Integrand size = 76, antiderivative size = 28 \[ \int \frac {-58 x+19 x^2+64 x^3+20 x^4+\left (4 x-3 x^2-x^3\right ) \log (x)+\left (-40 x+30 x^3+10 x^4\right ) \log \left (\frac {-8+8 x}{x}\right )}{-20+15 x^2+5 x^3} \, dx=x^2 \left (2-\frac {\log (x)}{5 (2+x)}+\log \left (\frac {4 (-2+2 x)}{x}\right )\right ) \] Output:
x^2*(2+ln(4*(-2+2*x)/x)-ln(x)/(5*x+10))
Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-58 x+19 x^2+64 x^3+20 x^4+\left (4 x-3 x^2-x^3\right ) \log (x)+\left (-40 x+30 x^3+10 x^4\right ) \log \left (\frac {-8+8 x}{x}\right )}{-20+15 x^2+5 x^3} \, dx=\frac {1}{5} x^2 \left (10+5 \log \left (8-\frac {8}{x}\right )-\frac {\log (x)}{2+x}\right ) \] Input:
Integrate[(-58*x + 19*x^2 + 64*x^3 + 20*x^4 + (4*x - 3*x^2 - x^3)*Log[x] + (-40*x + 30*x^3 + 10*x^4)*Log[(-8 + 8*x)/x])/(-20 + 15*x^2 + 5*x^3),x]
Output:
(x^2*(10 + 5*Log[8 - 8/x] - Log[x]/(2 + x)))/5
Leaf count is larger than twice the leaf count of optimal. \(238\) vs. \(2(28)=56\).
Time = 1.37 (sec) , antiderivative size = 238, normalized size of antiderivative = 8.50, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2463, 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.
\(\displaystyle \int \frac {20 x^4+64 x^3+19 x^2+\left (10 x^4+30 x^3-40 x\right ) \log \left (\frac {8 x-8}{x}\right )+\left (-x^3-3 x^2+4 x\right ) \log (x)-58 x}{5 x^3+15 x^2-20} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {20 x^4+64 x^3+19 x^2+\left (10 x^4+30 x^3-40 x\right ) \log \left (\frac {8 x-8}{x}\right )+\left (-x^3-3 x^2+4 x\right ) \log (x)-58 x}{45 (x-1)}-\frac {20 x^4+64 x^3+19 x^2+\left (10 x^4+30 x^3-40 x\right ) \log \left (\frac {8 x-8}{x}\right )+\left (-x^3-3 x^2+4 x\right ) \log (x)-58 x}{45 (x+2)}-\frac {20 x^4+64 x^3+19 x^2+\left (10 x^4+30 x^3-40 x\right ) \log \left (\frac {8 x-8}{x}\right )+\left (-x^3-3 x^2+4 x\right ) \log (x)-58 x}{15 (x+2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{18} x^4 \log \left (8-\frac {8}{x}\right )+\frac {1}{18} x^4 \log (x)-\frac {1}{18} x^4 \left (\log \left (-\frac {8 (1-x)}{x}\right )+\log (x)-\log (8 x-8)\right )-\frac {1}{18} x^4 \log (8 x-8)+\frac {2}{27} x^3 \log \left (8-\frac {8}{x}\right )+\frac {11}{135} x^3 \log (x)-\frac {2}{27} x^3 \left (\log \left (-\frac {8 (1-x)}{x}\right )+\log (x)-\log (8 x-8)\right )-\frac {2}{27} x^3 \log (8 x-8)+2 x^2+\frac {7}{9} x^2 \log \left (8-\frac {8}{x}\right )-\frac {8}{45} x^2 \log (x)+\frac {2}{9} x^2 \left (\log \left (-\frac {8 (1-x)}{x}\right )+\log (x)-\log (8 x-8)\right )+\frac {2}{9} x^2 \log (8 x-8)-\frac {1}{135} \left (x^3+6 x^2\right ) \log (x)+\frac {2 x \log (x)}{5 (x+2)}-\frac {1}{5} x \log (x)\) |
