Integrand size = 80, antiderivative size = 31 \[ \int \frac {15 x^2-3 x^3+\left (45 x-45 x^2+10 x^3\right ) \log \left (\frac {-3+2 x}{-3+x}\right )}{\left (-2250+3600 x-2120 x^2+588 x^3-78 x^4+4 x^5\right ) \log ^3\left (\frac {-3+2 x}{-3+x}\right )} \, dx=\log (2)-\frac {x^2}{4 (5-x)^2 \log ^2\left (\frac {-3+2 x}{-3+x}\right )} \]
Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {15 x^2-3 x^3+\left (45 x-45 x^2+10 x^3\right ) \log \left (\frac {-3+2 x}{-3+x}\right )}{\left (-2250+3600 x-2120 x^2+588 x^3-78 x^4+4 x^5\right ) \log ^3\left (\frac {-3+2 x}{-3+x}\right )} \, dx=-\frac {x^2}{4 (-5+x)^2 \log ^2\left (\frac {-3+2 x}{-3+x}\right )} \]
Integrate[(15*x^2 - 3*x^3 + (45*x - 45*x^2 + 10*x^3)*Log[(-3 + 2*x)/(-3 + x)])/((-2250 + 3600*x - 2120*x^2 + 588*x^3 - 78*x^4 + 4*x^5)*Log[(-3 + 2*x )/(-3 + x)]^3),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.
\(\displaystyle \int \frac {-3 x^3+15 x^2+\left (10 x^3-45 x^2+45 x\right ) \log \left (\frac {2 x-3}{x-3}\right )}{\left (4 x^5-78 x^4+588 x^3-2120 x^2+3600 x-2250\right ) \log ^3\left (\frac {2 x-3}{x-3}\right )} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {93 \left (-3 x^3+15 x^2+\left (10 x^3-45 x^2+45 x\right ) \log \left (\frac {2 x-3}{x-3}\right )\right )}{5488 (x-5) \log ^3\left (\frac {2 x-3}{x-3}\right )}-\frac {-3 x^3+15 x^2+\left (10 x^3-45 x^2+45 x\right ) \log \left (\frac {2 x-3}{x-3}\right )}{48 (x-3) \log ^3\left (\frac {2 x-3}{x-3}\right )}+\frac {8 \left (-3 x^3+15 x^2+\left (10 x^3-45 x^2+45 x\right ) \log \left (\frac {2 x-3}{x-3}\right )\right )}{1029 (2 x-3) \log ^3\left (\frac {2 x-3}{x-3}\right )}-\frac {11 \left (-3 x^3+15 x^2+\left (10 x^3-45 x^2+45 x\right ) \log \left (\frac {2 x-3}{x-3}\right )\right )}{392 (x-5)^2 \log ^3\left (\frac {2 x-3}{x-3}\right )}+\frac {-3 x^3+15 x^2+\left (10 x^3-45 x^2+45 x\right ) \log \left (\frac {2 x-3}{x-3}\right )}{28 (x-5)^3 \log ^3\left (\frac {2 x-3}{x-3}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {279 \int \frac {x^2}{\log ^3\left (\frac {2 x-3}{x-3}\right )}dx}{5488}-\frac {3}{28} \int \frac {x^2}{(x-5)^2 \log ^3\left (\frac {2 x-3}{x-3}\right )}dx+\frac {33}{392} \int \frac {x^2}{(x-5) \log ^3\left (\frac {2 x-3}{x-3}\right )}dx+\frac {1}{16} \int \frac {(x-5) x^2}{(x-3) \log ^3\left (\frac {2 x-3}{x-3}\right )}dx-\frac {8}{343} \int \frac {(x-5) x^2}{(2 x-3) \log ^3\left (\frac {2 x-3}{x-3}\right )}dx+\frac {40 \int \frac {(x-3) x}{\log ^2\left (\frac {2 x-3}{x-3}\right )}dx}{1029}-\frac {5}{48} \int \frac {x (2 x-3)}{\log ^2\left (\frac {2 x-3}{x-3}\right )}dx+\frac {5}{28} \int \frac {(x-3) x (2 x-3)}{(x-5)^3 \log ^2\left (\frac {2 x-3}{x-3}\right )}dx-\frac {55}{392} \int \frac {(x-3) x (2 x-3)}{(x-5)^2 \log ^2\left (\frac {2 x-3}{x-3}\right )}dx+\frac {465 \int \frac {(x-3) x (2 x-3)}{(x-5) \log ^2\left (\frac {2 x-3}{x-3}\right )}dx}{5488}\) |
