Integrand size = 184, antiderivative size = 30 \[ \int \frac {15 x^2-5 x^3-10 x^4+\left (20 x^2+80 x^3\right ) \log (x)-160 x^2 \log ^2(x)+\left (70 x^2-15 x^3+20 x^4+\left (70 x-15 x^2+20 x^3\right ) \log (5)+\left (-40 x^2-160 x^3+\left (-40 x-160 x^2\right ) \log (5)\right ) \log (x)+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)\right ) \log (x+\log (5))}{\left (x-2 x^2+x^3+\left (1-2 x+x^2\right ) \log (5)+\left (8 x-8 x^2+(8-8 x) \log (5)\right ) \log (x)+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx=1+\frac {5 x \left (2 x-\frac {5 x}{1-x+4 \log (x)}\right )}{\log (x+\log (5))} \] Output:
1+5*x/ln(ln(5)+x)*(2*x-x/(1/5-1/5*x+4/5*ln(x)))
Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {15 x^2-5 x^3-10 x^4+\left (20 x^2+80 x^3\right ) \log (x)-160 x^2 \log ^2(x)+\left (70 x^2-15 x^3+20 x^4+\left (70 x-15 x^2+20 x^3\right ) \log (5)+\left (-40 x^2-160 x^3+\left (-40 x-160 x^2\right ) \log (5)\right ) \log (x)+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)\right ) \log (x+\log (5))}{\left (x-2 x^2+x^3+\left (1-2 x+x^2\right ) \log (5)+\left (8 x-8 x^2+(8-8 x) \log (5)\right ) \log (x)+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx=\frac {5 x^2 (3+2 x-8 \log (x))}{(-1+x-4 \log (x)) \log (x+\log (5))} \] Input:
Integrate[(15*x^2 - 5*x^3 - 10*x^4 + (20*x^2 + 80*x^3)*Log[x] - 160*x^2*Lo g[x]^2 + (70*x^2 - 15*x^3 + 20*x^4 + (70*x - 15*x^2 + 20*x^3)*Log[5] + (-4 0*x^2 - 160*x^3 + (-40*x - 160*x^2)*Log[5])*Log[x] + (320*x^2 + 320*x*Log[ 5])*Log[x]^2)*Log[x + Log[5]])/((x - 2*x^2 + x^3 + (1 - 2*x + x^2)*Log[5] + (8*x - 8*x^2 + (8 - 8*x)*Log[5])*Log[x] + (16*x + 16*Log[5])*Log[x]^2)*L og[x + Log[5]]^2),x]
Output:
(5*x^2*(3 + 2*x - 8*Log[x]))/((-1 + x - 4*Log[x])*Log[x + Log[5]])
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.
| \(\displaystyle \int \frac {-10 x^4-5 x^3+15 x^2-160 x^2 \log ^2(x)+\left (80 x^3+20 x^2\right ) \log (x)+\left (20 x^4-15 x^3+70 x^2+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)+\left (-160 x^3-40 x^2+\left (-160 x^2-40 x\right ) \log (5)\right ) \log (x)+\left (20 x^3-15 x^2+70 x\right ) \log (5)\right ) \log (x+\log (5))}{\left (x^3-2 x^2+\left (-8 x^2+8 x+(8-8 x) \log (5)\right ) \log (x)+\left (x^2-2 x+1\right ) \log (5)+x+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {5 x \left (-x \left (2 x^2+x-3\right )+\left (4 x^2-3 x+14\right ) (x+\log (5)) \log (x+\log (5))+32 (2 (x+\log (5)) \log (x+\log (5))-x) \log ^2(x)-4 (4 x+1) (2 (x+\log (5)) \log (x+\log (5))-x) \log (x)\right )}{(x+\log (5)) (-x+4 \log (x)+1)^2 \log ^2(x+\log (5))}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 5 \int \frac {x \left (-32 (x-2 (x+\log (5)) \log (x+\log (5))) \log ^2(x)+4 (4 x+1) (x-2 (x+\log (5)) \log (x+\log (5))) \log (x)+x \left (-2 x^2-x+3\right )+\left (4 x^2-3 x+14\right ) (x+\log (5)) \log (x+\log (5))\right )}{(x+\log (5)) (-x+4 \log (x)+1)^2 \log ^2(x+\log (5))}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 5 \int \left (\frac {x \left (4 x^2-32 \log (x) x-3 x+64 \log ^2(x)-8 \log (x)+14\right )}{(x-4 \log (x)-1)^2 \log (x+\log (5))}-\frac {x^2 (2 x-8 \log (x)+3)}{(x+\log (5)) (x-4 \log (x)-1) \log ^2(x+\log (5))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 5 \left (4 \int \frac {x^3}{(x-4 \log (x)-1)^2 \log (x+\log (5))}dx-2 \int \frac {x^2}{(x-4 \log (x)-1) \log ^2(x+\log (5))}dx-3 \int \frac {x^2}{(x-4 \log (x)-1)^2 \log (x+\log (5))}dx-32 \int \frac {x^2 \log (x)}{(x-4 \log (x)-1)^2 \log (x+\log (5))}dx-2 \log ^2(5) \int \frac {1}{(x-4 \log (x)-1) \log ^2(x+\log (5))}dx+3 \log (5) \int \frac {1}{(x-4 \log (x)-1) \log ^2(x+\log (5))}dx+2 \log (5) \int \frac {x}{(x-4 \log (x)-1) \log ^2(x+\log (5))}dx-3 \int \frac {x}{(x-4 \log (x)-1) \log ^2(x+\log (5))}dx-3 \log ^2(5) \int \frac {1}{(x+\log (5)) (x-4 \log (x)-1) \log ^2(x+\log (5))}dx-8 \log (5) \int \frac {\log (x)}{(x-4 \log (x)-1) \log ^2(x+\log (5))}dx+8 \int \frac {x \log (x)}{(x-4 \log (x)-1) \log ^2(x+\log (5))}dx+8 \log ^2(5) \int \frac {\log (x)}{(x+\log (5)) (x-4 \log (x)-1) \log ^2(x+\log (5))}dx+64 \int \frac {x \log ^2(x)}{(x-4 \log (x)-1)^2 \log (x+\log (5))}dx+2 \log ^3(5) \int \frac {1}{(x+\log (5)) (x-4 \log (x)-1) \log ^2(x+\log (5))}dx+14 \int \frac {x}{(x-4 \log (x)-1)^2 \log (x+\log (5))}dx-8 \int \frac {x \log (x)}{(x-4 \log (x)-1)^2 \log (x+\log (5))}dx\right )\) |
Input:
Int[(15*x^2 - 5*x^3 - 10*x^4 + (20*x^2 + 80*x^3)*Log[x] - 160*x^2*Log[x]^2 + (70*x^2 - 15*x^3 + 20*x^4 + (70*x - 15*x^2 + 20*x^3)*Log[5] + (-40*x^2 - 160*x^3 + (-40*x - 160*x^2)*Log[5])*Log[x] + (320*x^2 + 320*x*Log[5])*Lo g[x]^2)*Log[x + Log[5]])/((x - 2*x^2 + x^3 + (1 - 2*x + x^2)*Log[5] + (8*x - 8*x^2 + (8 - 8*x)*Log[5])*Log[x] + (16*x + 16*Log[5])*Log[x]^2)*Log[x + Log[5]]^2),x]
Output:
$Aborted
Time = 106.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03
| method | result | size |
| risch | \(\frac {5 \left (2 x -8 \ln \left (x \right )+3\right ) x^{2}}{\left (x -4 \ln \left (x \right )-1\right ) \ln \left (\ln \left (5\right )+x \right )}\) | \(31\) |
| parallelrisch | \(-\frac {-40 x^{3}+160 x^{2} \ln \left (x \right )-60 x^{2}}{4 \left (x -4 \ln \left (x \right )-1\right ) \ln \left (\ln \left (5\right )+x \right )}\) | \(37\) |
Input:
int((((320*x*ln(5)+320*x^2)*ln(x)^2+((-160*x^2-40*x)*ln(5)-160*x^3-40*x^2) *ln(x)+(20*x^3-15*x^2+70*x)*ln(5)+20*x^4-15*x^3+70*x^2)*ln(ln(5)+x)-160*x^ 2*ln(x)^2+(80*x^3+20*x^2)*ln(x)-10*x^4-5*x^3+15*x^2)/((16*ln(5)+16*x)*ln(x )^2+((-8*x+8)*ln(5)-8*x^2+8*x)*ln(x)+(x^2-2*x+1)*ln(5)+x^3-2*x^2+x)/ln(ln( 5)+x)^2,x,method=_RETURNVERBOSE)
