Integrand size = 127, antiderivative size = 26 \[ \int \frac {500-5 x^2+125 \log (3)+(-625-125 \log (3)) \log (x)+(-125+125 \log (x)) \log (\log (x))}{25 x^2+10 x^3+x^4+\left (-1250 x-250 x^2+\left (-250 x-50 x^2\right ) \log (3)\right ) \log (x)+\left (15625+6250 \log (3)+625 \log ^2(3)\right ) \log ^2(x)+\left (\left (250 x+50 x^2\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+625 \log ^2(x) \log ^2(\log (x))} \, dx=\frac {x}{\frac {1}{5} x (5+x)+5 \log (x) (-5-\log (3)+\log (\log (x)))} \]
\[ \int \frac {500-5 x^2+125 \log (3)+(-625-125 \log (3)) \log (x)+(-125+125 \log (x)) \log (\log (x))}{25 x^2+10 x^3+x^4+\left (-1250 x-250 x^2+\left (-250 x-50 x^2\right ) \log (3)\right ) \log (x)+\left (15625+6250 \log (3)+625 \log ^2(3)\right ) \log ^2(x)+\left (\left (250 x+50 x^2\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+625 \log ^2(x) \log ^2(\log (x))} \, dx=\int \frac {500-5 x^2+125 \log (3)+(-625-125 \log (3)) \log (x)+(-125+125 \log (x)) \log (\log (x))}{25 x^2+10 x^3+x^4+\left (-1250 x-250 x^2+\left (-250 x-50 x^2\right ) \log (3)\right ) \log (x)+\left (15625+6250 \log (3)+625 \log ^2(3)\right ) \log ^2(x)+\left (\left (250 x+50 x^2\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+625 \log ^2(x) \log ^2(\log (x))} \, dx \]
Integrate[(500 - 5*x^2 + 125*Log[3] + (-625 - 125*Log[3])*Log[x] + (-125 + 125*Log[x])*Log[Log[x]])/(25*x^2 + 10*x^3 + x^4 + (-1250*x - 250*x^2 + (- 250*x - 50*x^2)*Log[3])*Log[x] + (15625 + 6250*Log[3] + 625*Log[3]^2)*Log[ x]^2 + ((250*x + 50*x^2)*Log[x] + (-6250 - 1250*Log[3])*Log[x]^2)*Log[Log[ x]] + 625*Log[x]^2*Log[Log[x]]^2),x]
Integrate[(500 - 5*x^2 + 125*Log[3] + (-625 - 125*Log[3])*Log[x] + (-125 + 125*Log[x])*Log[Log[x]])/(25*x^2 + 10*x^3 + x^4 + (-1250*x - 250*x^2 + (- 250*x - 50*x^2)*Log[3])*Log[x] + (15625 + 6250*Log[3] + 625*Log[3]^2)*Log[ x]^2 + ((250*x + 50*x^2)*Log[x] + (-6250 - 1250*Log[3])*Log[x]^2)*Log[Log[ x]] + 625*Log[x]^2*Log[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.
\(\displaystyle \int \frac {-5 x^2+(-625-125 \log (3)) \log (x)+(125 \log (x)-125) \log (\log (x))+500+125 \log (3)}{x^4+10 x^3+25 x^2+\left (\left (50 x^2+250 x\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+\left (-250 x^2+\left (-50 x^2-250 x\right ) \log (3)-1250 x\right ) \log (x)+\left (15625+625 \log ^2(3)+6250 \log (3)\right ) \log ^2(x)+625 \log ^2(x) \log ^2(\log (x))} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 \left (-x^2-25 \log (x) (-\log (\log (x))+5+\log (3))-25 \log (\log (x))+100 \left (1+\frac {\log (3)}{4}\right )\right )}{\left (x (x+5)+25 \log (x) \left (\log \left (\frac {\log (x)}{3}\right )-5\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int -\frac {x^2+25 \log (x) (-\log (\log (x))+\log (3)+5)+25 \log (\log (x))-25 (4+\log (3))}{\left (x (x+5)-25 \log (x) \left (5-\log \left (\frac {\log (x)}{3}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {x^2+25 \log (x) (-\log (\log (x))+\log (3)+5)+25 \log (\log (x))-25 (4+\log (3))}{\left (x (x+5)-25 \log (x) \left (5-\log \left (\frac {\log (x)}{3}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (\frac {x^2}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}-\frac {25 (\log (x)-1) \log (\log (x))}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}+\frac {125 \left (1+\frac {\log (3)}{5}\right ) \log (x)}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}-\frac {25 (4+\log (3))}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left (-25 (4+\log (3)) \int \frac {1}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}dx+\int \frac {x^2}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}dx+25 (5+\log (3)) \int \frac {\log (x)}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}dx+25 \int \frac {\log (\log (x))}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}dx-25 \int \frac {\log (x) \log (\log (x))}{\left (x^2+5 x-125 \log (x)+25 \log (x) \log \left (\frac {\log (x)}{3}\right )\right )^2}dx\right )\) |
