Integrand size = 129, antiderivative size = 33 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log ^2\left (\frac {1}{5} \left (-1-x^2-x^2 \log (4)+\frac {2 x \log (x)}{5-x}\right )\right ) \]
Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log ^2\left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right ) \]
Integrate[((-20 + 104*x - 40*x^2 + 4*x^3 + (100*x - 40*x^2 + 4*x^3)*Log[4] - 20*Log[x])*Log[(5 - x + 5*x^2 - x^3 + (5*x^2 - x^3)*Log[4] - 2*x*Log[x] )/(-25 + 5*x)])/(25 - 10*x + 26*x^2 - 10*x^3 + x^4 + (25*x^2 - 10*x^3 + x^ 4)*Log[4] + (-10*x + 2*x^2)*Log[x]),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.
| \(\displaystyle \int \frac {\left (4 x^3-40 x^2+\left (4 x^3-40 x^2+100 x\right ) \log (4)+104 x-20 \log (x)-20\right ) \log \left (\frac {-x^3+5 x^2+\left (5 x^2-x^3\right ) \log (4)-x-2 x \log (x)+5}{5 x-25}\right )}{x^4-10 x^3+26 x^2+\left (2 x^2-10 x\right ) \log (x)+\left (x^4-10 x^3+25 x^2\right ) \log (4)-10 x+25} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {4 \left (x^3 (1+\log (4))-10 x^2 (1+\log (4))+26 x \left (1+\frac {25 \log (4)}{26}\right )-5 \log (x)-5\right ) \log \left (\frac {-x^3+5 x^2+\left (5 x^2-x^3\right ) \log (4)-x-2 x \log (x)+5}{5 x-25}\right )}{(5-x) \left (-\left (x^3 (1+\log (4))\right )+5 x^2 (1+\log (4))-x-2 x \log (x)+5\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \int -\frac {\left (-\left ((1+\log (4)) x^3\right )+10 (1+\log (4)) x^2-(26+25 \log (4)) x+5 \log (x)+5\right ) \log \left (-\frac {-x^3+5 x^2-2 \log (x) x-x+\left (5 x^2-x^3\right ) \log (4)+5}{5 (5-x)}\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -4 \int \frac {\left (-\left ((1+\log (4)) x^3\right )+10 (1+\log (4)) x^2-(26+25 \log (4)) x+5 \log (x)+5\right ) \log \left (-\frac {-x^3+5 x^2-2 \log (x) x-x+\left (5 x^2-x^3\right ) \log (4)+5}{5 (5-x)}\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -4 \int \left (\frac {(-1-\log (4)) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right ) x^3}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}+\frac {10 (1+\log (4)) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right ) x^2}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}+\frac {(-26-25 \log (4)) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right ) x}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}+\frac {5 \log (x) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}+\frac {5 \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 \left (-5 (26+25 \log (4)) \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx+125 (1+\log (4)) \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx+5 \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx+5 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(5-x) \left (-\left ((1+\log (4)) x^3\right )+5 (1+\log (4)) x^2-2 \log (x) x-x+5\right )}dx-(26+25 \log (4)) \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(1+\log (4)) x^3-5 (1+\log (4)) x^2+2 \log (x) x+x-5}dx+25 (1+\log (4)) \int \frac {\log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(1+\log (4)) x^3-5 (1+\log (4)) x^2+2 \log (x) x+x-5}dx+5 (1+\log (4)) \int \frac {x \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(1+\log (4)) x^3-5 (1+\log (4)) x^2+2 \log (x) x+x-5}dx-(1+\log (4)) \int \frac {x^2 \log \left (\frac {1}{5} \left (-\left ((1+\log (4)) x^2\right )-\frac {2 \log (x) x}{x-5}-1\right )\right )}{(1+\log (4)) x^3-5 (1+\log (4)) x^2+2 \log (x) x+x-5}dx\right )\) |
Int[((-20 + 104*x - 40*x^2 + 4*x^3 + (100*x - 40*x^2 + 4*x^3)*Log[4] - 20* Log[x])*Log[(5 - x + 5*x^2 - x^3 + (5*x^2 - x^3)*Log[4] - 2*x*Log[x])/(-25 + 5*x)])/(25 - 10*x + 26*x^2 - 10*x^3 + x^4 + (25*x^2 - 10*x^3 + x^4)*Log [4] + (-10*x + 2*x^2)*Log[x]),x]
3.24.33.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 7.93 (sec) , antiderivative size = 1052, normalized size of antiderivative = 31.88
int((-20*ln(x)+2*(4*x^3-40*x^2+100*x)*ln(2)+4*x^3-40*x^2+104*x-20)*ln((-2* x*ln(x)+2*(-x^3+5*x^2)*ln(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-10*x)*ln(x)+ 2*(x^4-10*x^3+25*x^2)*ln(2)+x^4-10*x^3+26*x^2-10*x+25),x,method=_RETURNVER BOSE)
4*ln(5)*(1/2*ln(-5+x)-1/2*ln(2*x^3*ln(2)-10*x^2*ln(2)+x^3+2*x*ln(x)-5*x^2+ x-5))+ln((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2)^2-2*ln(-5+x) *ln((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2)+ln(-5+x)^2+I*ln(x ^3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x-5/(1+2*ln(2)))*Pi*csgn(I/(-5+x)*((x^3-5 *x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^3-2*ln(-5+x)*ln(2)+2*ln(x^ 3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x-5/(1+2*ln(2)))*ln(2)+2*I*ln(-5+x)*Pi*csg n(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^2+I*ln(- 5+x)*Pi*csgn(I/(-5+x))*csgn(I*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1 /2*x-5/2))*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x- 5/2))-I*ln(-5+x)*Pi*csgn(I*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2* x-5/2))*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2 ))^2-I*ln(-5+x)*Pi*csgn(I/(-5+x))*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3 +x*ln(x)-5/2*x^2+1/2*x-5/2))^2-2*I*ln(x^3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x- 5/(1+2*ln(2)))*Pi*csgn(I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2 +1/2*x-5/2))^2+I*ln(x^3-5*x^2+(2*ln(x)+1)/(1+2*ln(2))*x-5/(1+2*ln(2)))*Pi* csgn(I*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))*csgn(I/(-5+x )*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^2+I*ln(x^3-5*x^2+ (2*ln(x)+1)/(1+2*ln(2))*x-5/(1+2*ln(2)))*Pi*csgn(I/(-5+x))*csgn(I/(-5+x)*( (x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^2-I*ln(-5+x)*Pi*csgn (I/(-5+x)*((x^3-5*x^2)*ln(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^3-I*ln...
