Integrand size = 83, antiderivative size = 21 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=x+x^2+\frac {x^2 \log (\log (x))}{e^2-x} \] Output:
x^2+x^2/(exp(2)-x)*ln(ln(x))+x
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x \left (-\left (\left (e^2-x\right ) (1+x)\right )-x \log (\log (x))\right )}{-e^2+x} \] Input:
Integrate[(E^2*x - x^2 + (x^2 + 2*x^3 + E^4*(1 + 2*x) + E^2*(-2*x - 4*x^2) )*Log[x] + (2*E^2*x - x^2)*Log[x]*Log[Log[x]])/((E^4 - 2*E^2*x + x^2)*Log[ x]),x]
Output:
(x*(-((E^2 - x)*(1 + x)) - x*Log[Log[x]]))/(-E^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.
\(\displaystyle \int \frac {-x^2+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))+\left (2 x^3+x^2+e^2 \left (-4 x^2-2 x\right )+e^4 (2 x+1)\right ) \log (x)+e^2 x}{\left (x^2-2 e^2 x+e^4\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {-x^2+e^2 x+\left (2 x^3+x^2+e^4 (2 x+1)-2 e^2 \left (2 x^2+x\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{4 \left (e^2-x\right )^2 \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {-x^2+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))+\left (2 x^3+x^2-2 e^2 \left (2 x^2+x\right )+e^4 (2 x+1)\right ) \log (x)+e^2 x}{\left (e^2-x\right )^2 \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-2 x^2 \log (x)+x-\left (1-2 e^2\right ) x \log (x)+e^2 \log (x)}{\left (e^2-x\right ) \log (x)}+\frac {\left (2 e^2-x\right ) x \log (\log (x))}{\left (e^2-x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {x}{\left (e^2-x\right ) \log (x)}dx+e^4 \int \frac {\log (\log (x))}{\left (e^2-x\right )^2}dx+\operatorname {LogIntegral}(x)+x^2+x-x \log (\log (x))\) |
Input:
Int[(E^2*x - x^2 + (x^2 + 2*x^3 + E^4*(1 + 2*x) + E^2*(-2*x - 4*x^2))*Log[ x] + (2*E^2*x - x^2)*Log[x]*Log[Log[x]])/((E^4 - 2*E^2*x + x^2)*Log[x]),x]
Output:
$Aborted
Time = 2.82 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{2}+x^{2} \ln \left (\ln \left (x \right )\right )-x^{3}+{\mathrm e}^{2} x -x^{2}}{{\mathrm e}^{2}-x}\) | \(38\) |
parallelrisch | \(\frac {x^{2} {\mathrm e}^{2}-x^{3}+x^{2} \ln \left (\ln \left (x \right )\right )+{\mathrm e}^{4}-x^{2}}{{\mathrm e}^{2}-x}\) | \(38\) |
Input:
int(((2*exp(2)*x-x^2)*ln(x)*ln(ln(x))+((1+2*x)*exp(2)^2+(-4*x^2-2*x)*exp(2 )+2*x^3+x^2)*ln(x)+exp(2)*x-x^2)/(exp(2)^2-2*exp(2)*x+x^2)/ln(x),x,method= _RETURNVERBOSE)
Output:
(x^2*exp(2)+x^2*ln(ln(x))-x^3+exp(2)*x-x^2)/(exp(2)-x)
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x^{3} - x^{2} \log \left (\log \left (x\right )\right ) + x^{2} - {\left (x^{2} + x\right )} e^{2}}{x - e^{2}} \] Input:
integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((1+2*x)*exp(2)^2+(-4*x^2-2 *x)*exp(2)+2*x^3+x^2)*log(x)+exp(2)*x-x^2)/(exp(2)^2-2*exp(2)*x+x^2)/log(x ),x, algorithm="fricas")
Output:
(x^3 - x^2*log(log(x)) + x^2 - (x^2 + x)*e^2)/(x - e^2)
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=x^{2} + x - e^{2} \log {\left (\log {\left (x \right )} \right )} + \frac {\left (- x^{2} + x e^{2} - e^{4}\right ) \log {\left (\log {\left (x \right )} \right )}}{x - e^{2}} \] Input:
integrate(((2*exp(2)*x-x**2)*ln(x)*ln(ln(x))+((1+2*x)*exp(2)**2+(-4*x**2-2 *x)*exp(2)+2*x**3+x**2)*ln(x)+exp(2)*x-x**2)/(exp(2)**2-2*exp(2)*x+x**2)/l n(x),x)
Output:
x**2 + x - exp(2)*log(log(x)) + (-x**2 + x*exp(2) - exp(4))*log(log(x))/(x - exp(2))
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x^{3} - x^{2} {\left (e^{2} - 1\right )} - x^{2} \log \left (\log \left (x\right )\right ) - x e^{2}}{x - e^{2}} \] Input:
integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((1+2*x)*exp(2)^2+(-4*x^2-2 *x)*exp(2)+2*x^3+x^2)*log(x)+exp(2)*x-x^2)/(exp(2)^2-2*exp(2)*x+x^2)/log(x ),x, algorithm="maxima")
Output:
(x^3 - x^2*(e^2 - 1) - x^2*log(log(x)) - x*e^2)/(x - e^2)
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x^{3} - x^{2} e^{2} - x^{2} \log \left (\log \left (x\right )\right ) + x^{2} - x e^{2}}{x - e^{2}} \] Input:
integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((1+2*x)*exp(2)^2+(-4*x^2-2 *x)*exp(2)+2*x^3+x^2)*log(x)+exp(2)*x-x^2)/(exp(2)^2-2*exp(2)*x+x^2)/log(x ),x, algorithm="giac")
Output:
(x^3 - x^2*e^2 - x^2*log(log(x)) + x^2 - x*e^2)/(x - e^2)
Time = 3.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=x+x^2-\frac {x^2\,\ln \left (\ln \left (x\right )\right )}{x-{\mathrm {e}}^2} \] Input:
int((x*exp(2) - x^2 + log(x)*(x^2 - exp(2)*(2*x + 4*x^2) + 2*x^3 + exp(4)* (2*x + 1)) + log(log(x))*log(x)*(2*x*exp(2) - x^2))/(log(x)*(exp(4) - 2*x* exp(2) + x^2)),x)
Output:
x + x^2 - (x^2*log(log(x)))/(x - exp(2))
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x \left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x +e^{2} x +e^{2}-x^{2}-x \right )}{e^{2}-x} \] Input:
int(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((1+2*x)*exp(2)^2+(-4*x^2-2*x)*ex p(2)+2*x^3+x^2)*log(x)+exp(2)*x-x^2)/(exp(2)^2-2*exp(2)*x+x^2)/log(x),x)
Output:
(x*(log(log(x))*x + e**2*x + e**2 - x**2 - x))/(e**2 - x)