Integrand size = 103, antiderivative size = 23 \[ \int \frac {e^{\frac {1+e^{12} x}{x}} \left (x-x^2-4 x^5\right )+\left (e^{\frac {1+e^{12} x}{x}} \left (x-2 x^2+x^4-2 x^5\right )+e^{\frac {1+e^{12} x}{x}} (-1+2 x) \log (x)\right ) \log \left (x+x^4-\log (x)\right )}{-x-x^4+\log (x)} \, dx=e^{e^{12}+\frac {1}{x}} x^2 \log \left (x+x^4-\log (x)\right ) \]
Time = 0.43 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1+e^{12} x}{x}} \left (x-x^2-4 x^5\right )+\left (e^{\frac {1+e^{12} x}{x}} \left (x-2 x^2+x^4-2 x^5\right )+e^{\frac {1+e^{12} x}{x}} (-1+2 x) \log (x)\right ) \log \left (x+x^4-\log (x)\right )}{-x-x^4+\log (x)} \, dx=e^{e^{12}+\frac {1}{x}} x^2 \log \left (x+x^4-\log (x)\right ) \]
Integrate[(E^((1 + E^12*x)/x)*(x - x^2 - 4*x^5) + (E^((1 + E^12*x)/x)*(x - 2*x^2 + x^4 - 2*x^5) + E^((1 + E^12*x)/x)*(-1 + 2*x)*Log[x])*Log[x + x^4 - Log[x]])/(-x - x^4 + Log[x]),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.
\(\displaystyle \int \frac {e^{\frac {e^{12} x+1}{x}} \left (-4 x^5-x^2+x\right )+\left (e^{\frac {e^{12} x+1}{x}} \left (-2 x^5+x^4-2 x^2+x\right )+e^{\frac {e^{12} x+1}{x}} (2 x-1) \log (x)\right ) \log \left (x^4+x-\log (x)\right )}{-x^4-x+\log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {1}{x}+e^{12}} \left (x \left (4 x^4+x-1\right )+(2 x-1) \left (x^4+x-\log (x)\right ) \log \left (x^4+x-\log (x)\right )\right )}{x^4+x-\log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {1}{x}+e^{12}} x \left (4 x^4+x-1\right )}{x^4+x-\log (x)}+e^{\frac {1}{x}+e^{12}} (2 x-1) \log \left (x^4+x-\log (x)\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{e^{12}+\frac {1}{x}} x}{x^4+x-\log (x)}dx-\int e^{e^{12}+\frac {1}{x}} \log \left (x^4+x-\log (x)\right )dx+2 \int e^{e^{12}+\frac {1}{x}} x \log \left (x^4+x-\log (x)\right )dx+4 \int \frac {e^{e^{12}+\frac {1}{x}} x^5}{x^4+x-\log (x)}dx+\int \frac {e^{e^{12}+\frac {1}{x}} x^2}{x^4+x-\log (x)}dx\) |
Int[(E^((1 + E^12*x)/x)*(x - x^2 - 4*x^5) + (E^((1 + E^12*x)/x)*(x - 2*x^2 + x^4 - 2*x^5) + E^((1 + E^12*x)/x)*(-1 + 2*x)*Log[x])*Log[x + x^4 - Log[ x]])/(-x - x^4 + Log[x]),x]
3.16.71.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 11.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13
method | result | size |
risch | \({\mathrm e}^{\frac {x \,{\mathrm e}^{12}+1}{x}} \ln \left (-\ln \left (x \right )+x^{4}+x \right ) x^{2}\) | \(26\) |
parallelrisch | \({\mathrm e}^{\frac {x \,{\mathrm e}^{12}+1}{x}} \ln \left (-\ln \left (x \right )+x^{4}+x \right ) x^{2}\) | \(28\) |
int((((-1+2*x)*exp((x*exp(6)^2+1)/x)*ln(x)+(-2*x^5+x^4-2*x^2+x)*exp((x*exp (6)^2+1)/x))*ln(-ln(x)+x^4+x)+(-4*x^5-x^2+x)*exp((x*exp(6)^2+1)/x))/(ln(x) -x^4-x),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {1+e^{12} x}{x}} \left (x-x^2-4 x^5\right )+\left (e^{\frac {1+e^{12} x}{x}} \left (x-2 x^2+x^4-2 x^5\right )+e^{\frac {1+e^{12} x}{x}} (-1+2 x) \log (x)\right ) \log \left (x+x^4-\log (x)\right )}{-x-x^4+\log (x)} \, dx=x^{2} e^{\left (\frac {x e^{12} + 1}{x}\right )} \log \left (x^{4} + x - \log \left (x\right )\right ) \]
