Integrand size = 153, antiderivative size = 32 \[ \int \frac {\left (-1-7 x+8 x^2\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )+\left (-x-16 x^2+(1+16 x) \log (x)+\left (x+8 x^2+(-1-8 x) \log (x)\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )\right ) \log (-25 x+25 \log (x))+\left (-2 x^3-16 x^4+\left (2 x^2+16 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))}{\left (-x^3-8 x^4+\left (x^2+8 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))} \, dx=-3+2 x+\frac {\log \left (\frac {x}{4}+2 x^2\right )}{x \log (25 (-x+\log (x)))} \]
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {\left (-1-7 x+8 x^2\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )+\left (-x-16 x^2+(1+16 x) \log (x)+\left (x+8 x^2+(-1-8 x) \log (x)\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )\right ) \log (-25 x+25 \log (x))+\left (-2 x^3-16 x^4+\left (2 x^2+16 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))}{\left (-x^3-8 x^4+\left (x^2+8 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))} \, dx=2 x+\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x \log (25 (-x+\log (x)))} \]
Integrate[((-1 - 7*x + 8*x^2)*Log[(x + 8*x^2)/4] + (-x - 16*x^2 + (1 + 16* x)*Log[x] + (x + 8*x^2 + (-1 - 8*x)*Log[x])*Log[(x + 8*x^2)/4])*Log[-25*x + 25*Log[x]] + (-2*x^3 - 16*x^4 + (2*x^2 + 16*x^3)*Log[x])*Log[-25*x + 25* Log[x]]^2)/((-x^3 - 8*x^4 + (x^2 + 8*x^3)*Log[x])*Log[-25*x + 25*Log[x]]^2 ),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.
| \(\displaystyle \int \frac {\left (-16 x^2+\left (8 x^2+x+(-8 x-1) \log (x)\right ) \log \left (\frac {1}{4} \left (8 x^2+x\right )\right )-x+(16 x+1) \log (x)\right ) \log (25 \log (x)-25 x)+\left (8 x^2-7 x-1\right ) \log \left (\frac {1}{4} \left (8 x^2+x\right )\right )+\left (-16 x^4-2 x^3+\left (16 x^3+2 x^2\right ) \log (x)\right ) \log ^2(25 \log (x)-25 x)}{\left (-8 x^4-x^3+\left (8 x^3+x^2\right ) \log (x)\right ) \log ^2(25 \log (x)-25 x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-\left (-16 x^2+\left (8 x^2+x+(-8 x-1) \log (x)\right ) \log \left (\frac {1}{4} \left (8 x^2+x\right )\right )-x+(16 x+1) \log (x)\right ) \log (25 \log (x)-25 x)-\left (8 x^2-7 x-1\right ) \log \left (\frac {1}{4} \left (8 x^2+x\right )\right )-\left (\left (-16 x^4-2 x^3+\left (16 x^3+2 x^2\right ) \log (x)\right ) \log ^2(25 \log (x)-25 x)\right )}{x^2 (8 x+1) (x-\log (x)) \log ^2(-25 (x-\log (x)))}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {(x-1) \log \left (\frac {1}{4} x (8 x+1)\right )}{x^2 (x-\log (x)) \log ^2(25 (\log (x)-x))}+\frac {16 x-8 x \log \left (\frac {1}{4} x (8 x+1)\right )-\log \left (\frac {1}{4} x (8 x+1)\right )+1}{x^2 (8 x+1) \log (25 (\log (x)-x))}+2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {\log \left (\frac {1}{4} x (8 x+1)\right )}{x^2 (x-\log (x)) \log ^2(25 (\log (x)-x))}dx+\int \frac {1}{x^2 \log (25 (\log (x)-x))}dx-\int \frac {\log \left (\frac {1}{4} x (8 x+1)\right )}{x^2 \log (25 (\log (x)-x))}dx-\int \frac {\log \left (\frac {1}{4} x (8 x+1)\right )}{x (x-\log (x)) \log ^2(25 (\log (x)-x))}dx+8 \int \frac {1}{x \log (25 (\log (x)-x))}dx-64 \int \frac {1}{(8 x+1) \log (25 (\log (x)-x))}dx+2 x\) |
Int[((-1 - 7*x + 8*x^2)*Log[(x + 8*x^2)/4] + (-x - 16*x^2 + (1 + 16*x)*Log [x] + (x + 8*x^2 + (-1 - 8*x)*Log[x])*Log[(x + 8*x^2)/4])*Log[-25*x + 25*L og[x]] + (-2*x^3 - 16*x^4 + (2*x^2 + 16*x^3)*Log[x])*Log[-25*x + 25*Log[x] ]^2)/((-x^3 - 8*x^4 + (x^2 + 8*x^3)*Log[x])*Log[-25*x + 25*Log[x]]^2),x]
3.30.24.3.1 Defintions of rubi rules used
