Integrand size = 84, antiderivative size = 34 \[ \int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{\left (9 x-12 x^2+4 x^3\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=-2-x-5 (1+x)+\frac {2 x}{-3+2 x}+\frac {-2+2 x}{\log \left (\frac {x}{5}\right )} \] Output:
2/(-3+2*x)*x-7-6*x+(-2+2*x)/ln(1/5*x)
Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{\left (9 x-12 x^2+4 x^3\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=2 \left (-\frac {3}{2 (3-2 x)}-\frac {3}{2} (-3+2 x)+\frac {-1+x}{\log \left (\frac {x}{5}\right )}\right ) \] Input:
Integrate[(18 - 42*x + 32*x^2 - 8*x^3 + (18*x - 24*x^2 + 8*x^3)*Log[x/5] + (-60*x + 72*x^2 - 24*x^3)*Log[x/5]^2)/((9*x - 12*x^2 + 4*x^3)*Log[x/5]^2) ,x]
Output:
2*(-3/(2*(3 - 2*x)) - (3*(-3 + 2*x))/2 + (-1 + x)/Log[x/5])
Time = 0.69 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2026, 7239, 27, 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.
\(\displaystyle \int \frac {-8 x^3+32 x^2+\left (-24 x^3+72 x^2-60 x\right ) \log ^2\left (\frac {x}{5}\right )+\left (8 x^3-24 x^2+18 x\right ) \log \left (\frac {x}{5}\right )-42 x+18}{\left (4 x^3-12 x^2+9 x\right ) \log ^2\left (\frac {x}{5}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-8 x^3+32 x^2+\left (-24 x^3+72 x^2-60 x\right ) \log ^2\left (\frac {x}{5}\right )+\left (8 x^3-24 x^2+18 x\right ) \log \left (\frac {x}{5}\right )-42 x+18}{x \left (4 x^2-12 x+9\right ) \log ^2\left (\frac {x}{5}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int 2 \left (-\frac {6 \left (2 x^2-6 x+5\right )}{(3-2 x)^2}+\frac {\frac {1}{x}-1}{\log ^2\left (\frac {x}{5}\right )}+\frac {1}{\log \left (\frac {x}{5}\right )}\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \left (-\frac {1-\frac {1}{x}}{\log ^2\left (\frac {x}{5}\right )}-\frac {6 \left (2 x^2-6 x+5\right )}{(3-2 x)^2}+\frac {1}{\log \left (\frac {x}{5}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (-3 x-\frac {3}{2 (3-2 x)}+\frac {x}{\log \left (\frac {x}{5}\right )}-\frac {1}{\log \left (\frac {x}{5}\right )}\right )\) |
Input:
Int[(18 - 42*x + 32*x^2 - 8*x^3 + (18*x - 24*x^2 + 8*x^3)*Log[x/5] + (-60* x + 72*x^2 - 24*x^3)*Log[x/5]^2)/((9*x - 12*x^2 + 4*x^3)*Log[x/5]^2),x]
Output:
2*(-3/(2*(3 - 2*x)) - 3*x - Log[x/5]^(-1) + x/Log[x/5])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(-6 x +\frac {3}{-3+2 x}+\frac {2 x}{\ln \left (\frac {x}{5}\right )}-\frac {2}{\ln \left (\frac {x}{5}\right )}\) | \(31\) |
default | \(-6 x +\frac {3}{-3+2 x}+\frac {2 x}{\ln \left (\frac {x}{5}\right )}-\frac {2}{\ln \left (\frac {x}{5}\right )}\) | \(31\) |
parts | \(-6 x +\frac {3}{-3+2 x}+\frac {2 x}{\ln \left (\frac {x}{5}\right )}-\frac {2}{\ln \left (\frac {x}{5}\right )}\) | \(31\) |
risch | \(-\frac {3 \left (4 x^{2}-6 x -1\right )}{-3+2 x}+\frac {-2+2 x}{\ln \left (\frac {x}{5}\right )}\) | \(32\) |
norman | \(\frac {6+30 \ln \left (\frac {x}{5}\right )-10 x +4 x^{2}-12 \ln \left (\frac {x}{5}\right ) x^{2}}{\left (-3+2 x \right ) \ln \left (\frac {x}{5}\right )}\) | \(40\) |
parallelrisch | \(\frac {12-24 \ln \left (\frac {x}{5}\right ) x^{2}+8 x^{2}-20 x +60 \ln \left (\frac {x}{5}\right )}{2 \ln \left (\frac {x}{5}\right ) \left (-3+2 x \right )}\) | \(41\) |
Input:
int(((-24*x^3+72*x^2-60*x)*ln(1/5*x)^2+(8*x^3-24*x^2+18*x)*ln(1/5*x)-8*x^3 +32*x^2-42*x+18)/(4*x^3-12*x^2+9*x)/ln(1/5*x)^2,x,method=_RETURNVERBOSE)
Output:
-6*x+3/(-3+2*x)+2/ln(1/5*x)*x-2/ln(1/5*x)
Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{\left (9 x-12 x^2+4 x^3\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {4 \, x^{2} - 3 \, {\left (4 \, x^{2} - 6 \, x - 1\right )} \log \left (\frac {1}{5} \, x\right ) - 10 \, x + 6}{{\left (2 \, x - 3\right )} \log \left (\frac {1}{5} \, x\right )} \] Input:
integrate(((-24*x^3+72*x^2-60*x)*log(1/5*x)^2+(8*x^3-24*x^2+18*x)*log(1/5* x)-8*x^3+32*x^2-42*x+18)/(4*x^3-12*x^2+9*x)/log(1/5*x)^2,x, algorithm="fri cas")
Output:
(4*x^2 - 3*(4*x^2 - 6*x - 1)*log(1/5*x) - 10*x + 6)/((2*x - 3)*log(1/5*x))
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{\left (9 x-12 x^2+4 x^3\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=- 6 x + \frac {2 x - 2}{\log {\left (\frac {x}{5} \right )}} + \frac {3}{2 x - 3} \] Input:
integrate(((-24*x**3+72*x**2-60*x)*ln(1/5*x)**2+(8*x**3-24*x**2+18*x)*ln(1 /5*x)-8*x**3+32*x**2-42*x+18)/(4*x**3-12*x**2+9*x)/ln(1/5*x)**2,x)
Output:
-6*x + (2*x - 2)/log(x/5) + 3/(2*x - 3)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (26) = 52\).