Input:
Int[(-58*x + 19*x^2 + 64*x^3 + 20*x^4 + (4*x - 3*x^2 - x^3)*Log[x] + (-40* x + 30*x^3 + 10*x^4)*Log[(-8 + 8*x)/x])/(-20 + 15*x^2 + 5*x^3),x]
Output:
2*x^2 + (7*x^2*Log[8 - 8/x])/9 + (2*x^3*Log[8 - 8/x])/27 + (x^4*Log[8 - 8/ x])/18 - (x*Log[x])/5 - (8*x^2*Log[x])/45 + (11*x^3*Log[x])/135 + (x^4*Log [x])/18 + (2*x*Log[x])/(5*(2 + x)) - ((6*x^2 + x^3)*Log[x])/135 + (2*x^2*( Log[(-8*(1 - x))/x] + Log[x] - Log[-8 + 8*x]))/9 - (2*x^3*(Log[(-8*(1 - x) )/x] + Log[x] - Log[-8 + 8*x]))/27 - (x^4*(Log[(-8*(1 - x))/x] + Log[x] - Log[-8 + 8*x]))/18 + (2*x^2*Log[-8 + 8*x])/9 - (2*x^3*Log[-8 + 8*x])/27 - (x^4*Log[-8 + 8*x])/18
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 = 0.92 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.93
method | result | size |
parallelrisch | \(\frac {15 \ln \left (\frac {8 x -8}{x}\right ) x^{3}+30 x^{3}-3 x^{2} \ln \left (x \right )+30 \ln \left (\frac {8 x -8}{x}\right ) x^{2}+60 x^{2}}{30+15 x}\) | \(54\) |
parts | \(\ln \left (-\frac {8}{x}\right )-\frac {\ln \left (8-\frac {8}{x}\right ) \left (8-\frac {8}{x}\right ) \left (-\frac {8}{x}-8\right ) x^{2}}{64}+2 x^{2}+\ln \left (-1+x \right )-\frac {x \ln \left (x \right )}{5}+\frac {2 \ln \left (x \right ) x}{5 \left (2+x \right )}\) | \(59\) |
default | \(2 x^{2}+\ln \left (-1+x \right )+3 x^{2} \ln \left (2\right )+\ln \left (-\frac {1}{x}\right )-\ln \left (1-\frac {1}{x}\right ) \left (1-\frac {1}{x}\right ) \left (-1-\frac {1}{x}\right ) x^{2}-\frac {x \ln \left (x \right )}{5}+\frac {2 \ln \left (x \right ) x}{5 \left (2+x \right )}\) | \(66\) |
risch | \(x^{2} \ln \left (-1+x \right )+\frac {-5 i \pi \,x^{3} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )-10 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )-10 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{3}+10 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}+30 x^{3} \ln \left (2\right )+60 x^{2} \ln \left (2\right )-10 x^{3} \ln \left (x \right )-22 x^{2} \ln \left (x \right )+40 x^{2}+20 x^{3}+5 i \pi \,x^{3} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}+10 i \pi \,x^{2} \operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}+5 i \pi \,x^{3} \operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}-5 i \pi \,x^{3} \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{3}}{10 x +20}\) | \(260\) |
