\(\int \frac {(-20+104 x-40 x^2+4 x^3+(100 x-40 x^2+4 x^3) \log (4)-20 \log (x)) \log (\frac {5-x+5 x^2-x^3+(5 x^2-x^3) \log (4)-2 x \log (x)}{-25+5 x})}{25-10 x+26 x^2-10 x^3+x^4+(25 x^2-10 x^3+x^4) \log (4)+(-10 x+2 x^2) \log (x)} \, dx\) [8333]

Optimal result

Integrand size = 129, antiderivative size = 33 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log ^2\left (\frac {1}{5} \left (-1-x^2-x^2 \log (4)+\frac {2 x \log (x)}{5-x}\right )\right ) \]

[Out]

ln(2/5*ln(x)/(5-x)*x-2/5*x^2*ln(2)-1/5*x^2-1/5)^2

Rubi [F]

\[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx \]

[In]

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)*L
og[4] + (-10*x + 2*x^2)*Log[x]),x]

[Out]

20*Defer[Int][Log[(-1 - x^2*(1 + Log[4]) - (2*x*Log[x])/(-5 + x))/5]/((-5 + x)*(5 - x + 5*x^2*(1 + Log[4]) - x
^3*(1 + Log[4]) - 2*x*Log[x])), x] + 20*Defer[Int][(Log[x]*Log[(-1 - x^2*(1 + Log[4]) - (2*x*Log[x])/(-5 + x))
/5])/((-5 + x)*(5 - x + 5*x^2*(1 + Log[4]) - x^3*(1 + Log[4]) - 2*x*Log[x])), x] - 100*(1 + Log[4])*Defer[Int]
[Log[(-1 - x^2*(1 + Log[4]) - (2*x*Log[x])/(-5 + x))/5]/(-5 + x - 5*x^2*(1 + Log[4]) + x^3*(1 + Log[4]) + 2*x*
Log[x]), x] + 4*(26 + 25*Log[4])*Defer[Int][Log[(-1 - x^2*(1 + Log[4]) - (2*x*Log[x])/(-5 + x))/5]/(-5 + x - 5
*x^2*(1 + Log[4]) + x^3*(1 + Log[4]) + 2*x*Log[x]), x] - 20*(1 + Log[4])*Defer[Int][(x*Log[(-1 - x^2*(1 + Log[
4]) - (2*x*Log[x])/(-5 + x))/5])/(-5 + x - 5*x^2*(1 + Log[4]) + x^3*(1 + Log[4]) + 2*x*Log[x]), x] + 4*(1 + Lo
g[4])*Defer[Int][(x^2*Log[(-1 - x^2*(1 + Log[4]) - (2*x*Log[x])/(-5 + x))/5])/(-5 + x - 5*x^2*(1 + Log[4]) + x
^3*(1 + Log[4]) + 2*x*Log[x]), x] - 500*(1 + Log[4])*Defer[Int][Log[(-1 - x^2*(1 + Log[4]) - (2*x*Log[x])/(-5
+ x))/5]/((-5 + x)*((-5 + x)*(1 + x^2*(1 + Log[4])) + 2*x*Log[x])), x] + 20*(26 + 25*Log[4])*Defer[Int][Log[(-
1 - x^2*(1 + Log[4]) - (2*x*Log[x])/(-5 + x))/5]/((-5 + x)*((-5 + x)*(1 + x^2*(1 + Log[4])) + 2*x*Log[x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 \left (-\left ((-5+x) \left (1-5 x (1+\log (4))+x^2 (1+\log (4))\right )\right )+5 \log (x)\right ) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx \\ & = 4 \int \frac {\left (-\left ((-5+x) \left (1-5 x (1+\log (4))+x^2 (1+\log (4))\right )\right )+5 \log (x)\right ) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx \\ & = 4 \int \left (\frac {5 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {10 x^2 (-1-\log (4)) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {x^3 (1+\log (4)) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {x (1+25 (1+\log (4))) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {5 \log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}\right ) \, dx \\ & = 20 \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+20 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (1+\log (4))) \int \frac {x^3 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx-(40 (1+\log (4))) \int \frac {x^2 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (26+25 \log (4))) \int \frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx \\ & = 20 \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+20 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (1+\log (4))) \int \frac {x^3 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx-(40 (1+\log (4))) \int \frac {x^2 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx+(4 (26+25 \log (4))) \int \frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx \\ & = 20 \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+20 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (1+\log (4))) \int \left (\frac {125 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {25 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}+\frac {5 x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}+\frac {x^2 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}\right ) \, dx-(40 (1+\log (4))) \int \left (\frac {25 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {5 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}+\frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}\right ) \, dx+(4 (26+25 \log (4))) \int \left (\frac {5 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}\right ) \, dx \\ & = 20 \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+20 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (1+\log (4))) \int \frac {x^2 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx+(20 (1+\log (4))) \int \frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx-(40 (1+\log (4))) \int \frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx+(100 (1+\log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx-(200 (1+\log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx+(500 (1+\log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx-(1000 (1+\log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (26+25 \log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx+(20 (26+25 \log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx \\ & = 20 \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+20 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (1+\log (4))) \int \frac {x^2 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx+(20 (1+\log (4))) \int \frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx-(40 (1+\log (4))) \int \frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx+(100 (1+\log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx-(200 (1+\log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx+(500 (1+\log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx-(1000 (1+\log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx+(4 (26+25 \log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)} \, dx+(20 (26+25 \log (4))) \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log ^2\left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right ) \]