integrate((((-1+2*x)*exp((x*exp(6)^2+1)/x)*log(x)+(-2*x^5+x^4-2*x^2+x)*exp ((x*exp(6)^2+1)/x))*log(-log(x)+x^4+x)+(-4*x^5-x^2+x)*exp((x*exp(6)^2+1)/x ))/(log(x)-x^4-x),x, algorithm=\
Time = 1.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1+e^{12} x}{x}} \left (x-x^2-4 x^5\right )+\left (e^{\frac {1+e^{12} x}{x}} \left (x-2 x^2+x^4-2 x^5\right )+e^{\frac {1+e^{12} x}{x}} (-1+2 x) \log (x)\right ) \log \left (x+x^4-\log (x)\right )}{-x-x^4+\log (x)} \, dx=x^{2} e^{\frac {x e^{12} + 1}{x}} \log {\left (x^{4} + x - \log {\left (x \right )} \right )} \]
integrate((((-1+2*x)*exp((x*exp(6)**2+1)/x)*ln(x)+(-2*x**5+x**4-2*x**2+x)* exp((x*exp(6)**2+1)/x))*ln(-ln(x)+x**4+x)+(-4*x**5-x**2+x)*exp((x*exp(6)** 2+1)/x))/(ln(x)-x**4-x),x)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1+e^{12} x}{x}} \left (x-x^2-4 x^5\right )+\left (e^{\frac {1+e^{12} x}{x}} \left (x-2 x^2+x^4-2 x^5\right )+e^{\frac {1+e^{12} x}{x}} (-1+2 x) \log (x)\right ) \log \left (x+x^4-\log (x)\right )}{-x-x^4+\log (x)} \, dx=x^{2} e^{\left (\frac {1}{x} + e^{12}\right )} \log \left (x^{4} + x - \log \left (x\right )\right ) \]
integrate((((-1+2*x)*exp((x*exp(6)^2+1)/x)*log(x)+(-2*x^5+x^4-2*x^2+x)*exp ((x*exp(6)^2+1)/x))*log(-log(x)+x^4+x)+(-4*x^5-x^2+x)*exp((x*exp(6)^2+1)/x ))/(log(x)-x^4-x),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {1+e^{12} x}{x}} \left (x-x^2-4 x^5\right )+\left (e^{\frac {1+e^{12} x}{x}} \left (x-2 x^2+x^4-2 x^5\right )+e^{\frac {1+e^{12} x}{x}} (-1+2 x) \log (x)\right ) \log \left (x+x^4-\log (x)\right )}{-x-x^4+\log (x)} \, dx=x^{2} e^{\left (\frac {x e^{12} + 1}{x}\right )} \log \left (x^{4} + x - \log \left (x\right )\right ) \]
integrate((((-1+2*x)*exp((x*exp(6)^2+1)/x)*log(x)+(-2*x^5+x^4-2*x^2+x)*exp ((x*exp(6)^2+1)/x))*log(-log(x)+x^4+x)+(-4*x^5-x^2+x)*exp((x*exp(6)^2+1)/x ))/(log(x)-x^4-x),x, algorithm=\
Timed out. \[ \int \frac {e^{\frac {1+e^{12} x}{x}} \left (x-x^2-4 x^5\right )+\left (e^{\frac {1+e^{12} x}{x}} \left (x-2 x^2+x^4-2 x^5\right )+e^{\frac {1+e^{12} x}{x}} (-1+2 x) \log (x)\right ) \log \left (x+x^4-\log (x)\right )}{-x-x^4+\log (x)} \, dx=-\int \frac {\ln \left (x-\ln \left (x\right )+x^4\right )\,\left ({\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{12}+1}{x}}\,\left (-2\,x^5+x^4-2\,x^2+x\right )+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{12}+1}{x}}\,\ln \left (x\right )\,\left (2\,x-1\right )\right )-{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{12}+1}{x}}\,\left (4\,x^5+x^2-x\right )}{x-\ln \left (x\right )+x^4} \,d x \]
int(-(log(x - log(x) + x^4)*(exp((x*exp(12) + 1)/x)*(x - 2*x^2 + x^4 - 2*x ^5) + exp((x*exp(12) + 1)/x)*log(x)*(2*x - 1)) - exp((x*exp(12) + 1)/x)*(x ^2 - x + 4*x^5))/(x - log(x) + x^4),x)