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.85 \[ \int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{\left (9 x-12 x^2+4 x^3\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {4 \, x^{2} {\left (3 \, \log \left (5\right ) + 1\right )} - 2 \, x {\left (9 \, \log \left (5\right ) + 5\right )} - 3 \, {\left (4 \, x^{2} - 6 \, x - 1\right )} \log \left (x\right ) - 3 \, \log \left (5\right ) + 6}{2 \, x \log \left (5\right ) - {\left (2 \, x - 3\right )} \log \left (x\right ) - 3 \, \log \left (5\right )} \] Input:
integrate(((-24*x^3+72*x^2-60*x)*log(1/5*x)^2+(8*x^3-24*x^2+18*x)*log(1/5* x)-8*x^3+32*x^2-42*x+18)/(4*x^3-12*x^2+9*x)/log(1/5*x)^2,x, algorithm="max ima")
Output:
-(4*x^2*(3*log(5) + 1) - 2*x*(9*log(5) + 5) - 3*(4*x^2 - 6*x - 1)*log(x) - 3*log(5) + 6)/(2*x*log(5) - (2*x - 3)*log(x) - 3*log(5))
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{\left (9 x-12 x^2+4 x^3\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=-6 \, x + \frac {2 \, {\left (x - 1\right )}}{\log \left (\frac {1}{5} \, x\right )} + \frac {3}{2 \, x - 3} \] Input:
integrate(((-24*x^3+72*x^2-60*x)*log(1/5*x)^2+(8*x^3-24*x^2+18*x)*log(1/5* x)-8*x^3+32*x^2-42*x+18)/(4*x^3-12*x^2+9*x)/log(1/5*x)^2,x, algorithm="gia c")
Output:
-6*x + 2*(x - 1)/log(1/5*x) + 3/(2*x - 3)
Time = 3.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{\left (9 x-12 x^2+4 x^3\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {20\,x-12\,x^2}{2\,x-3}+\frac {4\,x^2-10\,x+6}{\ln \left (\frac {x}{5}\right )\,\left (2\,x-3\right )} \] Input:
int(-(42*x - log(x/5)*(18*x - 24*x^2 + 8*x^3) + log(x/5)^2*(60*x - 72*x^2 + 24*x^3) - 32*x^2 + 8*x^3 - 18)/(log(x/5)^2*(9*x - 12*x^2 + 4*x^3)),x)
Output:
(20*x - 12*x^2)/(2*x - 3) + (4*x^2 - 10*x + 6)/(log(x/5)*(2*x - 3))
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {18-42 x+32 x^2-8 x^3+\left (18 x-24 x^2+8 x^3\right ) \log \left (\frac {x}{5}\right )+\left (-60 x+72 x^2-24 x^3\right ) \log ^2\left (\frac {x}{5}\right )}{\left (9 x-12 x^2+4 x^3\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {-12 \,\mathrm {log}\left (\frac {x}{5}\right ) x^{2}+20 \,\mathrm {log}\left (\frac {x}{5}\right ) x +4 x^{2}-10 x +6}{\mathrm {log}\left (\frac {x}{5}\right ) \left (2 x -3\right )} \] Input:
int(((-24*x^3+72*x^2-60*x)*log(1/5*x)^2+(8*x^3-24*x^2+18*x)*log(1/5*x)-8*x ^3+32*x^2-42*x+18)/(4*x^3-12*x^2+9*x)/log(1/5*x)^2,x)
Output:
(2*( - 6*log(x/5)*x**2 + 10*log(x/5)*x + 2*x**2 - 5*x + 3))/(log(x/5)*(2*x - 3))