Integrand size = 269, antiderivative size = 30 \[ \int \frac {-19 x-12 x^2+e^2 (15+10 x)+\left (-10 e^2+12 x\right ) \log ^2(5)+\left (-3 x-2 x^2+e^2 (3+2 x)+\left (-2 e^2+2 x\right ) \log ^2(5)\right ) \log \left (-e^2+x\right )+\left (2+x-\log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )}{-20 x^2-10 x^3+e^2 \left (20 x+10 x^2\right )+\left (-10 e^2 x+10 x^2\right ) \log ^2(5)+\left (e^2 (-10-5 x)+10 x+5 x^2+\left (5 e^2-5 x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )+\log \left (-e^2+x\right ) \left (-4 x^2-2 x^3+e^2 \left (4 x+2 x^2\right )+\left (-2 e^2 x+2 x^2\right ) \log ^2(5)+\left (e^2 (-2-x)+2 x+x^2+\left (e^2-x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )\right )} \, dx=\log \left (\left (5+\log \left (-e^2+x\right )\right ) \left (-x+\frac {1}{2} \log \left (2+x-\log ^2(5)\right )\right )\right ) \]
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-19 x-12 x^2+e^2 (15+10 x)+\left (-10 e^2+12 x\right ) \log ^2(5)+\left (-3 x-2 x^2+e^2 (3+2 x)+\left (-2 e^2+2 x\right ) \log ^2(5)\right ) \log \left (-e^2+x\right )+\left (2+x-\log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )}{-20 x^2-10 x^3+e^2 \left (20 x+10 x^2\right )+\left (-10 e^2 x+10 x^2\right ) \log ^2(5)+\left (e^2 (-10-5 x)+10 x+5 x^2+\left (5 e^2-5 x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )+\log \left (-e^2+x\right ) \left (-4 x^2-2 x^3+e^2 \left (4 x+2 x^2\right )+\left (-2 e^2 x+2 x^2\right ) \log ^2(5)+\left (e^2 (-2-x)+2 x+x^2+\left (e^2-x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )\right )} \, dx=\log \left (5+\log \left (-e^2+x\right )\right )+\log \left (2 x-\log \left (2+x-\log ^2(5)\right )\right ) \]
Integrate[(-19*x - 12*x^2 + E^2*(15 + 10*x) + (-10*E^2 + 12*x)*Log[5]^2 + (-3*x - 2*x^2 + E^2*(3 + 2*x) + (-2*E^2 + 2*x)*Log[5]^2)*Log[-E^2 + x] + ( 2 + x - Log[5]^2)*Log[2 + x - Log[5]^2])/(-20*x^2 - 10*x^3 + E^2*(20*x + 1 0*x^2) + (-10*E^2*x + 10*x^2)*Log[5]^2 + (E^2*(-10 - 5*x) + 10*x + 5*x^2 + (5*E^2 - 5*x)*Log[5]^2)*Log[2 + x - Log[5]^2] + Log[-E^2 + x]*(-4*x^2 - 2 *x^3 + E^2*(4*x + 2*x^2) + (-2*E^2*x + 2*x^2)*Log[5]^2 + (E^2*(-2 - x) + 2 *x + x^2 + (E^2 - x)*Log[5]^2)*Log[2 + x - Log[5]^2])),x]
Time = 3.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {7239, 7293, 2009}
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.