orering | \(\frac {\left (x^{6}+7 x^{5}-137 x^{4}+713 x^{3}-1082 x^{2}+972 x -324\right ) \left (\left (-x^{3}-3 x^{2}+4 x \right ) \ln \left (x \right )+\left (10 x^{4}+30 x^{3}-40 x \right ) \ln \left (\frac {8 x -8}{x}\right )+20 x^{4}+64 x^{3}+19 x^{2}-58 x \right )}{x \left (x^{4}+9 x^{3}-143 x^{2}+475 x -132\right ) \left (5 x^{3}+15 x^{2}-20\right )}-\frac {\left (x^{6}+5 x^{5}-131 x^{4}+951 x^{3}-2032 x^{2}+1944 x -648\right ) \left (\frac {\left (-3 x^{2}-6 x +4\right ) \ln \left (x \right )+\frac {-x^{3}-3 x^{2}+4 x}{x}+\left (40 x^{3}+90 x^{2}-40\right ) \ln \left (\frac {8 x -8}{x}\right )+\frac {\left (10 x^{4}+30 x^{3}-40 x \right ) \left (\frac {8}{x}-\frac {8 x -8}{x^{2}}\right ) x}{8 x -8}+80 x^{3}+192 x^{2}+38 x -58}{5 x^{3}+15 x^{2}-20}-\frac {\left (\left (-x^{3}-3 x^{2}+4 x \right ) \ln \left (x \right )+\left (10 x^{4}+30 x^{3}-40 x \right ) \ln \left (\frac {8 x -8}{x}\right )+20 x^{4}+64 x^{3}+19 x^{2}-58 x \right ) \left (15 x^{2}+30 x \right )}{\left (5 x^{3}+15 x^{2}-20\right )^{2}}\right )}{2 \left (x^{4}+9 x^{3}-143 x^{2}+475 x -132\right )}+\frac {x \left (x^{4}+119 x^{2}-297 x +162\right ) \left (-1+x \right ) \left (2+x \right ) \left (\frac {\left (-6 x -6\right ) \ln \left (x \right )+\frac {-6 x^{2}-12 x +8}{x}-\frac {-x^{3}-3 x^{2}+4 x}{x^{2}}+\left (120 x^{2}+180 x \right ) \ln \left (\frac {8 x -8}{x}\right )+\frac {2 \left (40 x^{3}+90 x^{2}-40\right ) \left (\frac {8}{x}-\frac {8 x -8}{x^{2}}\right ) x}{8 x -8}+\frac {\left (10 x^{4}+30 x^{3}-40 x \right ) \left (-\frac {16}{x^{2}}+\frac {16 x -16}{x^{3}}\right ) x}{8 x -8}-\frac {8 \left (10 x^{4}+30 x^{3}-40 x \right ) \left (\frac {8}{x}-\frac {8 x -8}{x^{2}}\right ) x}{\left (8 x -8\right )^{2}}+\frac {\left (10 x^{4}+30 x^{3}-40 x \right ) \left (\frac {8}{x}-\frac {8 x -8}{x^{2}}\right )}{8 x -8}+240 x^{2}+384 x +38}{5 x^{3}+15 x^{2}-20}-\frac {2 \left (\left (-3 x^{2}-6 x +4\right ) \ln \left (x \right )+\frac {-x^{3}-3 x^{2}+4 x}{x}+\left (40 x^{3}+90 x^{2}-40\right ) \ln \left (\frac {8 x -8}{x}\right )+\frac {\left (10 x^{4}+30 x^{3}-40 x \right ) \left (\frac {8}{x}-\frac {8 x -8}{x^{2}}\right ) x}{8 x -8}+80 x^{3}+192 x^{2}+38 x -58\right ) \left (15 x^{2}+30 x \right )}{\left (5 x^{3}+15 x^{2}-20\right )^{2}}+\frac {2 \left (\left (-x^{3}-3 x^{2}+4 x \right ) \ln \left (x \right )+\left (10 x^{4}+30 x^{3}-40 x \right ) \ln \left (\frac {8 x -8}{x}\right )+20 x^{4}+64 x^{3}+19 x^{2}-58 x \right ) \left (15 x^{2}+30 x \right )^{2}}{\left (5 x^{3}+15 x^{2}-20\right )^{3}}-\frac {\left (\left (-x^{3}-3 x^{2}+4 x \right ) \ln \left (x \right )+\left (10 x^{4}+30 x^{3}-40 x \right ) \ln \left (\frac {8 x -8}{x}\right )+20 x^{4}+64 x^{3}+19 x^{2}-58 x \right ) \left (30 x +30\right )}{\left (5 x^{3}+15 x^{2}-20\right )^{2}}\right )}{2 x^{4}+18 x^{3}-286 x^{2}+950 x -264}\) | \(977\) |