Output:
5*(2*x-8*ln(x)+3)*x^2/(x-4*ln(x)-1)/ln(ln(5)+x)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {15 x^2-5 x^3-10 x^4+\left (20 x^2+80 x^3\right ) \log (x)-160 x^2 \log ^2(x)+\left (70 x^2-15 x^3+20 x^4+\left (70 x-15 x^2+20 x^3\right ) \log (5)+\left (-40 x^2-160 x^3+\left (-40 x-160 x^2\right ) \log (5)\right ) \log (x)+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)\right ) \log (x+\log (5))}{\left (x-2 x^2+x^3+\left (1-2 x+x^2\right ) \log (5)+\left (8 x-8 x^2+(8-8 x) \log (5)\right ) \log (x)+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx=\frac {5 \, {\left (2 \, x^{3} - 8 \, x^{2} \log \left (x\right ) + 3 \, x^{2}\right )}}{{\left (x - 4 \, \log \left (x\right ) - 1\right )} \log \left (x + \log \left (5\right )\right )} \] Input:
integrate((((320*x*log(5)+320*x^2)*log(x)^2+((-160*x^2-40*x)*log(5)-160*x^ 3-40*x^2)*log(x)+(20*x^3-15*x^2+70*x)*log(5)+20*x^4-15*x^3+70*x^2)*log(log (5)+x)-160*x^2*log(x)^2+(80*x^3+20*x^2)*log(x)-10*x^4-5*x^3+15*x^2)/((16*l og(5)+16*x)*log(x)^2+((-8*x+8)*log(5)-8*x^2+8*x)*log(x)+(x^2-2*x+1)*log(5) +x^3-2*x^2+x)/log(log(5)+x)^2,x, algorithm="fricas")
Output:
5*(2*x^3 - 8*x^2*log(x) + 3*x^2)/((x - 4*log(x) - 1)*log(x + log(5)))
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {15 x^2-5 x^3-10 x^4+\left (20 x^2+80 x^3\right ) \log (x)-160 x^2 \log ^2(x)+\left (70 x^2-15 x^3+20 x^4+\left (70 x-15 x^2+20 x^3\right ) \log (5)+\left (-40 x^2-160 x^3+\left (-40 x-160 x^2\right ) \log (5)\right ) \log (x)+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)\right ) \log (x+\log (5))}{\left (x-2 x^2+x^3+\left (1-2 x+x^2\right ) \log (5)+\left (8 x-8 x^2+(8-8 x) \log (5)\right ) \log (x)+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx=\frac {10 x^{3} - 40 x^{2} \log {\left (x \right )} + 15 x^{2}}{\left (x - 4 \log {\left (x \right )} - 1\right ) \log {\left (x + \log {\left (5 \right )} \right )}} \] Input:
integrate((((320*x*ln(5)+320*x**2)*ln(x)**2+((-160*x**2-40*x)*ln(5)-160*x* *3-40*x**2)*ln(x)+(20*x**3-15*x**2+70*x)*ln(5)+20*x**4-15*x**3+70*x**2)*ln (ln(5)+x)-160*x**2*ln(x)**2+(80*x**3+20*x**2)*ln(x)-10*x**4-5*x**3+15*x**2 )/((16*ln(5)+16*x)*ln(x)**2+((-8*x+8)*ln(5)-8*x**2+8*x)*ln(x)+(x**2-2*x+1) *ln(5)+x**3-2*x**2+x)/ln(ln(5)+x)**2,x)
Output:
(10*x**3 - 40*x**2*log(x) + 15*x**2)/((x - 4*log(x) - 1)*log(x + log(5)))
