\(\int \frac {5-2 x+(125-40 x+53 x^2-52 x^3+12 x^4) \log (x)+5 \log (x) \log (\log (x))}{(125-100 x+20 x^2) \log (x)} \, dx\) [8207]

Optimal result

Integrand size = 52, antiderivative size = 24 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {1}{5} x \left (5+x+x^2+\frac {x+\log (\log (x))}{5-2 x}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx \]

[In]

Int[(5 - 2*x + (125 - 40*x + 53*x^2 - 52*x^3 + 12*x^4)*Log[x] + 5*Log[x]*Log[Log[x]])/((125 - 100*x + 20*x^2)*
Log[x]),x]

[Out]

5/(4*(5 - 2*x)) + (9*x)/10 + x^2/5 + x^3/5 - Defer[Int][1/((-5 + 2*x)*Log[x]), x]/5 + Defer[Int][Log[Log[x]]/(
-5 + 2*x)^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{5 (-5+2 x)^2 \log (x)} \, dx \\ & = \frac {1}{5} \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{(-5+2 x)^2 \log (x)} \, dx \\ & = \frac {1}{5} \int \left (\frac {5-2 x+125 \log (x)-40 x \log (x)+53 x^2 \log (x)-52 x^3 \log (x)+12 x^4 \log (x)}{(-5+2 x)^2 \log (x)}+\frac {5 \log (\log (x))}{(-5+2 x)^2}\right ) \, dx \\ & = \frac {1}{5} \int \frac {5-2 x+125 \log (x)-40 x \log (x)+53 x^2 \log (x)-52 x^3 \log (x)+12 x^4 \log (x)}{(-5+2 x)^2 \log (x)} \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {125}{(-5+2 x)^2}-\frac {40 x}{(-5+2 x)^2}+\frac {53 x^2}{(-5+2 x)^2}-\frac {52 x^3}{(-5+2 x)^2}+\frac {12 x^4}{(-5+2 x)^2}-\frac {1}{(-5+2 x) \log (x)}\right ) \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ & = \frac {25}{2 (5-2 x)}-\frac {1}{5} \int \frac {1}{(-5+2 x) \log (x)} \, dx+\frac {12}{5} \int \frac {x^4}{(-5+2 x)^2} \, dx-8 \int \frac {x}{(-5+2 x)^2} \, dx-\frac {52}{5} \int \frac {x^3}{(-5+2 x)^2} \, dx+\frac {53}{5} \int \frac {x^2}{(-5+2 x)^2} \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ & = \frac {25}{2 (5-2 x)}-\frac {1}{5} \int \frac {1}{(-5+2 x) \log (x)} \, dx+\frac {12}{5} \int \left (\frac {75}{16}+\frac {5 x}{4}+\frac {x^2}{4}+\frac {625}{16 (-5+2 x)^2}+\frac {125}{4 (-5+2 x)}\right ) \, dx-8 \int \left (\frac {5}{2 (-5+2 x)^2}+\frac {1}{2 (-5+2 x)}\right ) \, dx-\frac {52}{5} \int \left (\frac {5}{4}+\frac {x}{4}+\frac {125}{8 (-5+2 x)^2}+\frac {75}{8 (-5+2 x)}\right ) \, dx+\frac {53}{5} \int \left (\frac {1}{4}+\frac {25}{4 (-5+2 x)^2}+\frac {5}{2 (-5+2 x)}\right ) \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ & = \frac {5}{4 (5-2 x)}+\frac {9 x}{10}+\frac {x^2}{5}+\frac {x^3}{5}-\frac {1}{5} \int \frac {1}{(-5+2 x) \log (x)} \, dx+\int \frac {\log (\log (x))}{(-5+2 x)^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {25+90 x-16 x^2+12 x^3-8 x^4+4 x \log (\log (x))}{5 (20-8 x)} \]

[In]

Integrate[(5 - 2*x + (125 - 40*x + 53*x^2 - 52*x^3 + 12*x^4)*Log[x] + 5*Log[x]*Log[Log[x]])/((125 - 100*x + 20
*x^2)*Log[x]),x]

[Out]

(25 + 90*x - 16*x^2 + 12*x^3 - 8*x^4 + 4*x*Log[Log[x]])/(5*(20 - 8*x))

Maple [A] (verified)