| \(\displaystyle \int \frac {-12 x^2+\left (-2 x^2-3 x+e^2 (2 x+3)+\left (2 x-2 e^2\right ) \log ^2(5)\right ) \log \left (x-e^2\right )-19 x+e^2 (10 x+15)+\left (x+2-\log ^2(5)\right ) \log \left (x+2-\log ^2(5)\right )+\left (12 x-10 e^2\right ) \log ^2(5)}{-10 x^3-20 x^2+e^2 \left (10 x^2+20 x\right )+\left (5 x^2+10 x+e^2 (-5 x-10)+\left (5 e^2-5 x\right ) \log ^2(5)\right ) \log \left (x+2-\log ^2(5)\right )+\left (10 x^2-10 e^2 x\right ) \log ^2(5)+\log \left (x-e^2\right ) \left (-2 x^3-4 x^2+e^2 \left (2 x^2+4 x\right )+\left (x^2+2 x+e^2 (-x-2)+\left (e^2-x\right ) \log ^2(5)\right ) \log \left (x+2-\log ^2(5)\right )+\left (2 x^2-2 e^2 x\right ) \log ^2(5)\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-12 x^2-19 x \left (1-\frac {2}{19} \left (5 e^2+6 \log ^2(5)\right )\right )+\left (e^2-x\right ) \left (2 x+3-2 \log ^2(5)\right ) \log \left (x-e^2\right )+\left (x+2-\log ^2(5)\right ) \log \left (x+2-\log ^2(5)\right )+15 e^2 \left (1-\frac {2 \log ^2(5)}{3}\right )}{\left (e^2-x\right ) \left (x+2-\log ^2(5)\right ) \left (\log \left (x-e^2\right )+5\right ) \left (2 x-\log \left (x+2-\log ^2(5)\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 x+3-2 \log ^2(5)}{\left (x+2-\log ^2(5)\right ) \left (2 x-\log \left (x+2-\log ^2(5)\right )\right )}-\frac {1}{\left (e^2-x\right ) \left (\log \left (x-e^2\right )+5\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \log \left (2 x-\log \left (x+2-\log ^2(5)\right )\right )+\log \left (\log \left (x-e^2\right )+5\right )\) |
Int[(-19*x - 12*x^2 + E^2*(15 + 10*x) + (-10*E^2 + 12*x)*Log[5]^2 + (-3*x - 2*x^2 + E^2*(3 + 2*x) + (-2*E^2 + 2*x)*Log[5]^2)*Log[-E^2 + x] + (2 + x - Log[5]^2)*Log[2 + x - Log[5]^2])/(-20*x^2 - 10*x^3 + E^2*(20*x + 10*x^2) + (-10*E^2*x + 10*x^2)*Log[5]^2 + (E^2*(-10 - 5*x) + 10*x + 5*x^2 + (5*E^ 2 - 5*x)*Log[5]^2)*Log[2 + x - Log[5]^2] + Log[-E^2 + x]*(-4*x^2 - 2*x^3 + E^2*(4*x + 2*x^2) + (-2*E^2*x + 2*x^2)*Log[5]^2 + (E^2*(-2 - x) + 2*x + x ^2 + (E^2 - x)*Log[5]^2)*Log[2 + x - Log[5]^2])),x]
3.24.36.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.67 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
| method | result | size |
| risch | \(\ln \left (\ln \left (x -{\mathrm e}^{2}\right )+5\right )+\ln \left (-2 x +\ln \left (-\ln \left (5\right )^{2}+2+x \right )\right )\) | \(27\) |
| parallelrisch | \(\ln \left (x -\frac {\ln \left (-\ln \left (5\right )^{2}+2+x \right )}{2}\right )+\ln \left (\ln \left (x -{\mathrm e}^{2}\right )+5\right )\) | \(27\) |
| default | \(\ln \left (\ln \left (x -{\mathrm e}^{2}\right )+5\right )+\ln \left (2 x -\ln \left (-\ln \left (5\right )^{2}+2+x \right )\right )\) | \(29\) |
int((((-2*exp(2)+2*x)*ln(5)^2+(3+2*x)*exp(2)-2*x^2-3*x)*ln(x-exp(2))+(-ln( 5)^2+2+x)*ln(-ln(5)^2+2+x)+(-10*exp(2)+12*x)*ln(5)^2+(10*x+15)*exp(2)-12*x ^2-19*x)/((((exp(2)-x)*ln(5)^2+(-2-x)*exp(2)+x^2+2*x)*ln(-ln(5)^2+2+x)+(-2 *exp(2)*x+2*x^2)*ln(5)^2+(2*x^2+4*x)*exp(2)-2*x^3-4*x^2)*ln(x-exp(2))+((5* exp(2)-5*x)*ln(5)^2+(-5*x-10)*exp(2)+5*x^2+10*x)*ln(-ln(5)^2+2+x)+(-10*exp (2)*x+10*x^2)*ln(5)^2+(10*x^2+20*x)*exp(2)-10*x^3-20*x^2),x,method=_RETURN VERBOSE)