Int[(15*x^2 - 3*x^3 + (45*x - 45*x^2 + 10*x^3)*Log[(-3 + 2*x)/(-3 + x)])/( (-2250 + 3600*x - 2120*x^2 + 588*x^3 - 78*x^4 + 4*x^5)*Log[(-3 + 2*x)/(-3 + x)]^3),x]
3.13.23.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 23.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81
method | result | size |
norman | \(-\frac {x^{2}}{4 \left (-5+x \right )^{2} \ln \left (\frac {-3+2 x}{-3+x}\right )^{2}}\) | \(25\) |
risch | \(-\frac {x^{2}}{4 \left (x^{2}-10 x +25\right ) \ln \left (\frac {-3+2 x}{-3+x}\right )^{2}}\) | \(30\) |
parallelrisch | \(-\frac {x^{2}}{4 \left (x^{2}-10 x +25\right ) \ln \left (\frac {-3+2 x}{-3+x}\right )^{2}}\) | \(30\) |
derivativedivides | \(-\frac {45 \left (28 \left (2+\frac {3}{-3+x}\right ) \ln \left (2+\frac {3}{-3+x}\right )-63 \ln \left (2+\frac {3}{-3+x}\right )+30-\frac {60}{-3+x}\right )}{112 \left (-3+\frac {6}{-3+x}\right )^{2} \ln \left (2+\frac {3}{-3+x}\right )^{3}}-\frac {9}{16 \ln \left (2+\frac {3}{-3+x}\right )^{2}}-\frac {225}{56 \ln \left (2+\frac {3}{-3+x}\right )^{3} \left (-3+\frac {6}{-3+x}\right )}\) | \(108\) |
default | \(-\frac {45 \left (28 \left (2+\frac {3}{-3+x}\right ) \ln \left (2+\frac {3}{-3+x}\right )-63 \ln \left (2+\frac {3}{-3+x}\right )+30-\frac {60}{-3+x}\right )}{112 \left (-3+\frac {6}{-3+x}\right )^{2} \ln \left (2+\frac {3}{-3+x}\right )^{3}}-\frac {9}{16 \ln \left (2+\frac {3}{-3+x}\right )^{2}}-\frac {225}{56 \ln \left (2+\frac {3}{-3+x}\right )^{3} \left (-3+\frac {6}{-3+x}\right )}\) | \(108\) |
int(((10*x^3-45*x^2+45*x)*ln((-3+2*x)/(-3+x))-3*x^3+15*x^2)/(4*x^5-78*x^4+ 588*x^3-2120*x^2+3600*x-2250)/ln((-3+2*x)/(-3+x))^3,x,method=_RETURNVERBOS E)
Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {15 x^2-3 x^3+\left (45 x-45 x^2+10 x^3\right ) \log \left (\frac {-3+2 x}{-3+x}\right )}{\left (-2250+3600 x-2120 x^2+588 x^3-78 x^4+4 x^5\right ) \log ^3\left (\frac {-3+2 x}{-3+x}\right )} \, dx=-\frac {x^{2}}{4 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (\frac {2 \, x - 3}{x - 3}\right )^{2}} \]
integrate(((10*x^3-45*x^2+45*x)*log((-3+2*x)/(-3+x))-3*x^3+15*x^2)/(4*x^5- 78*x^4+588*x^3-2120*x^2+3600*x-2250)/log((-3+2*x)/(-3+x))^3,x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {15 x^2-3 x^3+\left (45 x-45 x^2+10 x^3\right ) \log \left (\frac {-3+2 x}{-3+x}\right )}{\left (-2250+3600 x-2120 x^2+588 x^3-78 x^4+4 x^5\right ) \log ^3\left (\frac {-3+2 x}{-3+x}\right )} \, dx=- \frac {x^{2}}{\left (4 x^{2} - 40 x + 100\right ) \log {\left (\frac {2 x - 3}{x - 3} \right )}^{2}} \]