[In]

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]

[Out]

Log[(-1 - x^2*(1 + Log[4]) - (2*x*Log[x])/(-5 + x))/5]^2

Maple [C] (warning: unable to verify)

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

method	result	size
default	\(\text {Expression too large to display}\)	\(1052\)

[In]

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=_RETURNV
ERBOSE)

[Out]

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
*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*((
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*l
n(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))*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)*((x^3-5*x^2)*l
n(2)+1/2*x^3+x*ln(x)-5/2*x^2+1/2*x-5/2))^3-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*((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-2*I*ln(-5+x)*Pi

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )^{2} \]

[In]

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^2-10*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, a
lgorithm="fricas")

[Out]

log(-1/5*(x^3 - 5*x^2 + 2*(x^3 - 5*x^2)*log(2) + 2*x*log(x) + x - 5)/(x - 5))^2

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\log {\left (\frac {- x^{3} + 5 x^{2} - 2 x \log {\left (x \right )} - x + \left (- 2 x^{3} + 10 x^{2}\right ) \log {\left (2 \right )} + 5}{5 x - 25} \right )}^{2} \]

[In]

integrate((-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)*l
n(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)

[Out]

log((-x**3 + 5*x**2 - 2*x*log(x) - x + (-2*x**3 + 10*x**2)*log(2) + 5)/(5*x - 25))**2

Maxima [F]

\[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\int { \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} \log \left (2\right ) + 26 \, x - 5 \, \log \left (x\right ) - 5\right )} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )}{x^{4} - 10 \, x^{3} + 26 \, x^{2} + 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} \log \left (2\right ) + 2 \, {\left (x^{2} - 5 \, x\right )} \log \left (x\right ) - 10 \, x + 25} \,d x } \]

[In]

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^2-10*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, a
lgorithm="maxima")

[Out]

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)

Giac [F]

\[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx=\int { \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} \log \left (2\right ) + 26 \, x - 5 \, \log \left (x\right ) - 5\right )} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \left (2\right ) + 2 \, x \log \left (x\right ) + x - 5}{5 \, {\left (x - 5\right )}}\right )}{x^{4} - 10 \, x^{3} + 26 \, x^{2} + 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} \log \left (2\right ) + 2 \, {\left (x^{2} - 5 \, x\right )} \log \left (x\right ) - 10 \, x + 25} \,d x } \]

[In]

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^2-10*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, a
lgorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 14.54 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {\left (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (x)} \, dx={\ln \left (-\frac {x-2\,\ln \left (2\right )\,\left (5\,x^2-x^3\right )+2\,x\,\ln \left (x\right )-5\,x^2+x^3-5}{5\,x-25}\right )}^2 \]

[In]

int((log(-(x - 2*log(2)*(5*x^2 - x^3) + 2*x*log(x) - 5*x^2 + x^3 - 5)/(5*x - 25))*(104*x - 20*log(x) + 2*log(2
)*(100*x - 40*x^2 + 4*x^3) - 40*x^2 + 4*x^3 - 20))/(2*log(2)*(25*x^2 - 10*x^3 + x^4) - 10*x - log(x)*(10*x - 2
*x^2) + 26*x^2 - 10*x^3 + x^4 + 25),x)

[Out]

log(-(x - 2*log(2)*(5*x^2 - x^3) + 2*x*log(x) - 5*x^2 + x^3 - 5)/(5*x - 25))^2