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-19 x-12 x^2+e^2 (15+10 x)+\left (-10 e^2+12 x\right ) \log ^2(5)+\left (-3 x-2 x^2+e^2 (3+2 x)+\left (-2 e^2+2 x\right ) \log ^2(5)\right ) \log \left (-e^2+x\right )+\left (2+x-\log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )}{-20 x^2-10 x^3+e^2 \left (20 x+10 x^2\right )+\left (-10 e^2 x+10 x^2\right ) \log ^2(5)+\left (e^2 (-10-5 x)+10 x+5 x^2+\left (5 e^2-5 x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )+\log \left (-e^2+x\right ) \left (-4 x^2-2 x^3+e^2 \left (4 x+2 x^2\right )+\left (-2 e^2 x+2 x^2\right ) \log ^2(5)+\left (e^2 (-2-x)+2 x+x^2+\left (e^2-x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )\right )} \, dx=\log \left (-2 \, x + \log \left (-\log \left (5\right )^{2} + x + 2\right )\right ) + \log \left (\log \left (x - e^{2}\right ) + 5\right ) \]
integrate((((-2*exp(2)+2*x)*log(5)^2+(3+2*x)*exp(2)-2*x^2-3*x)*log(x-exp(2 ))+(-log(5)^2+2+x)*log(-log(5)^2+2+x)+(-10*exp(2)+12*x)*log(5)^2+(10*x+15) *exp(2)-12*x^2-19*x)/((((exp(2)-x)*log(5)^2+(-2-x)*exp(2)+x^2+2*x)*log(-lo g(5)^2+2+x)+(-2*exp(2)*x+2*x^2)*log(5)^2+(2*x^2+4*x)*exp(2)-2*x^3-4*x^2)*l og(x-exp(2))+((5*exp(2)-5*x)*log(5)^2+(-5*x-10)*exp(2)+5*x^2+10*x)*log(-lo g(5)^2+2+x)+(-10*exp(2)*x+10*x^2)*log(5)^2+(10*x^2+20*x)*exp(2)-10*x^3-20* x^2),x, algorithm=\
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-19 x-12 x^2+e^2 (15+10 x)+\left (-10 e^2+12 x\right ) \log ^2(5)+\left (-3 x-2 x^2+e^2 (3+2 x)+\left (-2 e^2+2 x\right ) \log ^2(5)\right ) \log \left (-e^2+x\right )+\left (2+x-\log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )}{-20 x^2-10 x^3+e^2 \left (20 x+10 x^2\right )+\left (-10 e^2 x+10 x^2\right ) \log ^2(5)+\left (e^2 (-10-5 x)+10 x+5 x^2+\left (5 e^2-5 x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )+\log \left (-e^2+x\right ) \left (-4 x^2-2 x^3+e^2 \left (4 x+2 x^2\right )+\left (-2 e^2 x+2 x^2\right ) \log ^2(5)+\left (e^2 (-2-x)+2 x+x^2+\left (e^2-x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )\right )} \, dx=\log {\left (- 2 x + \log {\left (x - \log {\left (5 \right )}^{2} + 2 \right )} \right )} + \log {\left (\log {\left (x - e^{2} \right )} + 5 \right )} \]
integrate((((-2*exp(2)+2*x)*ln(5)**2+(3+2*x)*exp(2)-2*x**2-3*x)*ln(x-exp(2 ))+(-ln(5)**2+2+x)*ln(-ln(5)**2+2+x)+(-10*exp(2)+12*x)*ln(5)**2+(10*x+15)* exp(2)-12*x**2-19*x)/((((exp(2)-x)*ln(5)**2+(-2-x)*exp(2)+x**2+2*x)*ln(-ln (5)**2+2+x)+(-2*exp(2)*x+2*x**2)*ln(5)**2+(2*x**2+4*x)*exp(2)-2*x**3-4*x** 2)*ln(x-exp(2))+((5*exp(2)-5*x)*ln(5)**2+(-5*x-10)*exp(2)+5*x**2+10*x)*ln( -ln(5)**2+2+x)+(-10*exp(2)*x+10*x**2)*ln(5)**2+(10*x**2+20*x)*exp(2)-10*x* *3-20*x**2),x)
Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-19 x-12 x^2+e^2 (15+10 x)+\left (-10 e^2+12 x\right ) \log ^2(5)+\left (-3 x-2 x^2+e^2 (3+2 x)+\left (-2 e^2+2 x\right ) \log ^2(5)\right ) \log \left (-e^2+x\right )+\left (2+x-\log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )}{-20 x^2-10 x^3+e^2 \left (20 x+10 x^2\right )+\left (-10 e^2 x+10 x^2\right ) \log ^2(5)+\left (e^2 (-10-5 x)+10 x+5 x^2+\left (5 e^2-5 x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )+\log \left (-e^2+x\right ) \left (-4 x^2-2 x^3+e^2 \left (4 x+2 x^2\right )+\left (-2 e^2 x+2 x^2\right ) \log ^2(5)+\left (e^2 (-2-x)+2 x+x^2+\left (e^2-x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )\right )} \, dx=\log \left (-2 \, x + \log \left (-\log \left (5\right )^{2} + x + 2\right )\right ) + \log \left (\log \left (x - e^{2}\right ) + 5\right ) \]
