Integrand size = 106, antiderivative size = 25 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {(1+x)^2 \left (x-\frac {2}{3} \log (5) \log (\log (x))\right )^2}{e^5 x} \] Output:
1/exp(5)/x*(1+x)^2*(x-2/3*ln(ln(x))*ln(5))^2
Time = 0.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {(1+x)^2 (3 x-2 \log (5) \log (\log (x)))^2}{9 e^5 x} \] Input:
Integrate[((-12*x - 24*x^2 - 12*x^3)*Log[5] + (9*x^2 + 36*x^3 + 27*x^4)*Lo g[x] + ((8 + 16*x + 8*x^2)*Log[5]^2 + (-24*x^2 - 24*x^3)*Log[5]*Log[x])*Lo g[Log[x]] + (-4 + 4*x^2)*Log[5]^2*Log[x]*Log[Log[x]]^2)/(9*E^5*x^2*Log[x]) ,x]
Output:
((1 + x)^2*(3*x - 2*Log[5]*Log[Log[x]])^2)/(9*E^5*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.
| \(\displaystyle \int \frac {\left (4 x^2-4\right ) \log ^2(5) \log (x) \log ^2(\log (x))+\left (\left (8 x^2+16 x+8\right ) \log ^2(5)+\left (-24 x^3-24 x^2\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-12 x^3-24 x^2-12 x\right ) \log (5)+\left (27 x^4+36 x^3+9 x^2\right ) \log (x)}{9 e^5 x^2 \log (x)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -\frac {4 \left (1-x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))-8 \left (\left (x^2+2 x+1\right ) \log ^2(5)-3 \left (x^3+x^2\right ) \log (5) \log (x)\right ) \log (\log (x))-9 \left (3 x^4+4 x^3+x^2\right ) \log (x)+12 \left (x^3+2 x^2+x\right ) \log (5)}{x^2 \log (x)}dx}{9 e^5}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {4 \left (1-x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))-8 \left (\left (x^2+2 x+1\right ) \log ^2(5)-3 \left (x^3+x^2\right ) \log (5) \log (x)\right ) \log (\log (x))-9 \left (3 x^4+4 x^3+x^2\right ) \log (x)+12 \left (x^3+2 x^2+x\right ) \log (5)}{x^2 \log (x)}dx}{9 e^5}\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle -\frac {\int \frac {(x+1) (3 x-2 \log (5) \log (\log (x))) (4 (x+1) \log (5)-\log (x) (3 x (3 x+1)-2 (x-1) \log (5) \log (\log (x))))}{x^2 \log (x)}dx}{9 e^5}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {\int \left (-\frac {4 (x-1) (x+1) \log ^2(5) \log ^2(\log (x))}{x^2}+\frac {8 (x+1) \log (5) \left (3 \log (x) x^2-\log (5) x-\log (5)\right ) \log (\log (x))}{x^2 \log (x)}-\frac {3 (x+1) \left (9 \log (x) x^2+3 \log (x) x-4 \log (5) x-4 \log (5)\right )}{x \log (x)}\right )dx}{9 e^5}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-8 \log ^2(5) \int \frac {\log (\log (x))}{x^2 \log (x)}dx+4 \log ^2(5) \int \frac {\log ^2(\log (x))}{x^2}dx-8 \log ^2(5) \int \frac {\log (\log (x))}{\log (x)}dx-4 \log ^2(5) \int \log ^2(\log (x))dx-9 x^3-18 x^2+12 x^2 \log (5) \log (\log (x))-9 x-8 \log ^2(5) \log ^2(\log (x))+24 x \log (5) \log (\log (x))+12 \log (5) \log (\log (x))}{9 e^5}\) |
Input:
Int[((-12*x - 24*x^2 - 12*x^3)*Log[5] + (9*x^2 + 36*x^3 + 27*x^4)*Log[x] + ((8 + 16*x + 8*x^2)*Log[5]^2 + (-24*x^2 - 24*x^3)*Log[5]*Log[x])*Log[Log[ x]] + (-4 + 4*x^2)*Log[5]^2*Log[x]*Log[Log[x]]^2)/(9*E^5*x^2*Log[x]),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(64\) vs. \(2(24)=48\).