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )^{2} \]
integrate((-20*log(x)+2*(4*x^3-40*x^2+100*x)*log(2)+4*x^3-40*x^2+104*x-20) *log((-2*x*log(x)+2*(-x^3+5*x^2)*log(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-1 0*x)*log(x)+2*(x^4-10*x^3+25*x^2)*log(2)+x^4-10*x^3+26*x^2-10*x+25),x, alg orithm=\
Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log {\left (\frac {- x^{3} + 5 x^{2} - 2 x \log {\left (x \right )} - x + \left (- 2 x^{3} + 10 x^{2}\right ) \log {\left (2 \right )} + 5}{5 x - 25} \right )}^{2} \]
integrate((-20*ln(x)+2*(4*x**3-40*x**2+100*x)*ln(2)+4*x**3-40*x**2+104*x-2 0)*ln((-2*x*ln(x)+2*(-x**3+5*x**2)*ln(2)-x**3+5*x**2-x+5)/(5*x-25))/((2*x* *2-10*x)*ln(x)+2*(x**4-10*x**3+25*x**2)*ln(2)+x**4-10*x**3+26*x**2-10*x+25 ),x)
\[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\int { \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} \log \left (2\right ) + 26 \, x - 5 \, \log \left (x\right ) - 5\right )} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )}{x^{4} - 10 \, x^{3} + 26 \, x^{2} + 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} \log \left (2\right ) + 2 \, {\left (x^{2} - 5 \, x\right )} \log \left (x\right ) - 10 \, x + 25} \,d x } \]
integrate((-20*log(x)+2*(4*x^3-40*x^2+100*x)*log(2)+4*x^3-40*x^2+104*x-20) *log((-2*x*log(x)+2*(-x^3+5*x^2)*log(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-1 0*x)*log(x)+2*(x^4-10*x^3+25*x^2)*log(2)+x^4-10*x^3+26*x^2-10*x+25),x, alg orithm=\
4*integrate((x^3 - 10*x^2 + 2*(x^3 - 10*x^2 + 25*x)*log(2) + 26*x - 5*log( x) - 5)*log(-1/5*(x^3 - 5*x^2 + 2*(x^3 - 5*x^2)*log(2) + 2*x*log(x) + x - 5)/(x - 5))/(x^4 - 10*x^3 + 26*x^2 + 2*(x^4 - 10*x^3 + 25*x^2)*log(2) + 2* (x^2 - 5*x)*log(x) - 10*x + 25), x)
\[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\int { \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} \log \left (2\right ) + 26 \, x - 5 \, \log \left (x\right ) - 5\right )} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )}{x^{4} - 10 \, x^{3} + 26 \, x^{2} + 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} \log \left (2\right ) + 2 \, {\left (x^{2} - 5 \, x\right )} \log \left (x\right ) - 10 \, x + 25} \,d x } \]
integrate((-20*log(x)+2*(4*x^3-40*x^2+100*x)*log(2)+4*x^3-40*x^2+104*x-20) *log((-2*x*log(x)+2*(-x^3+5*x^2)*log(2)-x^3+5*x^2-x+5)/(5*x-25))/((2*x^2-1 0*x)*log(x)+2*(x^4-10*x^3+25*x^2)*log(2)+x^4-10*x^3+26*x^2-10*x+25),x, alg orithm=\
integrate(4*(x^3 - 10*x^2 + 2*(x^3 - 10*x^2 + 25*x)*log(2) + 26*x - 5*log( x) - 5)*log(-1/5*(x^3 - 5*x^2 + 2*(x^3 - 5*x^2)*log(2) + 2*x*log(x) + x - 5)/(x - 5))/(x^4 - 10*x^3 + 26*x^2 + 2*(x^4 - 10*x^3 + 25*x^2)*log(2) + 2* (x^2 - 5*x)*log(x) - 10*x + 25), x)
Time = 14.54 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx={\ln \left (-\frac {x-2\,\ln \left (2\right )\,\left (5\,x^2-x^3\right )+2\,x\,\ln \left (x\right )-5\,x^2+x^3-5}{5\,x-25}\right )}^2 \]