Integrand size = 74, antiderivative size = 27 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x \log \left (\frac {5 e^{-\frac {-4+x^2}{x}} (-3+x)}{6-3 x}\right ) \] Output:
ln(5/(-3*x+6)*(-3+x)/exp((x^2-4)/x))*x
Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(27)=54\).
Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=-4-2 \log (2-x)+3 \log (3-x)-3 \log (-3+x)+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (-3+x)}{3 (-2+x)}\right )+2 \log (-2+x) \] Input:
Integrate[(-24 + 20*x - 9*x^2 + 5*x^3 - x^4 + (6*x - 5*x^2 + x^3)*Log[(15 - 5*x)/(E^((-4 + x^2)/x)*(-6 + 3*x))])/(6*x - 5*x^2 + x^3),x]
Output:
-4 - 2*Log[2 - x] + 3*Log[3 - x] - 3*Log[-3 + x] + x*Log[(-5*E^(4/x - x)*( -3 + x))/(3*(-2 + x))] + 2*Log[-2 + x]
Time = 0.59 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {2026, 7279, 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.
\(\displaystyle \int \frac {-x^4+5 x^3-9 x^2+\left (x^3-5 x^2+6 x\right ) \log \left (\frac {e^{-\frac {x^2-4}{x}} (15-5 x)}{3 x-6}\right )+20 x-24}{x^3-5 x^2+6 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^4+5 x^3-9 x^2+\left (x^3-5 x^2+6 x\right ) \log \left (\frac {e^{-\frac {x^2-4}{x}} (15-5 x)}{3 x-6}\right )+20 x-24}{x \left (x^2-5 x+6\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {5 x^2}{x^2-5 x+6}-\frac {9 x}{x^2-5 x+6}+\frac {20}{x^2-5 x+6}-\frac {24}{\left (x^2-5 x+6\right ) x}-\frac {x^3}{x^2-5 x+6}+\log \left (-\frac {5 e^{\frac {4}{x}-x} (x-3)}{3 (x-2)}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right )\) |
Input:
Int[(-24 + 20*x - 9*x^2 + 5*x^3 - x^4 + (6*x - 5*x^2 + x^3)*Log[(15 - 5*x) /(E^((-4 + x^2)/x)*(-6 + 3*x))])/(6*x - 5*x^2 + x^3),x]
Output:
x*Log[(-5*E^(4/x - x)*(3 - x))/(3*(2 - x))]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.42 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(x \ln \left (-\frac {5 \,{\mathrm e}^{-\frac {x^{2}-4}{x}} \left (-3+x \right )}{-6+3 x}\right )\) | \(28\) |
default | \(x \ln \left (\frac {\left (15-5 x \right ) {\mathrm e}^{-\frac {x^{2}-4}{x}}}{-6+3 x}\right )\) | \(29\) |
norman | \(x \ln \left (\frac {\left (15-5 x \right ) {\mathrm e}^{-\frac {x^{2}-4}{x}}}{-6+3 x}\right )\) | \(29\) |
parts | \(x \ln \left (\frac {\left (15-5 x \right ) {\mathrm e}^{-\frac {x^{2}-4}{x}}}{-6+3 x}\right )\) | \(29\) |
risch | \(-\ln \left ({\mathrm e}^{\frac {\left (-2+x \right ) \left (2+x \right )}{x}}\right ) x -\ln \left (-2+x \right ) x +\ln \left (-3+x \right ) x +\frac {i \pi x \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )^{3}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (-3+x \right )\right ) \operatorname {csgn}\left (\frac {i}{-2+x}\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right )}{2}+i \pi x +\frac {i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )^{2}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{-2+x}\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right )^{2}}{2}-\frac {i \pi x \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right )^{3}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}}\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (-3+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-3+x \right )}{-2+x}\right )^{2}}{2}-i \pi x \operatorname {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (-2+x \right ) \left (2+x \right )}{x}} \left (-3+x \right )}{-2+x}\right )^{2}-x \ln \left (3\right )+x \ln \left (5\right )\) | \(359\) |
Input:
int(((x^3-5*x^2+6*x)*ln((15-5*x)/(-6+3*x)/exp((x^2-4)/x))-x^4+5*x^3-9*x^2+ 20*x-24)/(x^3-5*x^2+6*x),x,method=_RETURNVERBOSE)
Output:
x*ln(-5/(-6+3*x)/exp((x^2-4)/x)*(-3+x))