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {15 x^2-5 x^3-10 x^4+\left (20 x^2+80 x^3\right ) \log (x)-160 x^2 \log ^2(x)+\left (70 x^2-15 x^3+20 x^4+\left (70 x-15 x^2+20 x^3\right ) \log (5)+\left (-40 x^2-160 x^3+\left (-40 x-160 x^2\right ) \log (5)\right ) \log (x)+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)\right ) \log (x+\log (5))}{\left (x-2 x^2+x^3+\left (1-2 x+x^2\right ) \log (5)+\left (8 x-8 x^2+(8-8 x) \log (5)\right ) \log (x)+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx=\frac {5 \, {\left (2 \, x^{3} - 8 \, x^{2} \log \left (x\right ) + 3 \, x^{2}\right )}}{{\left (x - 4 \, \log \left (x\right ) - 1\right )} \log \left (x + \log \left (5\right )\right )} \] Input:
integrate((((320*x*log(5)+320*x^2)*log(x)^2+((-160*x^2-40*x)*log(5)-160*x^ 3-40*x^2)*log(x)+(20*x^3-15*x^2+70*x)*log(5)+20*x^4-15*x^3+70*x^2)*log(log (5)+x)-160*x^2*log(x)^2+(80*x^3+20*x^2)*log(x)-10*x^4-5*x^3+15*x^2)/((16*l og(5)+16*x)*log(x)^2+((-8*x+8)*log(5)-8*x^2+8*x)*log(x)+(x^2-2*x+1)*log(5) +x^3-2*x^2+x)/log(log(5)+x)^2,x, algorithm="maxima")
Output:
5*(2*x^3 - 8*x^2*log(x) + 3*x^2)/((x - 4*log(x) - 1)*log(x + log(5)))
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {15 x^2-5 x^3-10 x^4+\left (20 x^2+80 x^3\right ) \log (x)-160 x^2 \log ^2(x)+\left (70 x^2-15 x^3+20 x^4+\left (70 x-15 x^2+20 x^3\right ) \log (5)+\left (-40 x^2-160 x^3+\left (-40 x-160 x^2\right ) \log (5)\right ) \log (x)+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)\right ) \log (x+\log (5))}{\left (x-2 x^2+x^3+\left (1-2 x+x^2\right ) \log (5)+\left (8 x-8 x^2+(8-8 x) \log (5)\right ) \log (x)+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx=\frac {5 \, {\left (2 \, x^{3} - 8 \, x^{2} \log \left (x\right ) + 3 \, x^{2}\right )}}{x \log \left (x + \log \left (5\right )\right ) - 4 \, \log \left (x + \log \left (5\right )\right ) \log \left (x\right ) - \log \left (x + \log \left (5\right )\right )} \] Input:
integrate((((320*x*log(5)+320*x^2)*log(x)^2+((-160*x^2-40*x)*log(5)-160*x^ 3-40*x^2)*log(x)+(20*x^3-15*x^2+70*x)*log(5)+20*x^4-15*x^3+70*x^2)*log(log (5)+x)-160*x^2*log(x)^2+(80*x^3+20*x^2)*log(x)-10*x^4-5*x^3+15*x^2)/((16*l og(5)+16*x)*log(x)^2+((-8*x+8)*log(5)-8*x^2+8*x)*log(x)+(x^2-2*x+1)*log(5) +x^3-2*x^2+x)/log(log(5)+x)^2,x, algorithm="giac")
Output:
5*(2*x^3 - 8*x^2*log(x) + 3*x^2)/(x*log(x + log(5)) - 4*log(x + log(5))*lo g(x) - log(x + log(5)))
Timed out. \[ \int \frac {15 x^2-5 x^3-10 x^4+\left (20 x^2+80 x^3\right ) \log (x)-160 x^2 \log ^2(x)+\left (70 x^2-15 x^3+20 x^4+\left (70 x-15 x^2+20 x^3\right ) \log (5)+\left (-40 x^2-160 x^3+\left (-40 x-160 x^2\right ) \log (5)\right ) \log (x)+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)\right ) \log (x+\log (5))}{\left (x-2 x^2+x^3+\left (1-2 x+x^2\right ) \log (5)+\left (8 x-8 x^2+(8-8 x) \log (5)\right ) \log (x)+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx=-\int -\frac {\ln \left (x\right )\,\left (80\,x^3+20\,x^2\right )+\ln \left (x+\ln \left (5\right )\right )\,\left (\ln \left (5\right )\,\left (20\,x^3-15\,x^2+70\,x\right )-\ln \left (x\right )\,\left (\ln \left (5\right )\,\left (160\,x^2+40\,x\right )+40\,x^2+160\,x^3\right )+{\ln \left (x\right )}^2\,\left (320\,x^2+320\,\ln \left (5\right )\,x\right )+70\,x^2-15\,x^3+20\,x^4\right )-160\,x^2\,{\ln \left (x\right )}^2+15\,x^2-5\,x^3-10\,x^4}{{\ln \left (x+\ln \left (5\right )\right )}^2\,\left (x+{\ln \left (x\right )}^2\,\left (16\,x+16\,\ln \left (5\right )\right )-\ln \left (x\right )\,\left (\ln \left (5\right )\,\left (8\,x-8\right )-8\,x+8\,x^2\right )-2\,x^2+x^3+\ln \left (5\right )\,\left (x^2-2\,x+1\right )\right )} \,d x \] Input:
int((log(x)*(20*x^2 + 80*x^3) + log(x + log(5))*(log(5)*(70*x - 15*x^2 + 2 0*x^3) - log(x)*(log(5)*(40*x + 160*x^2) + 40*x^2 + 160*x^3) + log(x)^2*(3 20*x*log(5) + 320*x^2) + 70*x^2 - 15*x^3 + 20*x^4) - 160*x^2*log(x)^2 + 15 *x^2 - 5*x^3 - 10*x^4)/(log(x + log(5))^2*(x + log(x)^2*(16*x + 16*log(5)) - log(x)*(log(5)*(8*x - 8) - 8*x + 8*x^2) - 2*x^2 + x^3 + log(5)*(x^2 - 2 *x + 1))),x)
Output:
-int(-(log(x)*(20*x^2 + 80*x^3) + log(x + log(5))*(log(5)*(70*x - 15*x^2 + 20*x^3) - log(x)*(log(5)*(40*x + 160*x^2) + 40*x^2 + 160*x^3) + log(x)^2* (320*x*log(5) + 320*x^2) + 70*x^2 - 15*x^3 + 20*x^4) - 160*x^2*log(x)^2 + 15*x^2 - 5*x^3 - 10*x^4)/(log(x + log(5))^2*(x + log(x)^2*(16*x + 16*log(5 )) - log(x)*(log(5)*(8*x - 8) - 8*x + 8*x^2) - 2*x^2 + x^3 + log(5)*(x^2 - 2*x + 1))), x)
Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {15 x^2-5 x^3-10 x^4+\left (20 x^2+80 x^3\right ) \log (x)-160 x^2 \log ^2(x)+\left (70 x^2-15 x^3+20 x^4+\left (70 x-15 x^2+20 x^3\right ) \log (5)+\left (-40 x^2-160 x^3+\left (-40 x-160 x^2\right ) \log (5)\right ) \log (x)+\left (320 x^2+320 x \log (5)\right ) \log ^2(x)\right ) \log (x+\log (5))}{\left (x-2 x^2+x^3+\left (1-2 x+x^2\right ) \log (5)+\left (8 x-8 x^2+(8-8 x) \log (5)\right ) \log (x)+(16 x+16 \log (5)) \log ^2(x)\right ) \log ^2(x+\log (5))} \, dx=\frac {5 x^{2} \left (8 \,\mathrm {log}\left (x \right )-2 x -3\right )}{\mathrm {log}\left (\mathrm {log}\left (5\right )+x \right ) \left (4 \,\mathrm {log}\left (x \right )-x +1\right )} \] Input:
int((((320*x*log(5)+320*x^2)*log(x)^2+((-160*x^2-40*x)*log(5)-160*x^3-40*x ^2)*log(x)+(20*x^3-15*x^2+70*x)*log(5)+20*x^4-15*x^3+70*x^2)*log(log(5)+x) -160*x^2*log(x)^2+(80*x^3+20*x^2)*log(x)-10*x^4-5*x^3+15*x^2)/((16*log(5)+ 16*x)*log(x)^2+((-8*x+8)*log(5)-8*x^2+8*x)*log(x)+(x^2-2*x+1)*log(5)+x^3-2 *x^2+x)/log(log(5)+x)^2,x)
Output:
(5*x**2*(8*log(x) - 2*x - 3))/(log(log(5) + x)*(4*log(x) - x + 1))