Time = 1.51 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38

method	result	size
parallelrisch	\(\frac {20 x^{4}-30 x^{3}-10 x \ln \left (\ln \left (x \right )\right )+40 x^{2}-625}{-250+100 x}\)	\(33\)
risch	\(-\frac {\ln \left (\ln \left (x \right )\right )}{2 \left (-5+2 x \right )}-\frac {-8 x^{4}+12 x^{3}+4 x \ln \left (\ln \left (x \right )\right )-16 x^{2}-10 \ln \left (\ln \left (x \right )\right )+90 x +25}{20 \left (-5+2 x \right )}\)	\(54\)

[In]

int((5*ln(x)*ln(ln(x))+(12*x^4-52*x^3+53*x^2-40*x+125)*ln(x)-2*x+5)/(20*x^2-100*x+125)/ln(x),x,method=_RETURNV
ERBOSE)

[Out]

1/50*(20*x^4-30*x^3-10*x*ln(ln(x))+40*x^2-625)/(-5+2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {8 \, x^{4} - 12 \, x^{3} + 16 \, x^{2} - 4 \, x \log \left (\log \left (x\right )\right ) - 90 \, x - 25}{20 \, {\left (2 \, x - 5\right )}} \]

[In]

integrate((5*log(x)*log(log(x))+(12*x^4-52*x^3+53*x^2-40*x+125)*log(x)-2*x+5)/(20*x^2-100*x+125)/log(x),x, alg
orithm="fricas")

[Out]

1/20*(8*x^4 - 12*x^3 + 16*x^2 - 4*x*log(log(x)) - 90*x - 25)/(2*x - 5)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {x^{3}}{5} + \frac {x^{2}}{5} + \frac {9 x}{10} - \frac {\log {\left (\log {\left (x \right )} \right )}}{10} - \frac {5}{8 x - 20} - \frac {\log {\left (\log {\left (x \right )} \right )}}{4 x - 10} \]

[In]

integrate((5*ln(x)*ln(ln(x))+(12*x**4-52*x**3+53*x**2-40*x+125)*ln(x)-2*x+5)/(20*x**2-100*x+125)/ln(x),x)

[Out]

x**3/5 + x**2/5 + 9*x/10 - log(log(x))/10 - 5/(8*x - 20) - log(log(x))/(4*x - 10)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {8 \, x^{4} - 12 \, x^{3} + 16 \, x^{2} - 4 \, x \log \left (\log \left (x\right )\right ) - 90 \, x - 25}{20 \, {\left (2 \, x - 5\right )}} \]

[In]

integrate((5*log(x)*log(log(x))+(12*x^4-52*x^3+53*x^2-40*x+125)*log(x)-2*x+5)/(20*x^2-100*x+125)/log(x),x, alg
orithm="maxima")

[Out]

1/20*(8*x^4 - 12*x^3 + 16*x^2 - 4*x*log(log(x)) - 90*x - 25)/(2*x - 5)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {1}{5} \, x^{3} + \frac {1}{5} \, x^{2} + \frac {9}{10} \, x - \frac {\log \left (\log \left (x\right )\right )}{2 \, {\left (2 \, x - 5\right )}} - \frac {5}{4 \, {\left (2 \, x - 5\right )}} - \frac {1}{10} \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((5*log(x)*log(log(x))+(12*x^4-52*x^3+53*x^2-40*x+125)*log(x)-2*x+5)/(20*x^2-100*x+125)/log(x),x, alg
orithm="giac")

[Out]

1/5*x^3 + 1/5*x^2 + 9/10*x - 1/2*log(log(x))/(2*x - 5) - 5/4/(2*x - 5) - 1/10*log(log(x))

Mupad [B] (verification not implemented)

Time = 12.76 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {5-2 x+\left (125-40 x+53 x^2-52 x^3+12 x^4\right ) \log (x)+5 \log (x) \log (\log (x))}{\left (125-100 x+20 x^2\right ) \log (x)} \, dx=\frac {9\,x}{10}-\frac {\ln \left (\ln \left (x\right )\right )}{10}-\frac {5}{8\,\left (x-\frac {5}{2}\right )}+\frac {x^2}{5}+\frac {x^3}{5}-\frac {\ln \left (\ln \left (x\right )\right )}{4\,\left (x-\frac {5}{2}\right )} \]

[In]

int((log(x)*(53*x^2 - 40*x - 52*x^3 + 12*x^4 + 125) - 2*x + 5*log(log(x))*log(x) + 5)/(log(x)*(20*x^2 - 100*x
+ 125)),x)

[Out]

(9*x)/10 - log(log(x))/10 - 5/(8*(x - 5/2)) + x^2/5 + x^3/5 - log(log(x))/(4*(x - 5/2))