Time = 0.55 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.60
| method | result | size |
| risch | \(\frac {4 \,{\mathrm e}^{-5} \ln \left (5\right )^{2} \left (x^{2}+2 x +1\right ) \ln \left (\ln \left (x \right )\right )^{2}}{9 x}-\frac {4 \,{\mathrm e}^{-5} \ln \left (5\right ) x \left (2+x \right ) \ln \left (\ln \left (x \right )\right )}{3}+{\mathrm e}^{-5} x^{3}+2 x^{2} {\mathrm e}^{-5}+x \,{\mathrm e}^{-5}-\frac {4 \,{\mathrm e}^{-5} \ln \left (\ln \left (x \right )\right ) \ln \left (5\right )}{3}\) | \(65\) |
| parallelrisch | \(\frac {{\mathrm e}^{-5} \left (4 \ln \left (5\right )^{2} \ln \left (\ln \left (x \right )\right )^{2} x^{2}+8 \ln \left (5\right )^{2} \ln \left (\ln \left (x \right )\right )^{2} x -12 \ln \left (5\right ) x^{3} \ln \left (\ln \left (x \right )\right )+4 \ln \left (5\right )^{2} \ln \left (\ln \left (x \right )\right )^{2}-24 \ln \left (5\right ) x^{2} \ln \left (\ln \left (x \right )\right )+9 x^{4}-12 x \ln \left (5\right ) \ln \left (\ln \left (x \right )\right )+18 x^{3}+9 x^{2}\right )}{9 x}\) | \(91\) |
Input:
int(1/9*((4*x^2-4)*ln(5)^2*ln(x)*ln(ln(x))^2+((-24*x^3-24*x^2)*ln(5)*ln(x) +(8*x^2+16*x+8)*ln(5)^2)*ln(ln(x))+(27*x^4+36*x^3+9*x^2)*ln(x)+(-12*x^3-24 *x^2-12*x)*ln(5))/x^2/exp(5)/ln(x),x,method=_RETURNVERBOSE)
Output:
4/9*exp(-5)*ln(5)^2*(x^2+2*x+1)/x*ln(ln(x))^2-4/3*exp(-5)*ln(5)*x*(2+x)*ln (ln(x))+exp(-5)*x^3+2*x^2*exp(-5)+x*exp(-5)-4/3*exp(-5)*ln(ln(x))*ln(5)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.36 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {{\left (4 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (5\right )^{2} \log \left (\log \left (x\right )\right )^{2} + 9 \, x^{4} + 18 \, x^{3} - 12 \, {\left (x^{3} + 2 \, x^{2} + x\right )} \log \left (5\right ) \log \left (\log \left (x\right )\right ) + 9 \, x^{2}\right )} e^{\left (-5\right )}}{9 \, x} \] Input:
integrate(1/9*((4*x^2-4)*log(5)^2*log(x)*log(log(x))^2+((-24*x^3-24*x^2)*l og(5)*log(x)+(8*x^2+16*x+8)*log(5)^2)*log(log(x))+(27*x^4+36*x^3+9*x^2)*lo g(x)+(-12*x^3-24*x^2-12*x)*log(5))/x^2/exp(5)/log(x),x, algorithm="fricas" )
Output:
1/9*(4*(x^2 + 2*x + 1)*log(5)^2*log(log(x))^2 + 9*x^4 + 18*x^3 - 12*(x^3 + 2*x^2 + x)*log(5)*log(log(x)) + 9*x^2)*e^(-5)/x
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (24) = 48\).
Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.00 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {x^{3}}{e^{5}} + \frac {2 x^{2}}{e^{5}} + \frac {x}{e^{5}} + \frac {\left (- 4 x^{2} \log {\left (5 \right )} - 8 x \log {\left (5 \right )}\right ) \log {\left (\log {\left (x \right )} \right )}}{3 e^{5}} - \frac {4 \log {\left (5 \right )} \log {\left (\log {\left (x \right )} \right )}}{3 e^{5}} + \frac {\left (4 x^{2} \log {\left (5 \right )}^{2} + 8 x \log {\left (5 \right )}^{2} + 4 \log {\left (5 \right )}^{2}\right ) \log {\left (\log {\left (x \right )} \right )}^{2}}{9 x e^{5}} \] Input:
integrate(1/9*((4*x**2-4)*ln(5)**2*ln(x)*ln(ln(x))**2+((-24*x**3-24*x**2)* ln(5)*ln(x)+(8*x**2+16*x+8)*ln(5)**2)*ln(ln(x))+(27*x**4+36*x**3+9*x**2)*l n(x)+(-12*x**3-24*x**2-12*x)*ln(5))/x**2/exp(5)/ln(x),x)
Output:
x**3*exp(-5) + 2*x**2*exp(-5) + x*exp(-5) + (-4*x**2*log(5) - 8*x*log(5))* exp(-5)*log(log(x))/3 - 4*exp(-5)*log(5)*log(log(x))/3 + (4*x**2*log(5)**2 + 8*x*log(5)**2 + 4*log(5)**2)*exp(-5)*log(log(x))**2/(9*x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.20 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {1}{9} \, {\left (9 \, x^{3} + 18 \, x^{2} - 12 \, {\left (x^{2} \log \left (\log \left (x\right )\right ) - {\rm Ei}\left (2 \, \log \left (x\right )\right )\right )} \log \left (5\right ) - 24 \, {\left (x \log \left (\log \left (x\right )\right ) - {\rm Ei}\left (\log \left (x\right )\right )\right )} \log \left (5\right ) - 12 \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) \log \left (5\right ) - 24 \, {\rm Ei}\left (\log \left (x\right )\right ) \log \left (5\right ) - 12 \, \log \left (5\right ) \log \left (\log \left (x\right )\right ) + \frac {4 \, {\left (x^{2} \log \left (5\right )^{2} + 2 \, x \log \left (5\right )^{2} + \log \left (5\right )^{2}\right )} \log \left (\log \left (x\right )\right )^{2}}{x} + 9 \, x\right )} e^{\left (-5\right )} \] Input:
integrate(1/9*((4*x^2-4)*log(5)^2*log(x)*log(log(x))^2+((-24*x^3-24*x^2)*l og(5)*log(x)+(8*x^2+16*x+8)*log(5)^2)*log(log(x))+(27*x^4+36*x^3+9*x^2)*lo g(x)+(-12*x^3-24*x^2-12*x)*log(5))/x^2/exp(5)/log(x),x, algorithm="maxima" )
Output:
1/9*(9*x^3 + 18*x^2 - 12*(x^2*log(log(x)) - Ei(2*log(x)))*log(5) - 24*(x*l og(log(x)) - Ei(log(x)))*log(5) - 12*Ei(2*log(x))*log(5) - 24*Ei(log(x))*l og(5) - 12*log(5)*log(log(x)) + 4*(x^2*log(5)^2 + 2*x*log(5)^2 + log(5)^2) *log(log(x))^2/x + 9*x)*e^(-5)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.80 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {1}{9} \, {\left (9 \, x^{3} + 4 \, {\left (x \log \left (5\right )^{2} + 2 \, \log \left (5\right )^{2} + \frac {\log \left (5\right )^{2}}{x}\right )} \log \left (\log \left (x\right )\right )^{2} + 18 \, x^{2} - 12 \, {\left (x^{2} \log \left (5\right ) + 2 \, x \log \left (5\right )\right )} \log \left (\log \left (x\right )\right ) - 12 \, \log \left (5\right ) \log \left (\log \left (x\right )\right ) + 9 \, x\right )} e^{\left (-5\right )} \] Input:
integrate(1/9*((4*x^2-4)*log(5)^2*log(x)*log(log(x))^2+((-24*x^3-24*x^2)*l og(5)*log(x)+(8*x^2+16*x+8)*log(5)^2)*log(log(x))+(27*x^4+36*x^3+9*x^2)*lo g(x)+(-12*x^3-24*x^2-12*x)*log(5))/x^2/exp(5)/log(x),x, algorithm="giac")
Output:
1/9*(9*x^3 + 4*(x*log(5)^2 + 2*log(5)^2 + log(5)^2/x)*log(log(x))^2 + 18*x ^2 - 12*(x^2*log(5) + 2*x*log(5))*log(log(x)) - 12*log(5)*log(log(x)) + 9* x)*e^(-5)
Time = 3.87 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.76 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx={\ln \left (\ln \left (x\right )\right )}^2\,\left (\frac {8\,{\mathrm {e}}^{-5}\,{\ln \left (5\right )}^2}{9}+\frac {8\,x\,{\mathrm {e}}^{-5}\,{\ln \left (5\right )}^2}{9}-\frac {{\mathrm {e}}^{-5}\,\left (4\,x^2\,{\ln \left (5\right )}^2-4\,{\ln \left (5\right )}^2\right )}{9\,x}\right )+x\,{\mathrm {e}}^{-5}+2\,x^2\,{\mathrm {e}}^{-5}+x^3\,{\mathrm {e}}^{-5}-\frac {\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-5}\,\left (4\,\ln \left (5\right )\,x^2+8\,\ln \left (5\right )\,x\right )}{3}-\frac {4\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-5}\,\ln \left (5\right )}{3} \] Input:
int((exp(-5)*((log(log(x))*(log(5)^2*(16*x + 8*x^2 + 8) - log(5)*log(x)*(2 4*x^2 + 24*x^3)))/9 + (log(x)*(9*x^2 + 36*x^3 + 27*x^4))/9 - (log(5)*(12*x + 24*x^2 + 12*x^3))/9 + (log(log(x))^2*log(5)^2*log(x)*(4*x^2 - 4))/9))/( x^2*log(x)),x)
Output:
log(log(x))^2*((8*exp(-5)*log(5)^2)/9 + (8*x*exp(-5)*log(5)^2)/9 - (exp(-5 )*(4*x^2*log(5)^2 - 4*log(5)^2))/(9*x)) + x*exp(-5) + 2*x^2*exp(-5) + x^3* exp(-5) - (log(log(x))*exp(-5)*(8*x*log(5) + 4*x^2*log(5)))/3 - (4*log(log (x))*exp(-5)*log(5))/3
Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.56 \[ \int \frac {\left (-12 x-24 x^2-12 x^3\right ) \log (5)+\left (9 x^2+36 x^3+27 x^4\right ) \log (x)+\left (\left (8+16 x+8 x^2\right ) \log ^2(5)+\left (-24 x^2-24 x^3\right ) \log (5) \log (x)\right ) \log (\log (x))+\left (-4+4 x^2\right ) \log ^2(5) \log (x) \log ^2(\log (x))}{9 e^5 x^2 \log (x)} \, dx=\frac {4 \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2} \mathrm {log}\left (5\right )^{2} x^{2}+8 \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2} \mathrm {log}\left (5\right )^{2} x +4 \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2} \mathrm {log}\left (5\right )^{2}-12 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (5\right ) x^{3}-24 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (5\right ) x^{2}-12 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (5\right ) x +9 x^{4}+18 x^{3}+9 x^{2}}{9 e^{5} x} \] Input:
int(1/9*((4*x^2-4)*log(5)^2*log(x)*log(log(x))^2+((-24*x^3-24*x^2)*log(5)* log(x)+(8*x^2+16*x+8)*log(5)^2)*log(log(x))+(27*x^4+36*x^3+9*x^2)*log(x)+( -12*x^3-24*x^2-12*x)*log(5))/x^2/exp(5)/log(x),x)
Output:
(4*log(log(x))**2*log(5)**2*x**2 + 8*log(log(x))**2*log(5)**2*x + 4*log(lo g(x))**2*log(5)**2 - 12*log(log(x))*log(5)*x**3 - 24*log(log(x))*log(5)*x* *2 - 12*log(log(x))*log(5)*x + 9*x**4 + 18*x**3 + 9*x**2)/(9*e**5*x)