Time = 10.89 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75
| method | result | size |
| parallelrisch | \(\frac {512 x^{2} \ln \left (25 \ln \left (x \right )-25 x \right )-32 \ln \left (25 \ln \left (x \right )-25 x \right ) x +256 \ln \left (2 x^{2}+\frac {1}{4} x \right )}{256 x \ln \left (25 \ln \left (x \right )-25 x \right )}\) | \(56\) |
| risch | \(2 x -\frac {-i \pi \,\operatorname {csgn}\left (i \left (x +\frac {1}{8}\right )\right ) \operatorname {csgn}\left (i x \left (x +\frac {1}{8}\right )\right )^{2}+i \pi \,\operatorname {csgn}\left (i \left (x +\frac {1}{8}\right )\right ) \operatorname {csgn}\left (i x \left (x +\frac {1}{8}\right )\right ) \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x \left (x +\frac {1}{8}\right )\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x \left (x +\frac {1}{8}\right )\right )^{3}+4 \ln \left (2\right )-2 \ln \left (x \right )-2 \ln \left (x +\frac {1}{8}\right )}{2 x \ln \left (25 \ln \left (x \right )-25 x \right )}\) | \(114\) |
int((((16*x^3+2*x^2)*ln(x)-16*x^4-2*x^3)*ln(25*ln(x)-25*x)^2+(((-8*x-1)*ln (x)+8*x^2+x)*ln(2*x^2+1/4*x)+(16*x+1)*ln(x)-16*x^2-x)*ln(25*ln(x)-25*x)+(8 *x^2-7*x-1)*ln(2*x^2+1/4*x))/((8*x^3+x^2)*ln(x)-8*x^4-x^3)/ln(25*ln(x)-25* x)^2,x,method=_RETURNVERBOSE)
1/256/x*(512*x^2*ln(25*ln(x)-25*x)-32*ln(25*ln(x)-25*x)*x+256*ln(2*x^2+1/4 *x))/ln(25*ln(x)-25*x)
Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {\left (-1-7 x+8 x^2\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )+\left (-x-16 x^2+(1+16 x) \log (x)+\left (x+8 x^2+(-1-8 x) \log (x)\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )\right ) \log (-25 x+25 \log (x))+\left (-2 x^3-16 x^4+\left (2 x^2+16 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))}{\left (-x^3-8 x^4+\left (x^2+8 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))} \, dx=\frac {2 \, x^{2} \log \left (-25 \, x + 25 \, \log \left (x\right )\right ) + \log \left (2 \, x^{2} + \frac {1}{4} \, x\right )}{x \log \left (-25 \, x + 25 \, \log \left (x\right )\right )} \]
integrate((((16*x^3+2*x^2)*log(x)-16*x^4-2*x^3)*log(25*log(x)-25*x)^2+(((- 8*x-1)*log(x)+8*x^2+x)*log(2*x^2+1/4*x)+(16*x+1)*log(x)-16*x^2-x)*log(25*l og(x)-25*x)+(8*x^2-7*x-1)*log(2*x^2+1/4*x))/((8*x^3+x^2)*log(x)-8*x^4-x^3) /log(25*log(x)-25*x)^2,x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-1-7 x+8 x^2\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )+\left (-x-16 x^2+(1+16 x) \log (x)+\left (x+8 x^2+(-1-8 x) \log (x)\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )\right ) \log (-25 x+25 \log (x))+\left (-2 x^3-16 x^4+\left (2 x^2+16 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))}{\left (-x^3-8 x^4+\left (x^2+8 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))} \, dx=2 x + \frac {\log {\left (2 x^{2} + \frac {x}{4} \right )}}{x \log {\left (- 25 x + 25 \log {\left (x \right )} \right )}} \]
integrate((((16*x**3+2*x**2)*ln(x)-16*x**4-2*x**3)*ln(25*ln(x)-25*x)**2+(( (-8*x-1)*ln(x)+8*x**2+x)*ln(2*x**2+1/4*x)+(16*x+1)*ln(x)-16*x**2-x)*ln(25* ln(x)-25*x)+(8*x**2-7*x-1)*ln(2*x**2+1/4*x))/((8*x**3+x**2)*ln(x)-8*x**4-x **3)/ln(25*ln(x)-25*x)**2,x)
