Integrand size = 119, antiderivative size = 26 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=-4+\frac {\log (x)}{x \left (\frac {5}{e^2}+x \left (x+\log \left (e^5 x\right )\right )\right )} \] Output:
ln(x)/x/((ln(x*exp(5))+x)*x+5/exp(2))-4
Time = 0.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {e^2 \log (x)}{x \left (5+e^2 x (5+x)+e^2 x \log (x)\right )} \] Input:
Integrate[(5*E^2 + E^4*x^2 + (-5*E^2 + E^4*(-x - 3*x^2))*Log[x] + (E^4*x - 2*E^4*x*Log[x])*Log[E^5*x])/(25*x^2 + 10*E^2*x^4 + E^4*x^6 + (10*E^2*x^3 + 2*E^4*x^5)*Log[E^5*x] + E^4*x^4*Log[E^5*x]^2),x]
Output:
(E^2*Log[x])/(x*(5 + E^2*x*(5 + x) + E^2*x*Log[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.
\(\displaystyle \int \frac {e^4 x^2+\left (e^4 \left (-3 x^2-x\right )-5 e^2\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )+5 e^2}{e^4 x^6+10 e^2 x^4+e^4 x^4 \log ^2\left (e^5 x\right )+25 x^2+\left (2 e^4 x^5+10 e^2 x^3\right ) \log \left (e^5 x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^2 \left (e^2 x (x+5)-2 e^2 x \log ^2(x)-\left (e^2 x (3 x+10)+5\right ) \log (x)+5\right )}{x^2 \left (e^2 x (x+5)+e^2 x \log (x)+5\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^2 \int \frac {-2 e^2 x \log ^2(x)-\left (e^2 x (3 x+10)+5\right ) \log (x)+e^2 x (x+5)+5}{x^2 \left (e^2 x (x+5)+e^2 x \log (x)+5\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle e^2 \int \left (\frac {e^2 x^2+10 e^2 x+15}{e^2 x^3 \left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )}+\frac {e^4 x^4+6 e^4 x^3+5 e^4 x^2-20 e^2 x-25}{e^2 x^3 \left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )^2}-\frac {2}{e^2 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^2 \left (6 e^2 \int \frac {1}{\left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )^2}dx-20 \int \frac {1}{x^2 \left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )^2}dx+5 e^2 \int \frac {1}{x \left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )^2}dx+e^2 \int \frac {x}{\left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )^2}dx+10 \int \frac {1}{x^2 \left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )}dx+\int \frac {1}{x \left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )}dx-\frac {25 \int \frac {1}{x^3 \left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )^2}dx}{e^2}+\frac {15 \int \frac {1}{x^3 \left (e^2 x^2+e^2 \log (x) x+5 e^2 x+5\right )}dx}{e^2}+\frac {1}{e^2 x^2}\right )\) |
Input:
Int[(5*E^2 + E^4*x^2 + (-5*E^2 + E^4*(-x - 3*x^2))*Log[x] + (E^4*x - 2*E^4 *x*Log[x])*Log[E^5*x])/(25*x^2 + 10*E^2*x^4 + E^4*x^6 + (10*E^2*x^3 + 2*E^ 4*x^5)*Log[E^5*x] + E^4*x^4*Log[E^5*x]^2),x]
Output:
$Aborted
Time = 1.60 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{2} \ln \left (x \right )}{x \left ({\mathrm e}^{2} \ln \left (x \,{\mathrm e}^{5}\right ) x +x^{2} {\mathrm e}^{2}+5\right )}\) | \(28\) |
default | \(\frac {{\mathrm e}^{2} \ln \left (x \right )}{x \left (x^{2} {\mathrm e}^{2}+x \,{\mathrm e}^{2} \ln \left (x \right )+5 \,{\mathrm e}^{2} x +5\right )}\) | \(30\) |
risch | \(\frac {1}{x^{2}}-\frac {10+2 x^{2} {\mathrm e}^{2}+10 \,{\mathrm e}^{2} x}{x^{2} \left (10+2 x^{2} {\mathrm e}^{2}+2 x \,{\mathrm e}^{2} \ln \left (x \right )+10 \,{\mathrm e}^{2} x \right )}\) | \(47\) |
Input:
int(((-2*x*exp(2)^2*ln(x)+x*exp(2)^2)*ln(x*exp(5))+((-3*x^2-x)*exp(2)^2-5* exp(2))*ln(x)+x^2*exp(2)^2+5*exp(2))/(x^4*exp(2)^2*ln(x*exp(5))^2+(2*x^5*e xp(2)^2+10*x^3*exp(2))*ln(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2)+25*x^2),x,m ethod=_RETURNVERBOSE)
Output:
1/x*exp(2)*ln(x)/(exp(2)*ln(x*exp(5))*x+x^2*exp(2)+5)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {e^{2} \log \left (x\right )}{x^{2} e^{2} \log \left (x\right ) + {\left (x^{3} + 5 \, x^{2}\right )} e^{2} + 5 \, x} \] Input:
integrate(((-2*x*exp(2)^2*log(x)+x*exp(2)^2)*log(x*exp(5))+((-3*x^2-x)*exp (2)^2-5*exp(2))*log(x)+x^2*exp(2)^2+5*exp(2))/(x^4*exp(2)^2*log(x*exp(5))^ 2+(2*x^5*exp(2)^2+10*x^3*exp(2))*log(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2)+ 25*x^2),x, algorithm="fricas")
Output:
e^2*log(x)/(x^2*e^2*log(x) + (x^3 + 5*x^2)*e^2 + 5*x)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {- x^{2} e^{2} - 5 x e^{2} - 5}{x^{4} e^{2} + x^{3} e^{2} \log {\left (x \right )} + 5 x^{3} e^{2} + 5 x^{2}} + \frac {1}{x^{2}} \] Input:
integrate(((-2*x*exp(2)**2*ln(x)+x*exp(2)**2)*ln(x*exp(5))+((-3*x**2-x)*ex p(2)**2-5*exp(2))*ln(x)+x**2*exp(2)**2+5*exp(2))/(x**4*exp(2)**2*ln(x*exp( 5))**2+(2*x**5*exp(2)**2+10*x**3*exp(2))*ln(x*exp(5))+x**6*exp(2)**2+10*x* *4*exp(2)+25*x**2),x)
Output:
(-x**2*exp(2) - 5*x*exp(2) - 5)/(x**4*exp(2) + x**3*exp(2)*log(x) + 5*x**3 *exp(2) + 5*x**2) + x**(-2)
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {e^{2} \log \left (x\right )}{x^{3} e^{2} + x^{2} e^{2} \log \left (x\right ) + 5 \, x^{2} e^{2} + 5 \, x} \] Input:
integrate(((-2*x*exp(2)^2*log(x)+x*exp(2)^2)*log(x*exp(5))+((-3*x^2-x)*exp (2)^2-5*exp(2))*log(x)+x^2*exp(2)^2+5*exp(2))/(x^4*exp(2)^2*log(x*exp(5))^ 2+(2*x^5*exp(2)^2+10*x^3*exp(2))*log(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2)+ 25*x^2),x, algorithm="maxima")
Output:
e^2*log(x)/(x^3*e^2 + x^2*e^2*log(x) + 5*x^2*e^2 + 5*x)
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {e^{2} \log \left (x\right )}{x^{3} e^{2} + x^{2} e^{2} \log \left (x\right ) + 5 \, x^{2} e^{2} + 5 \, x} \] Input:
integrate(((-2*x*exp(2)^2*log(x)+x*exp(2)^2)*log(x*exp(5))+((-3*x^2-x)*exp (2)^2-5*exp(2))*log(x)+x^2*exp(2)^2+5*exp(2))/(x^4*exp(2)^2*log(x*exp(5))^ 2+(2*x^5*exp(2)^2+10*x^3*exp(2))*log(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2)+ 25*x^2),x, algorithm="giac")
Output:
e^2*log(x)/(x^3*e^2 + x^2*e^2*log(x) + 5*x^2*e^2 + 5*x)
Time = 2.79 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {{\mathrm {e}}^2\,\ln \left (x\right )}{x\,\left (5\,x\,{\mathrm {e}}^2+x^2\,{\mathrm {e}}^2+x\,{\mathrm {e}}^2\,\ln \left (x\right )+5\right )} \] Input:
int((5*exp(2) - log(x)*(5*exp(2) + exp(4)*(x + 3*x^2)) + x^2*exp(4) + log( x*exp(5))*(x*exp(4) - 2*x*exp(4)*log(x)))/(10*x^4*exp(2) + x^6*exp(4) + lo g(x*exp(5))*(10*x^3*exp(2) + 2*x^5*exp(4)) + 25*x^2 + x^4*exp(4)*log(x*exp (5))^2),x)
Output:
(exp(2)*log(x))/(x*(5*x*exp(2) + x^2*exp(2) + x*exp(2)*log(x) + 5))
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx=\frac {\mathrm {log}\left (x \right ) e^{2}}{x \left (\mathrm {log}\left (e^{5} x \right ) e^{2} x +e^{2} x^{2}+5\right )} \] Input:
int(((-2*x*exp(2)^2*log(x)+x*exp(2)^2)*log(x*exp(5))+((-3*x^2-x)*exp(2)^2- 5*exp(2))*log(x)+x^2*exp(2)^2+5*exp(2))/(x^4*exp(2)^2*log(x*exp(5))^2+(2*x ^5*exp(2)^2+10*x^3*exp(2))*log(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2)+25*x^2 ),x)
Output:
(log(x)*e**2)/(x*(log(e**5*x)*e**2*x + e**2*x**2 + 5))