Integrand size = 48, antiderivative size = 22 \[ \int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+\left (45-30 x+5 x^2\right ) \log (25)} \, dx=-\frac {x \left (4+\frac {4}{-\frac {18}{5}+\log (25)}\right ) \log (x)}{-3+x} \] Output:
-ln(x)*(4/(-18/5+2*ln(5))+4)*x/(-3+x)
Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+\left (45-30 x+5 x^2\right ) \log (25)} \, dx=-\frac {4 x (-13+5 \log (25)) \log (x)}{(-3+x) (-18+5 \log (25))} \] Input:
Integrate[(-156 + 52*x + (60 - 20*x)*Log[25] + (-156 + 60*Log[25])*Log[x]) /(-162 + 108*x - 18*x^2 + (45 - 30*x + 5*x^2)*Log[25]),x]
Output:
(-4*x*(-13 + 5*Log[25])*Log[x])/((-3 + x)*(-18 + 5*Log[25]))
Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {7292, 27, 7239, 27, 7293, 2009}
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.
\(\displaystyle \int \frac {52 x+(60-20 x) \log (25)+(60 \log (25)-156) \log (x)-156}{-18 x^2+\left (5 x^2-30 x+45\right ) \log (25)+108 x-162} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 (13-5 \log (25)) (-x+3 \log (x)+3)}{x^2 (18-5 \log (25))-6 x (18-5 \log (25))+9 (18-5 \log (25))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 (13-5 \log (25)) \int \frac {-x+3 \log (x)+3}{(18-5 \log (25)) x^2-6 (18-5 \log (25)) x+9 (18-5 \log (25))}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle 4 (13-5 \log (25)) \int \frac {-x+3 \log (x)+3}{(3-x)^2 (18-5 \log (25))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 (13-5 \log (25)) \int \frac {-x+3 \log (x)+3}{(3-x)^2}dx}{18-5 \log (25)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {4 (13-5 \log (25)) \int \left (\frac {3 \log (x)}{(x-3)^2}+\frac {1}{3-x}\right )dx}{18-5 \log (25)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 x (13-5 \log (25)) \log (x)}{(3-x) (18-5 \log (25))}\) |
Input:
Int[(-156 + 52*x + (60 - 20*x)*Log[25] + (-156 + 60*Log[25])*Log[x])/(-162 + 108*x - 18*x^2 + (45 - 30*x + 5*x^2)*Log[25]),x]
Output:
(4*x*(13 - 5*Log[25])*Log[x])/((3 - x)*(18 - 5*Log[25]))
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 = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14
method | result | size |
norman | \(-\frac {2 \left (10 \ln \left (5\right )-13\right ) x \ln \left (x \right )}{\left (5 \ln \left (5\right )-9\right ) \left (-3+x \right )}\) | \(25\) |
parallelrisch | \(\frac {-20 \ln \left (x \right ) \ln \left (5\right ) x +26 x \ln \left (x \right )}{\left (5 \ln \left (5\right )-9\right ) \left (-3+x \right )}\) | \(28\) |
risch | \(-\frac {6 \left (10 \ln \left (5\right )-13\right ) \ln \left (x \right )}{5 x \ln \left (5\right )-15 \ln \left (5\right )-9 x +27}-\frac {20 \ln \left (x \right ) \ln \left (5\right )}{5 \ln \left (5\right )-9}+\frac {26 \ln \left (x \right )}{5 \ln \left (5\right )-9}\) | \(54\) |
