*x^2*log(x)^4 + x*log(x)^5), x)
Time = 3.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)+e^{e^{-1+e^{e^x}}} \left (-16-16 x+e^{-1+e^{e^x}+e^x} \left (4 e^x x^2+4 e^x x \log (x)\right )\right )}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx=x+\frac {4\,{\mathrm {e}}^{{\mathrm {e}}^{-1}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}}{{\left (x+\ln \left (x\right )\right )}^4} \] Input:
int((x*log(x)^5 + 5*x^5*log(x) + 5*x^2*log(x)^4 + 10*x^3*log(x)^3 + 10*x^4 *log(x)^2 - exp(exp(exp(exp(x)) - 1))*(16*x - exp(exp(exp(x)) - 1)*exp(exp (x))*(4*x^2*exp(x) + 4*x*exp(x)*log(x)) + 16) + x^6)/(x*log(x)^5 + 5*x^5*l og(x) + 5*x^2*log(x)^4 + 10*x^3*log(x)^3 + 10*x^4*log(x)^2 + x^6),x)
Output:
x + (4*exp(exp(-1)*exp(exp(exp(x)))))/(x + log(x))^4
Time = 0.16 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.82 \[ \int \frac {x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)+e^{e^{-1+e^{e^x}}} \left (-16-16 x+e^{-1+e^{e^x}+e^x} \left (4 e^x x^2+4 e^x x \log (x)\right )\right )}{x^6+5 x^5 \log (x)+10 x^4 \log ^2(x)+10 x^3 \log ^3(x)+5 x^2 \log ^4(x)+x \log ^5(x)} \, dx=\frac {4 e^{\frac {e^{e^{e^{x}}}}{e}}+\mathrm {log}\left (x \right )^{4} x +4 \mathrm {log}\left (x \right )^{3} x^{2}+6 \mathrm {log}\left (x \right )^{2} x^{3}+4 \,\mathrm {log}\left (x \right ) x^{4}+x^{5}}{\mathrm {log}\left (x \right )^{4}+4 \mathrm {log}\left (x \right )^{3} x +6 \mathrm {log}\left (x \right )^{2} x^{2}+4 \,\mathrm {log}\left (x \right ) x^{3}+x^{4}} \] Input:
int((((4*x*exp(x)*log(x)+4*exp(x)*x^2)*exp(exp(x))*exp(exp(exp(x))-1)-16*x -16)*exp(exp(exp(exp(x))-1))+x*log(x)^5+5*x^2*log(x)^4+10*x^3*log(x)^3+10* x^4*log(x)^2+5*x^5*log(x)+x^6)/(x*log(x)^5+5*x^2*log(x)^4+10*x^3*log(x)^3+ 10*x^4*log(x)^2+5*x^5*log(x)+x^6),x)
Output:
(4*e**(e**(e**(e**x))/e) + log(x)**4*x + 4*log(x)**3*x**2 + 6*log(x)**2*x* *3 + 4*log(x)*x**4 + x**5)/(log(x)**4 + 4*log(x)**3*x + 6*log(x)**2*x**2 + 4*log(x)*x**3 + x**4)