Integrand size = 112, antiderivative size = 23 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {-4+x}{-4-x+x^2-e^6 \log (3)+\log (x)} \]
Time = 1.82 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {-4+x}{-4-x+x^2-e^6 \log (3)+\log (x)} \]
Integrate[(4 - 9*x + 8*x^2 - x^3 - E^6*x*Log[3] + x*Log[x])/(16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5 + E^6*(8*x + 2*x^2 - 2*x^3)*Log[3] + E^12*x*Log[3]^2 + (-8*x - 2*x^2 + 2*x^3 - 2*E^6*x*Log[3])*Log[x] + x*Log[x]^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.
\(\displaystyle \int \frac {-x^3+8 x^2-9 x+x \log (x)-e^6 x \log (3)+4}{x^5-2 x^4-7 x^3+8 x^2+\left (2 x^3-2 x^2-8 x-2 e^6 x \log (3)\right ) \log (x)+e^6 \left (-2 x^3+2 x^2+8 x\right ) \log (3)+16 x+x \log ^2(x)+e^{12} x \log ^2(3)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x^3+8 x^2+x \log (x)+x \left (-9-e^6 \log (3)\right )+4}{x^5-2 x^4-7 x^3+8 x^2+\left (2 x^3-2 x^2-8 x-2 e^6 x \log (3)\right ) \log (x)+e^6 \left (-2 x^3+2 x^2+8 x\right ) \log (3)+16 x+x \log ^2(x)+e^{12} x \log ^2(3)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x^3+8 x^2+x \log (x)+x \left (-9-e^6 \log (3)\right )+4}{x^5-2 x^4-7 x^3+8 x^2+\left (2 x^3-2 x^2-8 x-2 e^6 x \log (3)\right ) \log (x)+e^6 \left (-2 x^3+2 x^2+8 x\right ) \log (3)+x \log ^2(x)+x \left (16+e^{12} \log ^2(3)\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-x^3+8 x^2+x \log (x)-x \left (9+e^6 \log (3)\right )+4}{x \left (-x^2+x-\log (x)+4 \left (1+\frac {1}{4} e^6 \log (3)\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{x^2-x+\log (x)-4 \left (1+\frac {1}{4} e^6 \log (3)\right )}+\frac {-2 x^3+9 x^2-5 x+4}{x \left (-x^2+x-\log (x)+4 \left (1+\frac {1}{4} e^6 \log (3)\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int \frac {1}{\left (-x^2+x-\log (x)+4 \left (1+\frac {1}{4} e^6 \log (3)\right )\right )^2}dx+4 \int \frac {1}{x \left (-x^2+x-\log (x)+4 \left (1+\frac {1}{4} e^6 \log (3)\right )\right )^2}dx+9 \int \frac {x}{\left (-x^2+x-\log (x)+4 \left (1+\frac {1}{4} e^6 \log (3)\right )\right )^2}dx-2 \int \frac {x^2}{\left (-x^2+x-\log (x)+4 \left (1+\frac {1}{4} e^6 \log (3)\right )\right )^2}dx+\int \frac {1}{x^2-x+\log (x)-4 \left (1+\frac {1}{4} e^6 \log (3)\right )}dx\) |
Int[(4 - 9*x + 8*x^2 - x^3 - E^6*x*Log[3] + x*Log[x])/(16*x + 8*x^2 - 7*x^ 3 - 2*x^4 + x^5 + E^6*(8*x + 2*x^2 - 2*x^3)*Log[3] + E^12*x*Log[3]^2 + (-8 *x - 2*x^2 + 2*x^3 - 2*E^6*x*Log[3])*Log[x] + x*Log[x]^2),x]
3.8.95.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 5.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {x -4}{{\mathrm e}^{6} \ln \left (3\right )-x^{2}+x -\ln \left (x \right )+4}\) | \(25\) |
parallelrisch | \(-\frac {x -4}{{\mathrm e}^{6} \ln \left (3\right )-x^{2}+x -\ln \left (x \right )+4}\) | \(27\) |
default | \(\frac {-x +4}{{\mathrm e}^{6} \ln \left (3\right )-x^{2}+x -\ln \left (x \right )+4}\) | \(28\) |
norman | \(\frac {-x +4}{{\mathrm e}^{6} \ln \left (3\right )-x^{2}+x -\ln \left (x \right )+4}\) | \(28\) |
int((x*ln(x)-x*exp(3)^2*ln(3)-x^3+8*x^2-9*x+4)/(x*ln(x)^2+(-2*x*exp(3)^2*l n(3)+2*x^3-2*x^2-8*x)*ln(x)+x*exp(3)^4*ln(3)^2+(-2*x^3+2*x^2+8*x)*exp(3)^2 *ln(3)+x^5-2*x^4-7*x^3+8*x^2+16*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x - 4}{x^{2} - e^{6} \log \left (3\right ) - x + \log \left (x\right ) - 4} \]
integrate((x*log(x)-x*exp(3)^2*log(3)-x^3+8*x^2-9*x+4)/(x*log(x)^2+(-2*x*e xp(3)^2*log(3)+2*x^3-2*x^2-8*x)*log(x)+x*exp(3)^4*log(3)^2+(-2*x^3+2*x^2+8 *x)*exp(3)^2*log(3)+x^5-2*x^4-7*x^3+8*x^2+16*x),x, algorithm=\
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x - 4}{x^{2} - x + \log {\left (x \right )} - e^{6} \log {\left (3 \right )} - 4} \]
integrate((x*ln(x)-x*exp(3)**2*ln(3)-x**3+8*x**2-9*x+4)/(x*ln(x)**2+(-2*x* exp(3)**2*ln(3)+2*x**3-2*x**2-8*x)*ln(x)+x*exp(3)**4*ln(3)**2+(-2*x**3+2*x **2+8*x)*exp(3)**2*ln(3)+x**5-2*x**4-7*x**3+8*x**2+16*x),x)
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x - 4}{x^{2} - e^{6} \log \left (3\right ) - x + \log \left (x\right ) - 4} \]
integrate((x*log(x)-x*exp(3)^2*log(3)-x^3+8*x^2-9*x+4)/(x*log(x)^2+(-2*x*e xp(3)^2*log(3)+2*x^3-2*x^2-8*x)*log(x)+x*exp(3)^4*log(3)^2+(-2*x^3+2*x^2+8 *x)*exp(3)^2*log(3)+x^5-2*x^4-7*x^3+8*x^2+16*x),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x - 4}{x^{2} - e^{6} \log \left (3\right ) - x + \log \left (x\right ) - 4} \]
integrate((x*log(x)-x*exp(3)^2*log(3)-x^3+8*x^2-9*x+4)/(x*log(x)^2+(-2*x*e xp(3)^2*log(3)+2*x^3-2*x^2-8*x)*log(x)+x*exp(3)^4*log(3)^2+(-2*x^3+2*x^2+8 *x)*exp(3)^2*log(3)+x^5-2*x^4-7*x^3+8*x^2+16*x),x, algorithm=\
Time = 9.68 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=-\frac {x-4}{x-\ln \left (x\right )+{\mathrm {e}}^6\,\ln \left (3\right )-x^2+4} \]