integrate((((-2*exp(2)+2*x)*log(5)^2+(3+2*x)*exp(2)-2*x^2-3*x)*log(x-exp(2 ))+(-log(5)^2+2+x)*log(-log(5)^2+2+x)+(-10*exp(2)+12*x)*log(5)^2+(10*x+15) *exp(2)-12*x^2-19*x)/((((exp(2)-x)*log(5)^2+(-2-x)*exp(2)+x^2+2*x)*log(-lo g(5)^2+2+x)+(-2*exp(2)*x+2*x^2)*log(5)^2+(2*x^2+4*x)*exp(2)-2*x^3-4*x^2)*l og(x-exp(2))+((5*exp(2)-5*x)*log(5)^2+(-5*x-10)*exp(2)+5*x^2+10*x)*log(-lo g(5)^2+2+x)+(-10*exp(2)*x+10*x^2)*log(5)^2+(10*x^2+20*x)*exp(2)-10*x^3-20* x^2),x, algorithm=\
Time = 0.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-19 x-12 x^2+e^2 (15+10 x)+\left (-10 e^2+12 x\right ) \log ^2(5)+\left (-3 x-2 x^2+e^2 (3+2 x)+\left (-2 e^2+2 x\right ) \log ^2(5)\right ) \log \left (-e^2+x\right )+\left (2+x-\log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )}{-20 x^2-10 x^3+e^2 \left (20 x+10 x^2\right )+\left (-10 e^2 x+10 x^2\right ) \log ^2(5)+\left (e^2 (-10-5 x)+10 x+5 x^2+\left (5 e^2-5 x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )+\log \left (-e^2+x\right ) \left (-4 x^2-2 x^3+e^2 \left (4 x+2 x^2\right )+\left (-2 e^2 x+2 x^2\right ) \log ^2(5)+\left (e^2 (-2-x)+2 x+x^2+\left (e^2-x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )\right )} \, dx=\log \left (2 \, x - \log \left (-\log \left (5\right )^{2} + x + 2\right )\right ) + \log \left (\log \left (x - e^{2}\right ) + 5\right ) \]
integrate((((-2*exp(2)+2*x)*log(5)^2+(3+2*x)*exp(2)-2*x^2-3*x)*log(x-exp(2 ))+(-log(5)^2+2+x)*log(-log(5)^2+2+x)+(-10*exp(2)+12*x)*log(5)^2+(10*x+15) *exp(2)-12*x^2-19*x)/((((exp(2)-x)*log(5)^2+(-2-x)*exp(2)+x^2+2*x)*log(-lo g(5)^2+2+x)+(-2*exp(2)*x+2*x^2)*log(5)^2+(2*x^2+4*x)*exp(2)-2*x^3-4*x^2)*l og(x-exp(2))+((5*exp(2)-5*x)*log(5)^2+(-5*x-10)*exp(2)+5*x^2+10*x)*log(-lo g(5)^2+2+x)+(-10*exp(2)*x+10*x^2)*log(5)^2+(10*x^2+20*x)*exp(2)-10*x^3-20* x^2),x, algorithm=\
Time = 0.80 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-19 x-12 x^2+e^2 (15+10 x)+\left (-10 e^2+12 x\right ) \log ^2(5)+\left (-3 x-2 x^2+e^2 (3+2 x)+\left (-2 e^2+2 x\right ) \log ^2(5)\right ) \log \left (-e^2+x\right )+\left (2+x-\log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )}{-20 x^2-10 x^3+e^2 \left (20 x+10 x^2\right )+\left (-10 e^2 x+10 x^2\right ) \log ^2(5)+\left (e^2 (-10-5 x)+10 x+5 x^2+\left (5 e^2-5 x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )+\log \left (-e^2+x\right ) \left (-4 x^2-2 x^3+e^2 \left (4 x+2 x^2\right )+\left (-2 e^2 x+2 x^2\right ) \log ^2(5)+\left (e^2 (-2-x)+2 x+x^2+\left (e^2-x\right ) \log ^2(5)\right ) \log \left (2+x-\log ^2(5)\right )\right )} \, dx=\ln \left (\ln \left (x-{\ln \left (5\right )}^2+2\right )-2\,x\right )+\ln \left (\ln \left (x-{\mathrm {e}}^2\right )+5\right ) \]
int((19*x - log(5)^2*(12*x - 10*exp(2)) - log(x - log(5)^2 + 2)*(x - log(5 )^2 + 2) + 12*x^2 + log(x - exp(2))*(3*x - log(5)^2*(2*x - 2*exp(2)) + 2*x ^2 - exp(2)*(2*x + 3)) - exp(2)*(10*x + 15))/(log(x - exp(2))*(4*x^2 - log (x - log(5)^2 + 2)*(2*x - exp(2)*(x + 2) - log(5)^2*(x - exp(2)) + x^2) - exp(2)*(4*x + 2*x^2) + 2*x^3 + log(5)^2*(2*x*exp(2) - 2*x^2)) - log(x - lo g(5)^2 + 2)*(10*x - log(5)^2*(5*x - 5*exp(2)) + 5*x^2 - exp(2)*(5*x + 10)) - exp(2)*(20*x + 10*x^2) + 20*x^2 + 10*x^3 + log(5)^2*(10*x*exp(2) - 10*x ^2)),x)