parts | \(\frac {\left (-20 \ln \left (5\right )+26\right ) \ln \left (\left (5 \ln \left (5\right )-9\right ) x -15 \ln \left (5\right )+27\right )}{5 \ln \left (5\right )-9}+\frac {\left (60 \ln \left (5\right )-78\right ) \left (\frac {\ln \left (-3+x \right )}{3}-\frac {\ln \left (x \right ) x}{3 \left (-3+x \right )}\right )}{5 \ln \left (5\right )-9}\) | \(64\) |
default | \(\frac {\left (-40 \ln \left (5\right )+52\right ) \ln \left (\left (5 \ln \left (5\right )-9\right ) x -15 \ln \left (5\right )+27\right )}{10 \ln \left (5\right )-18}+\frac {\left (120 \ln \left (5\right )-156\right ) \left (\frac {\ln \left (-3+x \right )}{3}-\frac {\ln \left (x \right ) x}{3 \left (-3+x \right )}\right )}{10 \ln \left (5\right )-18}\) | \(66\) |
orering | \(\frac {\left (\frac {1}{3} x^{2}+x \right ) \left (\left (120 \ln \left (5\right )-156\right ) \ln \left (x \right )+2 \left (-20 x +60\right ) \ln \left (5\right )+52 x -156\right )}{2 \left (5 x^{2}-30 x +45\right ) \ln \left (5\right )-18 x^{2}+108 x -162}+\frac {x^{2} \left (-3+x \right ) \left (\frac {\frac {120 \ln \left (5\right )-156}{x}-40 \ln \left (5\right )+52}{2 \left (5 x^{2}-30 x +45\right ) \ln \left (5\right )-18 x^{2}+108 x -162}-\frac {\left (\left (120 \ln \left (5\right )-156\right ) \ln \left (x \right )+2 \left (-20 x +60\right ) \ln \left (5\right )+52 x -156\right ) \left (2 \left (10 x -30\right ) \ln \left (5\right )-36 x +108\right )}{{\left (2 \left (5 x^{2}-30 x +45\right ) \ln \left (5\right )-18 x^{2}+108 x -162\right )}^{2}}\right )}{3}\) | \(176\) |
Input:
int(((120*ln(5)-156)*ln(x)+2*(-20*x+60)*ln(5)+52*x-156)/(2*(5*x^2-30*x+45) *ln(5)-18*x^2+108*x-162),x,method=_RETURNVERBOSE)
Output:
-2*(10*ln(5)-13)/(5*ln(5)-9)*x*ln(x)/(-3+x)
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+\left (45-30 x+5 x^2\right ) \log (25)} \, dx=-\frac {2 \, {\left (10 \, x \log \left (5\right ) - 13 \, x\right )} \log \left (x\right )}{5 \, {\left (x - 3\right )} \log \left (5\right ) - 9 \, x + 27} \] Input:
integrate(((120*log(5)-156)*log(x)+2*(-20*x+60)*log(5)+52*x-156)/(2*(5*x^2 -30*x+45)*log(5)-18*x^2+108*x-162),x, algorithm="fricas")
Output:
-2*(10*x*log(5) - 13*x)*log(x)/(5*(x - 3)*log(5) - 9*x + 27)
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+\left (45-30 x+5 x^2\right ) \log (25)} \, dx=\frac {\left (26 - 20 \log {\left (5 \right )}\right ) \log {\left (x \right )}}{-9 + 5 \log {\left (5 \right )}} + \frac {\left (78 - 60 \log {\left (5 \right )}\right ) \log {\left (x \right )}}{- 9 x + 5 x \log {\left (5 \right )} - 15 \log {\left (5 \right )} + 27} \] Input:
integrate(((120*ln(5)-156)*ln(x)+2*(-20*x+60)*ln(5)+52*x-156)/(2*(5*x**2-3 0*x+45)*ln(5)-18*x**2+108*x-162),x)
Output:
(26 - 20*log(5))*log(x)/(-9 + 5*log(5)) + (78 - 60*log(5))*log(x)/(-9*x + 5*x*log(5) - 15*log(5) + 27)
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (22) = 44\).