integrate(((10*x**3-45*x**2+45*x)*ln((-3+2*x)/(-3+x))-3*x**3+15*x**2)/(4*x **5-78*x**4+588*x**3-2120*x**2+3600*x-2250)/ln((-3+2*x)/(-3+x))**3,x)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.94 \[ \int \frac {15 x^2-3 x^3+\left (45 x-45 x^2+10 x^3\right ) \log \left (\frac {-3+2 x}{-3+x}\right )}{\left (-2250+3600 x-2120 x^2+588 x^3-78 x^4+4 x^5\right ) \log ^3\left (\frac {-3+2 x}{-3+x}\right )} \, dx=-\frac {x^{2}}{4 \, {\left ({\left (x^{2} - 10 \, x + 25\right )} \log \left (2 \, x - 3\right )^{2} - 2 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (2 \, x - 3\right ) \log \left (x - 3\right ) + {\left (x^{2} - 10 \, x + 25\right )} \log \left (x - 3\right )^{2}\right )}} \]
integrate(((10*x^3-45*x^2+45*x)*log((-3+2*x)/(-3+x))-3*x^3+15*x^2)/(4*x^5- 78*x^4+588*x^3-2120*x^2+3600*x-2250)/log((-3+2*x)/(-3+x))^3,x, algorithm=\
-1/4*x^2/((x^2 - 10*x + 25)*log(2*x - 3)^2 - 2*(x^2 - 10*x + 25)*log(2*x - 3)*log(x - 3) + (x^2 - 10*x + 25)*log(x - 3)^2)
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (27) = 54\).
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.29 \[ \int \frac {15 x^2-3 x^3+\left (45 x-45 x^2+10 x^3\right ) \log \left (\frac {-3+2 x}{-3+x}\right )}{\left (-2250+3600 x-2120 x^2+588 x^3-78 x^4+4 x^5\right ) \log ^3\left (\frac {-3+2 x}{-3+x}\right )} \, dx=-\frac {9 \, {\left (\frac {{\left (2 \, x - 3\right )}^{2}}{{\left (x - 3\right )}^{2}} - \frac {2 \, {\left (2 \, x - 3\right )}}{x - 3} + 1\right )}}{4 \, {\left (\frac {4 \, {\left (2 \, x - 3\right )}^{2} \log \left (\frac {2 \, x - 3}{x - 3}\right )^{2}}{{\left (x - 3\right )}^{2}} - \frac {28 \, {\left (2 \, x - 3\right )} \log \left (\frac {2 \, x - 3}{x - 3}\right )^{2}}{x - 3} + 49 \, \log \left (\frac {2 \, x - 3}{x - 3}\right )^{2}\right )}} \]
integrate(((10*x^3-45*x^2+45*x)*log((-3+2*x)/(-3+x))-3*x^3+15*x^2)/(4*x^5- 78*x^4+588*x^3-2120*x^2+3600*x-2250)/log((-3+2*x)/(-3+x))^3,x, algorithm=\
-9/4*((2*x - 3)^2/(x - 3)^2 - 2*(2*x - 3)/(x - 3) + 1)/(4*(2*x - 3)^2*log( (2*x - 3)/(x - 3))^2/(x - 3)^2 - 28*(2*x - 3)*log((2*x - 3)/(x - 3))^2/(x - 3) + 49*log((2*x - 3)/(x - 3))^2)
Time = 0.51 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {15 x^2-3 x^3+\left (45 x-45 x^2+10 x^3\right ) \log \left (\frac {-3+2 x}{-3+x}\right )}{\left (-2250+3600 x-2120 x^2+588 x^3-78 x^4+4 x^5\right ) \log ^3\left (\frac {-3+2 x}{-3+x}\right )} \, dx=\frac {875}{6\,{\left (x-5\right )}^2}-\frac {175\,x}{3\,{\left (x-5\right )}^2}+\frac {35\,x^2}{6\,{\left (x-5\right )}^2}-\frac {x^2}{4\,{\ln \left (\frac {2\,x-3}{x-3}\right )}^2\,{\left (x-5\right )}^2} \]
int((log((2*x - 3)/(x - 3))*(45*x - 45*x^2 + 10*x^3) + 15*x^2 - 3*x^3)/(lo g((2*x - 3)/(x - 3))^3*(3600*x - 2120*x^2 + 588*x^3 - 78*x^4 + 4*x^5 - 225 0)),x)