Integrand size = 158, antiderivative size = 31 \[ \int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )+\left (4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )\right ) \log \left (\frac {x}{5}\right )+\left (x^2-2 x^3+x^4+e^2 \left (x^2-2 x^3+x^4\right )\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {2-\frac {x}{2+\log \left (\frac {x}{5}\right )}}{\left (1+e^2\right ) (-1+x) x} \]
Time = 0.47 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )+\left (4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )\right ) \log \left (\frac {x}{5}\right )+\left (x^2-2 x^3+x^4+e^2 \left (x^2-2 x^3+x^4\right )\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {\frac {2}{-1+x}-\frac {2}{x}+\frac {1}{(1-x) \left (2+\log \left (\frac {x}{5}\right )\right )}}{1+e^2} \]
Integrate[(8 - 17*x + 3*x^2 + (8 - 16*x + x^2)*Log[x/5] + (2 - 4*x)*Log[x/ 5]^2)/(4*x^2 - 8*x^3 + 4*x^4 + E^2*(4*x^2 - 8*x^3 + 4*x^4) + (4*x^2 - 8*x^ 3 + 4*x^4 + E^2*(4*x^2 - 8*x^3 + 4*x^4))*Log[x/5] + (x^2 - 2*x^3 + x^4 + E ^2*(x^2 - 2*x^3 + x^4))*Log[x/5]^2),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.
\(\displaystyle \int \frac {3 x^2+\left (x^2-16 x+8\right ) \log \left (\frac {x}{5}\right )-17 x+(2-4 x) \log ^2\left (\frac {x}{5}\right )+8}{4 x^4-8 x^3+4 x^2+e^2 \left (4 x^4-8 x^3+4 x^2\right )+\left (x^4-2 x^3+x^2+e^2 \left (x^4-2 x^3+x^2\right )\right ) \log ^2\left (\frac {x}{5}\right )+\left (4 x^4-8 x^3+4 x^2+e^2 \left (4 x^4-8 x^3+4 x^2\right )\right ) \log \left (\frac {x}{5}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {3 x^2+\left (x^2-16 x+8\right ) \log \left (\frac {x}{5}\right )-17 x+(2-4 x) \log ^2\left (\frac {x}{5}\right )+8}{\left (1+e^2\right ) (1-x)^2 x^2 \left (\log \left (\frac {x}{5}\right )+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 x^2-17 x+2 (1-2 x) \log ^2\left (\frac {x}{5}\right )+\left (x^2-16 x+8\right ) \log \left (\frac {x}{5}\right )+8}{(1-x)^2 x^2 \left (\log \left (\frac {x}{5}\right )+2\right )^2}dx}{1+e^2}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (-\frac {2 (2 x-1)}{(x-1)^2 x^2}+\frac {1}{(x-1)^2 \left (\log \left (\frac {x}{5}\right )+2\right )}+\frac {1}{(x-1) x \left (\log \left (\frac {x}{5}\right )+2\right )^2}\right )dx}{1+e^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \frac {1}{(x-1) x \left (\log \left (\frac {x}{5}\right )+2\right )^2}dx+\int \frac {1}{(x-1)^2 \left (\log \left (\frac {x}{5}\right )+2\right )}dx-\frac {2}{(1-x) x}}{1+e^2}\) |
Int[(8 - 17*x + 3*x^2 + (8 - 16*x + x^2)*Log[x/5] + (2 - 4*x)*Log[x/5]^2)/ (4*x^2 - 8*x^3 + 4*x^4 + E^2*(4*x^2 - 8*x^3 + 4*x^4) + (4*x^2 - 8*x^3 + 4* x^4 + E^2*(4*x^2 - 8*x^3 + 4*x^4))*Log[x/5] + (x^2 - 2*x^3 + x^4 + E^2*(x^ 2 - 2*x^3 + x^4))*Log[x/5]^2),x]
3.21.56.3.1 Defintions of rubi rules used
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 = 0.70 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13
method | result | size |
parallelrisch | \(\frac {2 \ln \left (\frac {x}{5}\right )+4-x}{x \left ({\mathrm e}^{2}+1\right ) \left (\ln \left (\frac {x}{5}\right )+2\right ) \left (-1+x \right )}\) | \(35\) |