Time = 0.03 (sec) , antiderivative size = 141, normalized size of antiderivative = 6.41 \[ \int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+\left (45-30 x+5 x^2\right ) \log (25)} \, dx=20 \, {\left (\frac {\log \left (x - 3\right )}{5 \, \log \left (5\right ) - 9} - \frac {\log \left (x\right )}{5 \, \log \left (5\right ) - 9}\right )} \log \left (5\right ) - 20 \, {\left (\frac {\log \left (x - 3\right )}{5 \, \log \left (5\right ) - 9} - \frac {3}{x {\left (5 \, \log \left (5\right ) - 9\right )} - 15 \, \log \left (5\right ) + 27}\right )} \log \left (5\right ) - \frac {60 \, \log \left (5\right ) \log \left (x\right )}{x {\left (5 \, \log \left (5\right ) - 9\right )} - 15 \, \log \left (5\right ) + 27} - \frac {60 \, \log \left (5\right )}{x {\left (5 \, \log \left (5\right ) - 9\right )} - 15 \, \log \left (5\right ) + 27} + \frac {78 \, \log \left (x\right )}{x {\left (5 \, \log \left (5\right ) - 9\right )} - 15 \, \log \left (5\right ) + 27} + \frac {26 \, \log \left (x\right )}{5 \, \log \left (5\right ) - 9} \] Input:
integrate(((120*log(5)-156)*log(x)+2*(-20*x+60)*log(5)+52*x-156)/(2*(5*x^2 -30*x+45)*log(5)-18*x^2+108*x-162),x, algorithm="maxima")
Output:
20*(log(x - 3)/(5*log(5) - 9) - log(x)/(5*log(5) - 9))*log(5) - 20*(log(x - 3)/(5*log(5) - 9) - 3/(x*(5*log(5) - 9) - 15*log(5) + 27))*log(5) - 60*l og(5)*log(x)/(x*(5*log(5) - 9) - 15*log(5) + 27) - 60*log(5)/(x*(5*log(5) - 9) - 15*log(5) + 27) + 78*log(x)/(x*(5*log(5) - 9) - 15*log(5) + 27) + 2 6*log(x)/(5*log(5) - 9)
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05 \[ \int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+\left (45-30 x+5 x^2\right ) \log (25)} \, dx=-\frac {6 \, {\left (10 \, \log \left (5\right ) - 13\right )} \log \left (x\right )}{5 \, x \log \left (5\right ) - 9 \, x - 15 \, \log \left (5\right ) + 27} - \frac {2 \, {\left (10 \, \log \left (5\right ) - 13\right )} \log \left (x\right )}{5 \, \log \left (5\right ) - 9} \] Input:
integrate(((120*log(5)-156)*log(x)+2*(-20*x+60)*log(5)+52*x-156)/(2*(5*x^2 -30*x+45)*log(5)-18*x^2+108*x-162),x, algorithm="giac")
Output:
-6*(10*log(5) - 13)*log(x)/(5*x*log(5) - 9*x - 15*log(5) + 27) - 2*(10*log (5) - 13)*log(x)/(5*log(5) - 9)
Time = 4.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+\left (45-30 x+5 x^2\right ) \log (25)} \, dx=-\frac {2\,x\,\ln \left (x\right )\,\left (10\,\ln \left (5\right )-13\right )}{\left (5\,\ln \left (5\right )-9\right )\,\left (x-3\right )} \] Input:
int((52*x - 2*log(5)*(20*x - 60) + log(x)*(120*log(5) - 156) - 156)/(108*x + 2*log(5)*(5*x^2 - 30*x + 45) - 18*x^2 - 162),x)
Output:
-(2*x*log(x)*(10*log(5) - 13))/((5*log(5) - 9)*(x - 3))
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-156+52 x+(60-20 x) \log (25)+(-156+60 \log (25)) \log (x)}{-162+108 x-18 x^2+\left (45-30 x+5 x^2\right ) \log (25)} \, dx=\frac {2 \,\mathrm {log}\left (x \right ) x \left (-10 \,\mathrm {log}\left (5\right )+13\right )}{5 \,\mathrm {log}\left (5\right ) x -15 \,\mathrm {log}\left (5\right )-9 x +27} \] Input:
int(((120*log(5)-156)*log(x)+2*(-20*x+60)*log(5)+52*x-156)/(2*(5*x^2-30*x+ 45)*log(5)-18*x^2+108*x-162),x)
Output:
(2*log(x)*x*( - 10*log(5) + 13))/(5*log(5)*x - 15*log(5) - 9*x + 27)