Integrand size = 119, antiderivative size = 33 \[ \int \frac {-5+e^4-x+x^2+\left (-5+e^4-x^2\right ) \log (x)+(4-e) \log ^2(x)}{25+e^8+10 x-9 x^2-2 x^3+x^4+e^4 \left (-10-2 x+2 x^2\right )+\left (-40+(8-2 e) e^4-8 x+8 x^2+e \left (10+2 x-2 x^2\right )\right ) \log (x)+\left (16-8 e+e^2\right ) \log ^2(x)} \, dx=\frac {x^2}{2 x+x \left (2-e+\frac {-5+e^4-x+x^2}{\log (x)}\right )} \]
Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {-5+e^4-x+x^2+\left (-5+e^4-x^2\right ) \log (x)+(4-e) \log ^2(x)}{25+e^8+10 x-9 x^2-2 x^3+x^4+e^4 \left (-10-2 x+2 x^2\right )+\left (-40+(8-2 e) e^4-8 x+8 x^2+e \left (10+2 x-2 x^2\right )\right ) \log (x)+\left (16-8 e+e^2\right ) \log ^2(x)} \, dx=\frac {x \log (x)}{-5+e^4-x+x^2-(-4+e) \log (x)} \]
Integrate[(-5 + E^4 - x + x^2 + (-5 + E^4 - x^2)*Log[x] + (4 - E)*Log[x]^2 )/(25 + E^8 + 10*x - 9*x^2 - 2*x^3 + x^4 + E^4*(-10 - 2*x + 2*x^2) + (-40 + (8 - 2*E)*E^4 - 8*x + 8*x^2 + E*(10 + 2*x - 2*x^2))*Log[x] + (16 - 8*E + E^2)*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.
\(\displaystyle \int \frac {x^2+\left (-x^2+e^4-5\right ) \log (x)-x+(4-e) \log ^2(x)+e^4-5}{x^4-2 x^3-9 x^2+e^4 \left (2 x^2-2 x-10\right )+\left (8 x^2+e \left (-2 x^2+2 x+10\right )-8 x+(8-2 e) e^4-40\right ) \log (x)+10 x+\left (16-8 e+e^2\right ) \log ^2(x)+e^8+25} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x^2+\left (-x^2+e^4-5\right ) \log (x)-x-(e-4) \log ^2(x)-5 \left (1-\frac {e^4}{5}\right )}{\left (-x^2+x+(e-4) \log (x)+5 \left (1-\frac {e^4}{5}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 x^2-2 x+e^4-5}{(4-e) \left (-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )\right )}+\frac {2 x^4-3 x^3-\left (5+e-2 e^4\right ) x^2+\left (1+e-e^4\right ) x-(4-e) \left (5-e^4\right )}{(4-e) \left (-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )\right )^2}+\frac {1}{4-e}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\left (5-e^4\right ) \int \frac {1}{\left (-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )\right )^2}dx+\frac {\left (1+e-e^4\right ) \int \frac {x}{\left (-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )\right )^2}dx}{4-e}-\frac {\left (5+e-2 e^4\right ) \int \frac {x^2}{\left (-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )\right )^2}dx}{4-e}-\frac {\left (5-e^4\right ) \int \frac {1}{-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )}dx}{4-e}+\frac {3 \int \frac {x^2}{-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )}dx}{4-e}+\frac {2 \int \frac {x}{x^2-x+4 \left (1-\frac {e}{4}\right ) \log (x)-5 \left (1-\frac {e^4}{5}\right )}dx}{4-e}+\frac {2 \int \frac {x^4}{\left (-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )\right )^2}dx}{4-e}-\frac {3 \int \frac {x^3}{\left (-x^2+x-4 \left (1-\frac {e}{4}\right ) \log (x)+5 \left (1-\frac {e^4}{5}\right )\right )^2}dx}{4-e}+\frac {x}{4-e}\) |
Int[(-5 + E^4 - x + x^2 + (-5 + E^4 - x^2)*Log[x] + (4 - E)*Log[x]^2)/(25 + E^8 + 10*x - 9*x^2 - 2*x^3 + x^4 + E^4*(-10 - 2*x + 2*x^2) + (-40 + (8 - 2*E)*E^4 - 8*x + 8*x^2 + E*(10 + 2*x - 2*x^2))*Log[x] + (16 - 8*E + E^2)* Log[x]^2),x]
3.5.70.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {x \ln \left (x \right )}{{\mathrm e} \ln \left (x \right )-x^{2}-4 \ln \left (x \right )-{\mathrm e}^{4}+x +5}\) | \(29\) |
norman | \(-\frac {x \ln \left (x \right )}{{\mathrm e} \ln \left (x \right )-x^{2}-4 \ln \left (x \right )-{\mathrm e}^{4}+x +5}\) | \(29\) |
parallelrisch | \(-\frac {x \ln \left (x \right )}{{\mathrm e} \ln \left (x \right )-x^{2}-4 \ln \left (x \right )-{\mathrm e}^{4}+x +5}\) | \(29\) |
risch | \(-\frac {x}{{\mathrm e}-4}+\frac {x \left ({\mathrm e}^{4}+x^{2}-x -5\right )}{\left ({\mathrm e}-4\right ) \left ({\mathrm e}^{4}-{\mathrm e} \ln \left (x \right )+x^{2}-x +4 \ln \left (x \right )-5\right )}\) | \(51\) |
int(((4-exp(1))*ln(x)^2+(exp(4)-x^2-5)*ln(x)+exp(4)+x^2-x-5)/((exp(1)^2-8* exp(1)+16)*ln(x)^2+((-2*exp(1)+8)*exp(4)+(-2*x^2+2*x+10)*exp(1)+8*x^2-8*x- 40)*ln(x)+exp(4)^2+(2*x^2-2*x-10)*exp(4)+x^4-2*x^3-9*x^2+10*x+25),x,method =_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {-5+e^4-x+x^2+\left (-5+e^4-x^2\right ) \log (x)+(4-e) \log ^2(x)}{25+e^8+10 x-9 x^2-2 x^3+x^4+e^4 \left (-10-2 x+2 x^2\right )+\left (-40+(8-2 e) e^4-8 x+8 x^2+e \left (10+2 x-2 x^2\right )\right ) \log (x)+\left (16-8 e+e^2\right ) \log ^2(x)} \, dx=\frac {x \log \left (x\right )}{x^{2} - {\left (e - 4\right )} \log \left (x\right ) - x + e^{4} - 5} \]
integrate(((4-exp(1))*log(x)^2+(exp(4)-x^2-5)*log(x)+exp(4)+x^2-x-5)/((exp (1)^2-8*exp(1)+16)*log(x)^2+((-2*exp(1)+8)*exp(4)+(-2*x^2+2*x+10)*exp(1)+8 *x^2-8*x-40)*log(x)+exp(4)^2+(2*x^2-2*x-10)*exp(4)+x^4-2*x^3-9*x^2+10*x+25 ),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (26) = 52\).
Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.12 \[ \int \frac {-5+e^4-x+x^2+\left (-5+e^4-x^2\right ) \log (x)+(4-e) \log ^2(x)}{25+e^8+10 x-9 x^2-2 x^3+x^4+e^4 \left (-10-2 x+2 x^2\right )+\left (-40+(8-2 e) e^4-8 x+8 x^2+e \left (10+2 x-2 x^2\right )\right ) \log (x)+\left (16-8 e+e^2\right ) \log ^2(x)} \, dx=- \frac {x}{-4 + e} + \frac {- x^{3} + x^{2} - x e^{4} + 5 x}{- e x^{2} + 4 x^{2} - 4 x + e x + \left (- 8 e + e^{2} + 16\right ) \log {\left (x \right )} - e^{5} - 20 + 5 e + 4 e^{4}} \]
integrate(((4-exp(1))*ln(x)**2+(exp(4)-x**2-5)*ln(x)+exp(4)+x**2-x-5)/((ex p(1)**2-8*exp(1)+16)*ln(x)**2+((-2*exp(1)+8)*exp(4)+(-2*x**2+2*x+10)*exp(1 )+8*x**2-8*x-40)*ln(x)+exp(4)**2+(2*x**2-2*x-10)*exp(4)+x**4-2*x**3-9*x**2 +10*x+25),x)
-x/(-4 + E) + (-x**3 + x**2 - x*exp(4) + 5*x)/(-E*x**2 + 4*x**2 - 4*x + E* x + (-8*E + exp(2) + 16)*log(x) - exp(5) - 20 + 5*E + 4*exp(4))
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {-5+e^4-x+x^2+\left (-5+e^4-x^2\right ) \log (x)+(4-e) \log ^2(x)}{25+e^8+10 x-9 x^2-2 x^3+x^4+e^4 \left (-10-2 x+2 x^2\right )+\left (-40+(8-2 e) e^4-8 x+8 x^2+e \left (10+2 x-2 x^2\right )\right ) \log (x)+\left (16-8 e+e^2\right ) \log ^2(x)} \, dx=\frac {x \log \left (x\right )}{x^{2} - {\left (e - 4\right )} \log \left (x\right ) - x + e^{4} - 5} \]
integrate(((4-exp(1))*log(x)^2+(exp(4)-x^2-5)*log(x)+exp(4)+x^2-x-5)/((exp (1)^2-8*exp(1)+16)*log(x)^2+((-2*exp(1)+8)*exp(4)+(-2*x^2+2*x+10)*exp(1)+8 *x^2-8*x-40)*log(x)+exp(4)^2+(2*x^2-2*x-10)*exp(4)+x^4-2*x^3-9*x^2+10*x+25 ),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {-5+e^4-x+x^2+\left (-5+e^4-x^2\right ) \log (x)+(4-e) \log ^2(x)}{25+e^8+10 x-9 x^2-2 x^3+x^4+e^4 \left (-10-2 x+2 x^2\right )+\left (-40+(8-2 e) e^4-8 x+8 x^2+e \left (10+2 x-2 x^2\right )\right ) \log (x)+\left (16-8 e+e^2\right ) \log ^2(x)} \, dx=\frac {x \log \left (x\right )}{x^{2} - e \log \left (x\right ) - x + e^{4} + 4 \, \log \left (x\right ) - 5} \]
integrate(((4-exp(1))*log(x)^2+(exp(4)-x^2-5)*log(x)+exp(4)+x^2-x-5)/((exp (1)^2-8*exp(1)+16)*log(x)^2+((-2*exp(1)+8)*exp(4)+(-2*x^2+2*x+10)*exp(1)+8 *x^2-8*x-40)*log(x)+exp(4)^2+(2*x^2-2*x-10)*exp(4)+x^4-2*x^3-9*x^2+10*x+25 ),x, algorithm=\
Time = 13.44 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.94 \[ \int \frac {-5+e^4-x+x^2+\left (-5+e^4-x^2\right ) \log (x)+(4-e) \log ^2(x)}{25+e^8+10 x-9 x^2-2 x^3+x^4+e^4 \left (-10-2 x+2 x^2\right )+\left (-40+(8-2 e) e^4-8 x+8 x^2+e \left (10+2 x-2 x^2\right )\right ) \log (x)+\left (16-8 e+e^2\right ) \log ^2(x)} \, dx=\frac {x-{\mathrm {e}}^4-4\,\ln \left (x\right )+\mathrm {e}\,\ln \left (x\right )+16\,x\,\ln \left (x\right )-x^2-4\,x\,\mathrm {e}\,\ln \left (x\right )+5}{4\,\left (\mathrm {e}-4\right )\,\left (x-{\mathrm {e}}^4-4\,\ln \left (x\right )+\mathrm {e}\,\ln \left (x\right )-x^2+5\right )} \]
int(-(x - exp(4) + log(x)*(x^2 - exp(4) + 5) + log(x)^2*(exp(1) - 4) - x^2 + 5)/(10*x + exp(8) + log(x)^2*(exp(2) - 8*exp(1) + 16) - exp(4)*(2*x - 2 *x^2 + 10) - log(x)*(8*x - exp(1)*(2*x - 2*x^2 + 10) - 8*x^2 + exp(4)*(2*e xp(1) - 8) + 40) - 9*x^2 - 2*x^3 + x^4 + 25),x)