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x \log \left (-\frac {5 \, {\left (x - 3\right )} e^{\left (-\frac {x^{2} - 4}{x}\right )}}{3 \, {\left (x - 2\right )}}\right ) \] Input:
integrate(((x^3-5*x^2+6*x)*log((15-5*x)/(-6+3*x)/exp((x^2-4)/x))-x^4+5*x^3 -9*x^2+20*x-24)/(x^3-5*x^2+6*x),x, algorithm="fricas")
Output:
x*log(-5/3*(x - 3)*e^(-(x^2 - 4)/x)/(x - 2))
Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x \log {\left (\frac {\left (15 - 5 x\right ) e^{- \frac {x^{2} - 4}{x}}}{3 x - 6} \right )} \] Input:
integrate(((x**3-5*x**2+6*x)*ln((15-5*x)/(-6+3*x)/exp((x**2-4)/x))-x**4+5* x**3-9*x**2+20*x-24)/(x**3-5*x**2+6*x),x)
Output:
x*log((15 - 5*x)*exp(-(x**2 - 4)/x)/(3*x - 6))
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 200, normalized size of antiderivative = 7.41 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx={\left (i \, \pi + \log \left (5\right ) - \log \left (3\right )\right )} x - x^{2} + {\left (6 i \, \pi - x + 6 \, \log \left (5\right ) - 6 \, \log \left (3\right ) + 6 \, \log \left (x - 3\right ) + 2\right )} \log \left (x - 2\right ) - 3 \, \log \left (x - 2\right )^{2} + {\left (-6 i \, \pi + x - 6 \, \log \left (5\right ) + 6 \, \log \left (3\right ) + 7\right )} \log \left (x - 3\right ) - 3 \, \log \left (x - 3\right )^{2} - 6 \, {\left (\log \left (x - 2\right ) - \log \left (x - 3\right )\right )} \log \left (-\frac {5 \, x e^{\left (-x + \frac {4}{x}\right )}}{3 \, {\left (x - 2\right )}} + \frac {5 \, e^{\left (-x + \frac {4}{x}\right )}}{x - 2}\right ) - \frac {3 \, x \log \left (x - 2\right )^{2} + 3 \, x \log \left (x - 3\right )^{2} + 6 \, {\left (x^{2} - x \log \left (x - 3\right ) - 4\right )} \log \left (x - 2\right ) - 2 \, {\left (3 \, x^{2} - 5 \, x - 12\right )} \log \left (x - 3\right )}{x} - 2 \, \log \left (x - 2\right ) + 3 \, \log \left (x - 3\right ) \] Input:
integrate(((x^3-5*x^2+6*x)*log((15-5*x)/(-6+3*x)/exp((x^2-4)/x))-x^4+5*x^3 -9*x^2+20*x-24)/(x^3-5*x^2+6*x),x, algorithm="maxima")
Output:
(I*pi + log(5) - log(3))*x - x^2 + (6*I*pi - x + 6*log(5) - 6*log(3) + 6*l og(x - 3) + 2)*log(x - 2) - 3*log(x - 2)^2 + (-6*I*pi + x - 6*log(5) + 6*l og(3) + 7)*log(x - 3) - 3*log(x - 3)^2 - 6*(log(x - 2) - log(x - 3))*log(- 5/3*x*e^(-x + 4/x)/(x - 2) + 5*e^(-x + 4/x)/(x - 2)) - (3*x*log(x - 2)^2 + 3*x*log(x - 3)^2 + 6*(x^2 - x*log(x - 3) - 4)*log(x - 2) - 2*(3*x^2 - 5*x - 12)*log(x - 3))/x - 2*log(x - 2) + 3*log(x - 3)
Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=-x^{2} + x \log \left (-\frac {5 \, {\left (x - 3\right )}}{3 \, {\left (x - 2\right )}}\right ) \] Input:
integrate(((x^3-5*x^2+6*x)*log((15-5*x)/(-6+3*x)/exp((x^2-4)/x))-x^4+5*x^3 -9*x^2+20*x-24)/(x^3-5*x^2+6*x),x, algorithm="giac")
Output:
-x^2 + x*log(-5/3*(x - 3)/(x - 2))
Time = 3.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=x\,\ln \left (-\frac {5\,x-15}{3\,x-6}\right )-x^2 \] Input:
int((20*x + log(-(exp(-(x^2 - 4)/x)*(5*x - 15))/(3*x - 6))*(6*x - 5*x^2 + x^3) - 9*x^2 + 5*x^3 - x^4 - 24)/(6*x - 5*x^2 + x^3),x)
Output:
x*log(-(5*x - 15)/(3*x - 6)) - x^2
\[ \int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{6 x-5 x^2+x^3} \, dx=\int \frac {\left (x^{3}-5 x^{2}+6 x \right ) \mathrm {log}\left (\frac {15-5 x}{\left (-6+3 x \right ) {\mathrm e}^{\frac {x^{2}-4}{x}}}\right )-x^{4}+5 x^{3}-9 x^{2}+20 x -24}{x^{3}-5 x^{2}+6 x}d x \] Input:
int(((x^3-5*x^2+6*x)*log((15-5*x)/(-6+3*x)/exp((x^2-4)/x))-x^4+5*x^3-9*x^2 +20*x-24)/(x^3-5*x^2+6*x),x)
Output:
int(((x^3-5*x^2+6*x)*log((15-5*x)/(-6+3*x)/exp((x^2-4)/x))-x^4+5*x^3-9*x^2 +20*x-24)/(x^3-5*x^2+6*x),x)