Time = 0.36 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.56 \[ \int \frac {\left (-1-7 x+8 x^2\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )+\left (-x-16 x^2+(1+16 x) \log (x)+\left (x+8 x^2+(-1-8 x) \log (x)\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )\right ) \log (-25 x+25 \log (x))+\left (-2 x^3-16 x^4+\left (2 x^2+16 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))}{\left (-x^3-8 x^4+\left (x^2+8 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))} \, dx=\frac {4 \, x^{2} \log \left (5\right ) + 2 \, x^{2} \log \left (-x + \log \left (x\right )\right ) - 2 \, \log \left (2\right ) + \log \left (8 \, x + 1\right ) + \log \left (x\right )}{2 \, x \log \left (5\right ) + x \log \left (-x + \log \left (x\right )\right )} \]
integrate((((16*x^3+2*x^2)*log(x)-16*x^4-2*x^3)*log(25*log(x)-25*x)^2+(((- 8*x-1)*log(x)+8*x^2+x)*log(2*x^2+1/4*x)+(16*x+1)*log(x)-16*x^2-x)*log(25*l og(x)-25*x)+(8*x^2-7*x-1)*log(2*x^2+1/4*x))/((8*x^3+x^2)*log(x)-8*x^4-x^3) /log(25*log(x)-25*x)^2,x, algorithm=\
(4*x^2*log(5) + 2*x^2*log(-x + log(x)) - 2*log(2) + log(8*x + 1) + log(x)) /(2*x*log(5) + x*log(-x + log(x)))
Time = 0.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {\left (-1-7 x+8 x^2\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )+\left (-x-16 x^2+(1+16 x) \log (x)+\left (x+8 x^2+(-1-8 x) \log (x)\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )\right ) \log (-25 x+25 \log (x))+\left (-2 x^3-16 x^4+\left (2 x^2+16 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))}{\left (-x^3-8 x^4+\left (x^2+8 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))} \, dx=2 \, x - \frac {2 \, \log \left (2\right ) - \log \left (8 \, x + 1\right ) - \log \left (x\right )}{x \log \left (-25 \, x + 25 \, \log \left (x\right )\right )} \]
integrate((((16*x^3+2*x^2)*log(x)-16*x^4-2*x^3)*log(25*log(x)-25*x)^2+(((- 8*x-1)*log(x)+8*x^2+x)*log(2*x^2+1/4*x)+(16*x+1)*log(x)-16*x^2-x)*log(25*l og(x)-25*x)+(8*x^2-7*x-1)*log(2*x^2+1/4*x))/((8*x^3+x^2)*log(x)-8*x^4-x^3) /log(25*log(x)-25*x)^2,x, algorithm=\
Time = 12.83 (sec) , antiderivative size = 183, normalized size of antiderivative = 5.72 \[ \int \frac {\left (-1-7 x+8 x^2\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )+\left (-x-16 x^2+(1+16 x) \log (x)+\left (x+8 x^2+(-1-8 x) \log (x)\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )\right ) \log (-25 x+25 \log (x))+\left (-2 x^3-16 x^4+\left (2 x^2+16 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))}{\left (-x^3-8 x^4+\left (x^2+8 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))} \, dx=2\,x+\frac {\frac {\ln \left (2\,x^2+\frac {x}{4}\right )}{x}-\frac {\ln \left (25\,\ln \left (x\right )-25\,x\right )\,\left (x-\ln \left (x\right )\right )\,\left (16\,x-\ln \left (2\,x^2+\frac {x}{4}\right )-8\,x\,\ln \left (2\,x^2+\frac {x}{4}\right )+1\right )}{x\,\left (8\,x+1\right )\,\left (x-1\right )}}{\ln \left (25\,\ln \left (x\right )-25\,x\right )}+\ln \left (2\,x^2+\frac {x}{4}\right )\,\left (\frac {1}{x-x^2}+\frac {x-1}{x-x^2}-\frac {\ln \left (x\right )}{x-x^2}\right )-\frac {2\,x+\frac {1}{8}}{-x^2+\frac {7\,x}{8}+\frac {1}{8}}+\frac {\ln \left (x\right )\,\left (2\,x+\frac {1}{8}\right )}{-x^3+\frac {7\,x^2}{8}+\frac {x}{8}} \]
int((log(25*log(x) - 25*x)^2*(2*x^3 - log(x)*(2*x^2 + 16*x^3) + 16*x^4) + log(25*log(x) - 25*x)*(x - log(x)*(16*x + 1) - log(x/4 + 2*x^2)*(x - log(x )*(8*x + 1) + 8*x^2) + 16*x^2) + log(x/4 + 2*x^2)*(7*x - 8*x^2 + 1))/(log( 25*log(x) - 25*x)^2*(x^3 - log(x)*(x^2 + 8*x^3) + 8*x^4)),x)
2*x + (log(x/4 + 2*x^2)/x - (log(25*log(x) - 25*x)*(x - log(x))*(16*x - lo g(x/4 + 2*x^2) - 8*x*log(x/4 + 2*x^2) + 1))/(x*(8*x + 1)*(x - 1)))/log(25* log(x) - 25*x) + log(x/4 + 2*x^2)*(1/(x - x^2) + (x - 1)/(x - x^2) - log(x )/(x - x^2)) - (2*x + 1/8)/((7*x)/8 - x^2 + 1/8) + (log(x)*(2*x + 1/8))/(x /8 + (7*x^2)/8 - x^3)