Input:
int(((-x^3-3*x^2+4*x)*ln(x)+(10*x^4+30*x^3-40*x)*ln((8*x-8)/x)+20*x^4+64*x ^3+19*x^2-58*x)/(5*x^3+15*x^2-20),x,method=_RETURNVERBOSE)
Output:
1/15*(15*ln(8*(-1+x)/x)*x^3+30*x^3-3*x^2*ln(x)+30*ln(8*(-1+x)/x)*x^2+60*x^ 2)/(2+x)
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {-58 x+19 x^2+64 x^3+20 x^4+\left (4 x-3 x^2-x^3\right ) \log (x)+\left (-40 x+30 x^3+10 x^4\right ) \log \left (\frac {-8+8 x}{x}\right )}{-20+15 x^2+5 x^3} \, dx=\frac {10 \, x^{3} - x^{2} \log \left (x\right ) + 20 \, x^{2} + 5 \, {\left (x^{3} + 2 \, x^{2}\right )} \log \left (\frac {8 \, {\left (x - 1\right )}}{x}\right )}{5 \, {\left (x + 2\right )}} \] Input:
integrate(((-x^3-3*x^2+4*x)*log(x)+(10*x^4+30*x^3-40*x)*log((8*x-8)/x)+20* x^4+64*x^3+19*x^2-58*x)/(5*x^3+15*x^2-20),x, algorithm="fricas")
Output:
1/5*(10*x^3 - x^2*log(x) + 20*x^2 + 5*(x^3 + 2*x^2)*log(8*(x - 1)/x))/(x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).
Time = 0.35 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {-58 x+19 x^2+64 x^3+20 x^4+\left (4 x-3 x^2-x^3\right ) \log (x)+\left (-40 x+30 x^3+10 x^4\right ) \log \left (\frac {-8+8 x}{x}\right )}{-20+15 x^2+5 x^3} \, dx=2 x^{2} + \left (x^{2} - \frac {1}{6}\right ) \log {\left (\frac {8 x - 8}{x} \right )} + \frac {7 \log {\left (x \right )}}{30} + \frac {\log {\left (x - 1 \right )}}{6} + \frac {\left (- x^{2} - 2 x - 4\right ) \log {\left (x \right )}}{5 x + 10} \] Input:
integrate(((-x**3-3*x**2+4*x)*ln(x)+(10*x**4+30*x**3-40*x)*ln((8*x-8)/x)+2 0*x**4+64*x**3+19*x**2-58*x)/(5*x**3+15*x**2-20),x)
Output:
2*x**2 + (x**2 - 1/6)*log((8*x - 8)/x) + 7*log(x)/30 + log(x - 1)/6 + (-x* *2 - 2*x - 4)*log(x)/(5*x + 10)
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (26) = 52\).
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.86 \[ \int \frac {-58 x+19 x^2+64 x^3+20 x^4+\left (4 x-3 x^2-x^3\right ) \log (x)+\left (-40 x+30 x^3+10 x^4\right ) \log \left (\frac {-8+8 x}{x}\right )}{-20+15 x^2+5 x^3} \, dx=\frac {45 \, x^{3} {\left (3 \, \log \left (2\right ) + 2\right )} + 90 \, x^{2} {\left (3 \, \log \left (2\right ) + 2\right )} + {\left (45 \, x^{3} + 90 \, x^{2} + 58 \, x + 116\right )} \log \left (x - 1\right ) - 9 \, {\left (5 \, x^{3} + 11 \, x^{2}\right )} \log \left (x\right ) - 348}{45 \, {\left (x + 2\right )}} + \frac {116}{15 \, {\left (x + 2\right )}} - \frac {58}{45} \, \log \left (x - 1\right ) \] Input:
integrate(((-x^3-3*x^2+4*x)*log(x)+(10*x^4+30*x^3-40*x)*log((8*x-8)/x)+20* x^4+64*x^3+19*x^2-58*x)/(5*x^3+15*x^2-20),x, algorithm="maxima")
Output:
1/45*(45*x^3*(3*log(2) + 2) + 90*x^2*(3*log(2) + 2) + (45*x^3 + 90*x^2 + 5 8*x + 116)*log(x - 1) - 9*(5*x^3 + 11*x^2)*log(x) - 348)/(x + 2) + 116/15/ (x + 2) - 58/45*log(x - 1)
Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-58 x+19 x^2+64 x^3+20 x^4+\left (4 x-3 x^2-x^3\right ) \log (x)+\left (-40 x+30 x^3+10 x^4\right ) \log \left (\frac {-8+8 x}{x}\right )}{-20+15 x^2+5 x^3} \, dx=x^{2} \log \left (8 \, x - 8\right ) + 2 \, x^{2} - \frac {1}{5} \, {\left (5 \, x^{2} + x + \frac {4}{x + 2}\right )} \log \left (x\right ) + \frac {2}{5} \, \log \left (x\right ) \] Input:
integrate(((-x^3-3*x^2+4*x)*log(x)+(10*x^4+30*x^3-40*x)*log((8*x-8)/x)+20* x^4+64*x^3+19*x^2-58*x)/(5*x^3+15*x^2-20),x, algorithm="giac")
Output:
x^2*log(8*x - 8) + 2*x^2 - 1/5*(5*x^2 + x + 4/(x + 2))*log(x) + 2/5*log(x)
Time = 2.80 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {-58 x+19 x^2+64 x^3+20 x^4+\left (4 x-3 x^2-x^3\right ) \log (x)+\left (-40 x+30 x^3+10 x^4\right ) \log \left (\frac {-8+8 x}{x}\right )}{-20+15 x^2+5 x^3} \, dx=x^2\,\ln \left (\frac {8\,x-8}{x}\right )+2\,x^2-\frac {x^2\,\ln \left (x\right )}{5\,\left (x+2\right )} \] Input:
int((19*x^2 - log(x)*(3*x^2 - 4*x + x^3) - 58*x + 64*x^3 + 20*x^4 + log((8 *x - 8)/x)*(30*x^3 - 40*x + 10*x^4))/(15*x^2 + 5*x^3 - 20),x)
Output:
x^2*log((8*x - 8)/x) + 2*x^2 - (x^2*log(x))/(5*(x + 2))
Time = 0.51 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.68 \[ \int \frac {-58 x+19 x^2+64 x^3+20 x^4+\left (4 x-3 x^2-x^3\right ) \log (x)+\left (-40 x+30 x^3+10 x^4\right ) \log \left (\frac {-8+8 x}{x}\right )}{-20+15 x^2+5 x^3} \, dx=\frac {5 \,\mathrm {log}\left (x -1\right ) x +10 \,\mathrm {log}\left (x -1\right )+5 \,\mathrm {log}\left (\frac {8 x -8}{x}\right ) x^{3}+10 \,\mathrm {log}\left (\frac {8 x -8}{x}\right ) x^{2}-5 \,\mathrm {log}\left (\frac {8 x -8}{x}\right ) x -10 \,\mathrm {log}\left (\frac {8 x -8}{x}\right )-\mathrm {log}\left (x \right ) x^{2}-5 \,\mathrm {log}\left (x \right ) x -10 \,\mathrm {log}\left (x \right )+10 x^{3}+20 x^{2}}{5 x +10} \] Input:
int(((-x^3-3*x^2+4*x)*log(x)+(10*x^4+30*x^3-40*x)*log((8*x-8)/x)+20*x^4+64 *x^3+19*x^2-58*x)/(5*x^3+15*x^2-20),x)
Output:
(5*log(x - 1)*x + 10*log(x - 1) + 5*log((8*x - 8)/x)*x**3 + 10*log((8*x - 8)/x)*x**2 - 5*log((8*x - 8)/x)*x - 10*log((8*x - 8)/x) - log(x)*x**2 - 5* log(x)*x - 10*log(x) + 10*x**3 + 20*x**2)/(5*(x + 2))