Int[(500 - 5*x^2 + 125*Log[3] + (-625 - 125*Log[3])*Log[x] + (-125 + 125*L og[x])*Log[Log[x]])/(25*x^2 + 10*x^3 + x^4 + (-1250*x - 250*x^2 + (-250*x - 50*x^2)*Log[3])*Log[x] + (15625 + 6250*Log[3] + 625*Log[3]^2)*Log[x]^2 + ((250*x + 50*x^2)*Log[x] + (-6250 - 1250*Log[3])*Log[x]^2)*Log[Log[x]] + 625*Log[x]^2*Log[Log[x]]^2),x]
3.21.22.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.90 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23
method | result | size |
default | \(-\frac {5 x}{25 \ln \left (3\right ) \ln \left (x \right )-25 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )-x^{2}+125 \ln \left (x \right )-5 x}\) | \(32\) |
risch | \(-\frac {5 x}{25 \ln \left (3\right ) \ln \left (x \right )-25 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )-x^{2}+125 \ln \left (x \right )-5 x}\) | \(32\) |
parallelrisch | \(-\frac {5 x}{25 \ln \left (3\right ) \ln \left (x \right )-25 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )-x^{2}+125 \ln \left (x \right )-5 x}\) | \(32\) |
int(((125*ln(x)-125)*ln(ln(x))+(-125*ln(3)-625)*ln(x)+125*ln(3)-5*x^2+500) /(625*ln(x)^2*ln(ln(x))^2+((-1250*ln(3)-6250)*ln(x)^2+(50*x^2+250*x)*ln(x) )*ln(ln(x))+(625*ln(3)^2+6250*ln(3)+15625)*ln(x)^2+((-50*x^2-250*x)*ln(3)- 250*x^2-1250*x)*ln(x)+x^4+10*x^3+25*x^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {500-5 x^2+125 \log (3)+(-625-125 \log (3)) \log (x)+(-125+125 \log (x)) \log (\log (x))}{25 x^2+10 x^3+x^4+\left (-1250 x-250 x^2+\left (-250 x-50 x^2\right ) \log (3)\right ) \log (x)+\left (15625+6250 \log (3)+625 \log ^2(3)\right ) \log ^2(x)+\left (\left (250 x+50 x^2\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+625 \log ^2(x) \log ^2(\log (x))} \, dx=\frac {5 \, x}{x^{2} - 25 \, {\left (\log \left (3\right ) + 5\right )} \log \left (x\right ) + 25 \, \log \left (x\right ) \log \left (\log \left (x\right )\right ) + 5 \, x} \]
integrate(((125*log(x)-125)*log(log(x))+(-125*log(3)-625)*log(x)+125*log(3 )-5*x^2+500)/(625*log(x)^2*log(log(x))^2+((-1250*log(3)-6250)*log(x)^2+(50 *x^2+250*x)*log(x))*log(log(x))+(625*log(3)^2+6250*log(3)+15625)*log(x)^2+ ((-50*x^2-250*x)*log(3)-250*x^2-1250*x)*log(x)+x^4+10*x^3+25*x^2),x, algor ithm=\
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {500-5 x^2+125 \log (3)+(-625-125 \log (3)) \log (x)+(-125+125 \log (x)) \log (\log (x))}{25 x^2+10 x^3+x^4+\left (-1250 x-250 x^2+\left (-250 x-50 x^2\right ) \log (3)\right ) \log (x)+\left (15625+6250 \log (3)+625 \log ^2(3)\right ) \log ^2(x)+\left (\left (250 x+50 x^2\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+625 \log ^2(x) \log ^2(\log (x))} \, dx=\frac {5 x}{x^{2} + 5 x + 25 \log {\left (x \right )} \log {\left (\log {\left (x \right )} \right )} - 125 \log {\left (x \right )} - 25 \log {\left (3 \right )} \log {\left (x \right )}} \]
integrate(((125*ln(x)-125)*ln(ln(x))+(-125*ln(3)-625)*ln(x)+125*ln(3)-5*x* *2+500)/(625*ln(x)**2*ln(ln(x))**2+((-1250*ln(3)-6250)*ln(x)**2+(50*x**2+2 50*x)*ln(x))*ln(ln(x))+(625*ln(3)**2+6250*ln(3)+15625)*ln(x)**2+((-50*x**2 -250*x)*ln(3)-250*x**2-1250*x)*ln(x)+x**4+10*x**3+25*x**2),x)
Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {500-5 x^2+125 \log (3)+(-625-125 \log (3)) \log (x)+(-125+125 \log (x)) \log (\log (x))}{25 x^2+10 x^3+x^4+\left (-1250 x-250 x^2+\left (-250 x-50 x^2\right ) \log (3)\right ) \log (x)+\left (15625+6250 \log (3)+625 \log ^2(3)\right ) \log ^2(x)+\left (\left (250 x+50 x^2\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+625 \log ^2(x) \log ^2(\log (x))} \, dx=\frac {5 \, x}{x^{2} - 25 \, {\left (\log \left (3\right ) + 5\right )} \log \left (x\right ) + 25 \, \log \left (x\right ) \log \left (\log \left (x\right )\right ) + 5 \, x} \]
integrate(((125*log(x)-125)*log(log(x))+(-125*log(3)-625)*log(x)+125*log(3 )-5*x^2+500)/(625*log(x)^2*log(log(x))^2+((-1250*log(3)-6250)*log(x)^2+(50 *x^2+250*x)*log(x))*log(log(x))+(625*log(3)^2+6250*log(3)+15625)*log(x)^2+ ((-50*x^2-250*x)*log(3)-250*x^2-1250*x)*log(x)+x^4+10*x^3+25*x^2),x, algor ithm=\
Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {500-5 x^2+125 \log (3)+(-625-125 \log (3)) \log (x)+(-125+125 \log (x)) \log (\log (x))}{25 x^2+10 x^3+x^4+\left (-1250 x-250 x^2+\left (-250 x-50 x^2\right ) \log (3)\right ) \log (x)+\left (15625+6250 \log (3)+625 \log ^2(3)\right ) \log ^2(x)+\left (\left (250 x+50 x^2\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+625 \log ^2(x) \log ^2(\log (x))} \, dx=\frac {5 \, x}{x^{2} - 25 \, \log \left (3\right ) \log \left (x\right ) + 25 \, \log \left (x\right ) \log \left (\log \left (x\right )\right ) + 5 \, x - 125 \, \log \left (x\right )} \]
integrate(((125*log(x)-125)*log(log(x))+(-125*log(3)-625)*log(x)+125*log(3 )-5*x^2+500)/(625*log(x)^2*log(log(x))^2+((-1250*log(3)-6250)*log(x)^2+(50 *x^2+250*x)*log(x))*log(log(x))+(625*log(3)^2+6250*log(3)+15625)*log(x)^2+ ((-50*x^2-250*x)*log(3)-250*x^2-1250*x)*log(x)+x^4+10*x^3+25*x^2),x, algor ithm=\
Timed out. \[ \int \frac {500-5 x^2+125 \log (3)+(-625-125 \log (3)) \log (x)+(-125+125 \log (x)) \log (\log (x))}{25 x^2+10 x^3+x^4+\left (-1250 x-250 x^2+\left (-250 x-50 x^2\right ) \log (3)\right ) \log (x)+\left (15625+6250 \log (3)+625 \log ^2(3)\right ) \log ^2(x)+\left (\left (250 x+50 x^2\right ) \log (x)+(-6250-1250 \log (3)) \log ^2(x)\right ) \log (\log (x))+625 \log ^2(x) \log ^2(\log (x))} \, dx=\int \frac {125\,\ln \left (3\right )+\ln \left (\ln \left (x\right )\right )\,\left (125\,\ln \left (x\right )-125\right )-\ln \left (x\right )\,\left (125\,\ln \left (3\right )+625\right )-5\,x^2+500}{625\,{\ln \left (\ln \left (x\right )\right )}^2\,{\ln \left (x\right )}^2+\ln \left (\ln \left (x\right )\right )\,\left (\ln \left (x\right )\,\left (50\,x^2+250\,x\right )-{\ln \left (x\right )}^2\,\left (1250\,\ln \left (3\right )+6250\right )\right )+25\,x^2+10\,x^3+x^4-\ln \left (x\right )\,\left (1250\,x+\ln \left (3\right )\,\left (50\,x^2+250\,x\right )+250\,x^2\right )+{\ln \left (x\right )}^2\,\left (6250\,\ln \left (3\right )+625\,{\ln \left (3\right )}^2+15625\right )} \,d x \]
int((125*log(3) + log(log(x))*(125*log(x) - 125) - log(x)*(125*log(3) + 62 5) - 5*x^2 + 500)/(625*log(log(x))^2*log(x)^2 + log(log(x))*(log(x)*(250*x + 50*x^2) - log(x)^2*(1250*log(3) + 6250)) + 25*x^2 + 10*x^3 + x^4 - log( x)*(1250*x + log(3)*(250*x + 50*x^2) + 250*x^2) + log(x)^2*(6250*log(3) + 625*log(3)^2 + 15625)),x)
int((125*log(3) + log(log(x))*(125*log(x) - 125) - log(x)*(125*log(3) + 62 5) - 5*x^2 + 500)/(625*log(log(x))^2*log(x)^2 + log(log(x))*(log(x)*(250*x + 50*x^2) - log(x)^2*(1250*log(3) + 6250)) + 25*x^2 + 10*x^3 + x^4 - log( x)*(1250*x + log(3)*(250*x + 50*x^2) + 250*x^2) + log(x)^2*(6250*log(3) + 625*log(3)^2 + 15625)), x)