Integrand size = 160, antiderivative size = 29 \[ \int \frac {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx=\frac {x}{5 \left (-x+x^2+\frac {\log (x)}{\frac {-2+x}{x}+\log (4)}\right )} \] Output:
x/(5/(2*ln(2)+(-2+x)/x)*ln(x)+5*x^2-5*x)
Time = 5.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx=\frac {-2+x+x \log (4)}{5 ((-1+x) (-2+x+x \log (4))+\log (x))} \] Input:
Integrate[(2 - 5*x + 4*x^2 - x^3 + (-x + 4*x^2 - 2*x^3)*Log[4] - x^3*Log[4 ]^2 + (x + x*Log[4])*Log[x])/(20*x - 60*x^2 + 65*x^3 - 30*x^4 + 5*x^5 + (- 20*x^2 + 50*x^3 - 40*x^4 + 10*x^5)*Log[4] + (5*x^3 - 10*x^4 + 5*x^5)*Log[4 ]^2 + (20*x - 30*x^2 + 10*x^3 + (-10*x^2 + 10*x^3)*Log[4])*Log[x] + 5*x*Lo g[x]^2),x]
Output:
(-2 + x + x*Log[4])/(5*((-1 + x)*(-2 + x + x*Log[4]) + 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.
| \(\displaystyle \int \frac {-x^3+x^3 \left (-\log ^2(4)\right )+4 x^2+\left (-2 x^3+4 x^2-x\right ) \log (4)-5 x+(x+x \log (4)) \log (x)+2}{5 x^5-30 x^4+65 x^3-60 x^2+\left (10 x^3-30 x^2+\left (10 x^3-10 x^2\right ) \log (4)+20 x\right ) \log (x)+\left (5 x^5-10 x^4+5 x^3\right ) \log ^2(4)+\left (10 x^5-40 x^4+50 x^3-20 x^2\right ) \log (4)+20 x+5 x \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {x^3 \left (-1-\log ^2(4)\right )+4 x^2+\left (-2 x^3+4 x^2-x\right ) \log (4)-5 x+(x+x \log (4)) \log (x)+2}{5 x^5-30 x^4+65 x^3-60 x^2+\left (10 x^3-30 x^2+\left (10 x^3-10 x^2\right ) \log (4)+20 x\right ) \log (x)+\left (5 x^5-10 x^4+5 x^3\right ) \log ^2(4)+\left (10 x^5-40 x^4+50 x^3-20 x^2\right ) \log (4)+20 x+5 x \log ^2(x)}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {x^3 \left (-(1+\log (4))^2\right )+4 x^2 (1+\log (4))+x (1+\log (4)) \log (x)-x (5+\log (4))+2}{5 x ((x-1) (x+x \log (4)-2)+\log (x))^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {-(1+\log (4))^2 x^3+4 (1+\log (4)) x^2+(1+\log (4)) \log (x) x-(5+\log (4)) x+2}{x ((1-x) (-\log (4) x-x+2)+\log (x))^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{5} \int \left (\frac {(1-2 x) \left ((1+\log (4))^2 x^2-3 (1+\log (4)) x+2\right )}{x \left ((1+\log (4)) x^2-3 \left (1+\frac {2 \log (2)}{3}\right ) x+\log (x)+2\right )^2}+\frac {1+\log (4)}{(1+\log (4)) x^2-3 \left (1+\frac {2 \log (2)}{3}\right ) x+\log (x)+2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (-\left ((7+\log (64)) \int \frac {1}{\left ((1+\log (4)) x^2-3 \left (1+\frac {2 \log (2)}{3}\right ) x+\log (x)+2\right )^2}dx\right )+2 \int \frac {1}{x \left ((1+\log (4)) x^2-3 \left (1+\frac {2 \log (2)}{3}\right ) x+\log (x)+2\right )^2}dx+(1+\log (4)) (7+\log (4)) \int \frac {x}{\left ((1+\log (4)) x^2-3 \left (1+\frac {2 \log (2)}{3}\right ) x+\log (x)+2\right )^2}dx-2 (1+\log (4))^2 \int \frac {x^2}{\left ((1+\log (4)) x^2-3 \left (1+\frac {2 \log (2)}{3}\right ) x+\log (x)+2\right )^2}dx+(1+\log (4)) \int \frac {1}{(1+\log (4)) x^2-3 \left (1+\frac {2 \log (2)}{3}\right ) x+\log (x)+2}dx\right )\) |
Input:
Int[(2 - 5*x + 4*x^2 - x^3 + (-x + 4*x^2 - 2*x^3)*Log[4] - x^3*Log[4]^2 + (x + x*Log[4])*Log[x])/(20*x - 60*x^2 + 65*x^3 - 30*x^4 + 5*x^5 + (-20*x^2 + 50*x^3 - 40*x^4 + 10*x^5)*Log[4] + (5*x^3 - 10*x^4 + 5*x^5)*Log[4]^2 + (20*x - 30*x^2 + 10*x^3 + (-10*x^2 + 10*x^3)*Log[4])*Log[x] + 5*x*Log[x]^2 ),x]
Output:
$Aborted
Time = 0.82 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21
| method | result | size |
| risch | \(\frac {2 x \ln \left (2\right )+x -2}{10 x^{2} \ln \left (2\right )-10 x \ln \left (2\right )+5 x^{2}+5 \ln \left (x \right )-15 x +10}\) | \(35\) |
| parallelrisch | \(\frac {2 x \ln \left (2\right )+x -2}{10 x^{2} \ln \left (2\right )-10 x \ln \left (2\right )+5 x^{2}+5 \ln \left (x \right )-15 x +10}\) | \(35\) |
| norman | \(\frac {-\frac {2}{5}+\left (\frac {2 \ln \left (2\right )}{5}+\frac {1}{5}\right ) x}{2 x^{2} \ln \left (2\right )-2 x \ln \left (2\right )+x^{2}+\ln \left (x \right )-3 x +2}\) | \(36\) |
| default | \(-\frac {\left (-2 \ln \left (2\right )-1\right ) x +2}{5 \left (2 x^{2} \ln \left (2\right )-2 x \ln \left (2\right )+x^{2}+\ln \left (x \right )-3 x +2\right )}\) | \(37\) |
Input:
int(((x+2*x*ln(2))*ln(x)-4*x^3*ln(2)^2+2*(-2*x^3+4*x^2-x)*ln(2)-x^3+4*x^2- 5*x+2)/(5*x*ln(x)^2+(2*(10*x^3-10*x^2)*ln(2)+10*x^3-30*x^2+20*x)*ln(x)+4*( 5*x^5-10*x^4+5*x^3)*ln(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2)*ln(2)+5*x^5-30 *x^4+65*x^3-60*x^2+20*x),x,method=_RETURNVERBOSE)
Output:
1/5*(2*x*ln(2)+x-2)/(2*x^2*ln(2)-2*x*ln(2)+x^2+ln(x)-3*x+2)
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx=\frac {2 \, x \log \left (2\right ) + x - 2}{5 \, {\left (x^{2} + 2 \, {\left (x^{2} - x\right )} \log \left (2\right ) - 3 \, x + \log \left (x\right ) + 2\right )}} \] Input:
integrate(((x+2*x*log(2))*log(x)-4*x^3*log(2)^2+2*(-2*x^3+4*x^2-x)*log(2)- x^3+4*x^2-5*x+2)/(5*x*log(x)^2+(2*(10*x^3-10*x^2)*log(2)+10*x^3-30*x^2+20* x)*log(x)+4*(5*x^5-10*x^4+5*x^3)*log(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2)* log(2)+5*x^5-30*x^4+65*x^3-60*x^2+20*x),x, algorithm="fricas")
Output:
1/5*(2*x*log(2) + x - 2)/(x^2 + 2*(x^2 - x)*log(2) - 3*x + log(x) + 2)
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx=\frac {x + 2 x \log {\left (2 \right )} - 2}{5 x^{2} + 10 x^{2} \log {\left (2 \right )} - 15 x - 10 x \log {\left (2 \right )} + 5 \log {\left (x \right )} + 10} \] Input:
integrate(((x+2*x*ln(2))*ln(x)-4*x**3*ln(2)**2+2*(-2*x**3+4*x**2-x)*ln(2)- x**3+4*x**2-5*x+2)/(5*x*ln(x)**2+(2*(10*x**3-10*x**2)*ln(2)+10*x**3-30*x** 2+20*x)*ln(x)+4*(5*x**5-10*x**4+5*x**3)*ln(2)**2+2*(10*x**5-40*x**4+50*x** 3-20*x**2)*ln(2)+5*x**5-30*x**4+65*x**3-60*x**2+20*x),x)
Output:
(x + 2*x*log(2) - 2)/(5*x**2 + 10*x**2*log(2) - 15*x - 10*x*log(2) + 5*log (x) + 10)
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx=\frac {x {\left (2 \, \log \left (2\right ) + 1\right )} - 2}{5 \, {\left (x^{2} {\left (2 \, \log \left (2\right ) + 1\right )} - x {\left (2 \, \log \left (2\right ) + 3\right )} + \log \left (x\right ) + 2\right )}} \] Input:
integrate(((x+2*x*log(2))*log(x)-4*x^3*log(2)^2+2*(-2*x^3+4*x^2-x)*log(2)- x^3+4*x^2-5*x+2)/(5*x*log(x)^2+(2*(10*x^3-10*x^2)*log(2)+10*x^3-30*x^2+20* x)*log(x)+4*(5*x^5-10*x^4+5*x^3)*log(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2)* log(2)+5*x^5-30*x^4+65*x^3-60*x^2+20*x),x, algorithm="maxima")
Output:
1/5*(x*(2*log(2) + 1) - 2)/(x^2*(2*log(2) + 1) - x*(2*log(2) + 3) + log(x) + 2)
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx=\frac {2 \, x \log \left (2\right ) + x - 2}{5 \, {\left (2 \, x^{2} \log \left (2\right ) + x^{2} - 2 \, x \log \left (2\right ) - 3 \, x + \log \left (x\right ) + 2\right )}} \] Input:
integrate(((x+2*x*log(2))*log(x)-4*x^3*log(2)^2+2*(-2*x^3+4*x^2-x)*log(2)- x^3+4*x^2-5*x+2)/(5*x*log(x)^2+(2*(10*x^3-10*x^2)*log(2)+10*x^3-30*x^2+20* x)*log(x)+4*(5*x^5-10*x^4+5*x^3)*log(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2)* log(2)+5*x^5-30*x^4+65*x^3-60*x^2+20*x),x, algorithm="giac")
Output:
1/5*(2*x*log(2) + x - 2)/(2*x^2*log(2) + x^2 - 2*x*log(2) - 3*x + log(x) + 2)
Timed out. \[ \int \frac {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx=\int -\frac {5\,x+4\,x^3\,{\ln \left (2\right )}^2+2\,\ln \left (2\right )\,\left (2\,x^3-4\,x^2+x\right )-\ln \left (x\right )\,\left (x+2\,x\,\ln \left (2\right )\right )-4\,x^2+x^3-2}{20\,x-2\,\ln \left (2\right )\,\left (-10\,x^5+40\,x^4-50\,x^3+20\,x^2\right )+5\,x\,{\ln \left (x\right )}^2+4\,{\ln \left (2\right )}^2\,\left (5\,x^5-10\,x^4+5\,x^3\right )+\ln \left (x\right )\,\left (20\,x-2\,\ln \left (2\right )\,\left (10\,x^2-10\,x^3\right )-30\,x^2+10\,x^3\right )-60\,x^2+65\,x^3-30\,x^4+5\,x^5} \,d x \] Input:
int(-(5*x + 4*x^3*log(2)^2 + 2*log(2)*(x - 4*x^2 + 2*x^3) - log(x)*(x + 2* x*log(2)) - 4*x^2 + x^3 - 2)/(20*x - 2*log(2)*(20*x^2 - 50*x^3 + 40*x^4 - 10*x^5) + 5*x*log(x)^2 + 4*log(2)^2*(5*x^3 - 10*x^4 + 5*x^5) + log(x)*(20* x - 2*log(2)*(10*x^2 - 10*x^3) - 30*x^2 + 10*x^3) - 60*x^2 + 65*x^3 - 30*x ^4 + 5*x^5),x)
Output:
int(-(5*x + 4*x^3*log(2)^2 + 2*log(2)*(x - 4*x^2 + 2*x^3) - log(x)*(x + 2* x*log(2)) - 4*x^2 + x^3 - 2)/(20*x - 2*log(2)*(20*x^2 - 50*x^3 + 40*x^4 - 10*x^5) + 5*x*log(x)^2 + 4*log(2)^2*(5*x^3 - 10*x^4 + 5*x^5) + log(x)*(20* x - 2*log(2)*(10*x^2 - 10*x^3) - 30*x^2 + 10*x^3) - 60*x^2 + 65*x^3 - 30*x ^4 + 5*x^5), x)
Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx=\frac {2 \,\mathrm {log}\left (2\right ) x +x -2}{5 \,\mathrm {log}\left (x \right )+10 \,\mathrm {log}\left (2\right ) x^{2}-10 \,\mathrm {log}\left (2\right ) x +5 x^{2}-15 x +10} \] Input:
int(((x+2*x*log(2))*log(x)-4*x^3*log(2)^2+2*(-2*x^3+4*x^2-x)*log(2)-x^3+4* x^2-5*x+2)/(5*x*log(x)^2+(2*(10*x^3-10*x^2)*log(2)+10*x^3-30*x^2+20*x)*log (x)+4*(5*x^5-10*x^4+5*x^3)*log(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2)*log(2) +5*x^5-30*x^4+65*x^3-60*x^2+20*x),x)
Output:
(2*log(2)*x + x - 2)/(5*(log(x) + 2*log(2)*x**2 - 2*log(2)*x + x**2 - 3*x + 2))