risch | \(\frac {2}{x \left ({\mathrm e}^{2} x -{\mathrm e}^{2}+x -1\right )}-\frac {1}{\left ({\mathrm e}^{2} x -{\mathrm e}^{2}+x -1\right ) \left (\ln \left (\frac {x}{5}\right )+2\right )}\) | \(43\) |
derivativedivides | \(-\frac {\frac {x}{{\mathrm e}^{2}+1}-\frac {4}{{\mathrm e}^{2}+1}-\frac {2 \ln \left (\frac {x}{5}\right )}{{\mathrm e}^{2}+1}}{x \left (-1+x \right ) \left (\ln \left (\frac {x}{5}\right )+2\right )}\) | \(48\) |
default | \(-\frac {\frac {x}{{\mathrm e}^{2}+1}-\frac {4}{{\mathrm e}^{2}+1}-\frac {2 \ln \left (\frac {x}{5}\right )}{{\mathrm e}^{2}+1}}{x \left (-1+x \right ) \left (\ln \left (\frac {x}{5}\right )+2\right )}\) | \(48\) |
norman | \(\frac {-\frac {x}{{\mathrm e}^{2}+1}+\frac {4}{{\mathrm e}^{2}+1}+\frac {2 \ln \left (\frac {x}{5}\right )}{{\mathrm e}^{2}+1}}{x \left (\ln \left (\frac {x}{5}\right )+2\right ) \left (-1+x \right )}\) | \(48\) |
int(((-4*x+2)*ln(1/5*x)^2+(x^2-16*x+8)*ln(1/5*x)+3*x^2-17*x+8)/(((x^4-2*x^ 3+x^2)*exp(2)+x^4-2*x^3+x^2)*ln(1/5*x)^2+((4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4 -8*x^3+4*x^2)*ln(1/5*x)+(4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4*x^2),x,me thod=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77 \[ \int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )+\left (4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )\right ) \log \left (\frac {x}{5}\right )+\left (x^2-2 x^3+x^4+e^2 \left (x^2-2 x^3+x^4\right )\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {x - 2 \, \log \left (\frac {1}{5} \, x\right ) - 4}{2 \, x^{2} + 2 \, {\left (x^{2} - x\right )} e^{2} + {\left (x^{2} + {\left (x^{2} - x\right )} e^{2} - x\right )} \log \left (\frac {1}{5} \, x\right ) - 2 \, x} \]
integrate(((-4*x+2)*log(1/5*x)^2+(x^2-16*x+8)*log(1/5*x)+3*x^2-17*x+8)/((( x^4-2*x^3+x^2)*exp(2)+x^4-2*x^3+x^2)*log(1/5*x)^2+((4*x^4-8*x^3+4*x^2)*exp (2)+4*x^4-8*x^3+4*x^2)*log(1/5*x)+(4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4 *x^2),x, algorithm=\
-(x - 2*log(1/5*x) - 4)/(2*x^2 + 2*(x^2 - x)*e^2 + (x^2 + (x^2 - x)*e^2 - x)*log(1/5*x) - 2*x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )+\left (4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )\right ) \log \left (\frac {x}{5}\right )+\left (x^2-2 x^3+x^4+e^2 \left (x^2-2 x^3+x^4\right )\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=- \frac {1}{2 x + 2 x e^{2} + \left (x + x e^{2} - e^{2} - 1\right ) \log {\left (\frac {x}{5} \right )} - 2 e^{2} - 2} + \frac {2}{x^{2} \cdot \left (1 + e^{2}\right ) + x \left (- e^{2} - 1\right )} \]
integrate(((-4*x+2)*ln(1/5*x)**2+(x**2-16*x+8)*ln(1/5*x)+3*x**2-17*x+8)/(( (x**4-2*x**3+x**2)*exp(2)+x**4-2*x**3+x**2)*ln(1/5*x)**2+((4*x**4-8*x**3+4 *x**2)*exp(2)+4*x**4-8*x**3+4*x**2)*ln(1/5*x)+(4*x**4-8*x**3+4*x**2)*exp(2 )+4*x**4-8*x**3+4*x**2),x)
-1/(2*x + 2*x*exp(2) + (x + x*exp(2) - exp(2) - 1)*log(x/5) - 2*exp(2) - 2 ) + 2/(x**2*(1 + exp(2)) + x*(-exp(2) - 1))
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (28) = 56\).
Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.06 \[ \int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )+\left (4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )\right ) \log \left (\frac {x}{5}\right )+\left (x^2-2 x^3+x^4+e^2 \left (x^2-2 x^3+x^4\right )\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {x + 2 \, \log \left (5\right ) - 2 \, \log \left (x\right ) - 4}{{\left ({\left (\log \left (5\right ) - 2\right )} e^{2} + \log \left (5\right ) - 2\right )} x^{2} - {\left ({\left (\log \left (5\right ) - 2\right )} e^{2} + \log \left (5\right ) - 2\right )} x - {\left (x^{2} {\left (e^{2} + 1\right )} - x {\left (e^{2} + 1\right )}\right )} \log \left (x\right )} \]
integrate(((-4*x+2)*log(1/5*x)^2+(x^2-16*x+8)*log(1/5*x)+3*x^2-17*x+8)/((( x^4-2*x^3+x^2)*exp(2)+x^4-2*x^3+x^2)*log(1/5*x)^2+((4*x^4-8*x^3+4*x^2)*exp (2)+4*x^4-8*x^3+4*x^2)*log(1/5*x)+(4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4 *x^2),x, algorithm=\
(x + 2*log(5) - 2*log(x) - 4)/(((log(5) - 2)*e^2 + log(5) - 2)*x^2 - ((log (5) - 2)*e^2 + log(5) - 2)*x - (x^2*(e^2 + 1) - x*(e^2 + 1))*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (28) = 56\).
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.19 \[ \int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )+\left (4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )\right ) \log \left (\frac {x}{5}\right )+\left (x^2-2 x^3+x^4+e^2 \left (x^2-2 x^3+x^4\right )\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {x - 2 \, \log \left (\frac {1}{5} \, x\right ) - 4}{x^{2} e^{2} \log \left (\frac {1}{5} \, x\right ) + 2 \, x^{2} e^{2} + x^{2} \log \left (\frac {1}{5} \, x\right ) - x e^{2} \log \left (\frac {1}{5} \, x\right ) + 2 \, x^{2} - 2 \, x e^{2} - x \log \left (\frac {1}{5} \, x\right ) - 2 \, x} \]
integrate(((-4*x+2)*log(1/5*x)^2+(x^2-16*x+8)*log(1/5*x)+3*x^2-17*x+8)/((( x^4-2*x^3+x^2)*exp(2)+x^4-2*x^3+x^2)*log(1/5*x)^2+((4*x^4-8*x^3+4*x^2)*exp (2)+4*x^4-8*x^3+4*x^2)*log(1/5*x)+(4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4 *x^2),x, algorithm=\
-(x - 2*log(1/5*x) - 4)/(x^2*e^2*log(1/5*x) + 2*x^2*e^2 + x^2*log(1/5*x) - x*e^2*log(1/5*x) + 2*x^2 - 2*x*e^2 - x*log(1/5*x) - 2*x)
Time = 12.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )+\left (4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )\right ) \log \left (\frac {x}{5}\right )+\left (x^2-2 x^3+x^4+e^2 \left (x^2-2 x^3+x^4\right )\right ) \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {2}{x\,\left ({\mathrm {e}}^2+1\right )\,\left (x-1\right )}-\frac {1}{\left (\ln \left (\frac {x}{5}\right )+2\right )\,\left ({\mathrm {e}}^2+1\right )\,\left (x-1\right )} \]