Integrand size = 175, antiderivative size = 29 \[ \int \frac {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx=\frac {10 e^{-e^3+i \pi +x} \log (1+x)}{-5+x+\log (-5+x)} \] Output:
-exp(ln(2)+x-exp(3))/(1/5*x+1/5*ln(-5+x)-1)*ln(1+x)
Time = 5.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx=-\frac {10 e^{-e^3+x} \log (1+x)}{-5+x+\log (-5+x)} \] Input:
Integrate[(2*E^(-E^3 + I*Pi + x)*(125 - 50*x + 5*x^2) + 2*E^(-E^3 + I*Pi + x)*(-25 + 5*x)*Log[-5 + x] + (2*E^(-E^3 + I*Pi + x)*(145 + 90*x - 50*x^2 + 5*x^3) + 2*E^(-E^3 + I*Pi + x)*(-25 - 20*x + 5*x^2)*Log[-5 + x])*Log[1 + x])/(-125 - 50*x + 60*x^2 - 14*x^3 + x^4 + (50 + 30*x - 18*x^2 + 2*x^3)*L og[-5 + x] + (-5 - 4*x + x^2)*Log[-5 + x]^2),x]
Output:
(-10*E^(-E^3 + x)*Log[1 + x])/(-5 + x + Log[-5 + 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.
\(\displaystyle \int \frac {2 e^{x+i \pi -e^3} \left (5 x^2-50 x+125\right )+\left (2 e^{x+i \pi -e^3} \left (5 x^2-20 x-25\right ) \log (x-5)+2 e^{x+i \pi -e^3} \left (5 x^3-50 x^2+90 x+145\right )\right ) \log (x+1)+2 e^{x+i \pi -e^3} (5 x-25) \log (x-5)}{x^4-14 x^3+60 x^2+\left (x^2-4 x-5\right ) \log ^2(x-5)+\left (2 x^3-18 x^2+30 x+50\right ) \log (x-5)-50 x-125} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {10 e^{x-e^3} \left (\left (x^3-10 x^2+18 x+29\right ) \log (x+1)+(x-5)^2+(x-5) \log (x-5) ((x+1) \log (x+1)+1)\right )}{\left (-x^2+4 x+5\right ) (-x-\log (x-5)+5)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 10 \int \frac {e^{x-e^3} \left ((x-5)^2+\left (x^3-10 x^2+18 x+29\right ) \log (x+1)-(5-x) \log (x-5) ((x+1) \log (x+1)+1)\right )}{\left (-x^2+4 x+5\right ) (-x-\log (x-5)+5)^2}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle 10 \int \left (-\frac {e^{x-e^3} \left (x^2+\log (x-5) x-11 x-5 \log (x-5)+29\right ) \log (x+1)}{(x-5) (x+\log (x-5)-5)^2}-\frac {e^{x-e^3}}{(x+1) (x+\log (x-5)-5)}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle 10 \int \left (-\frac {e^{x-e^3} \left (x^2+\log (x-5) x-11 x-5 \log (x-5)+29\right ) \log (x+1)}{(x-5) (x+\log (x-5)-5)^2}-\frac {e^{x-e^3}}{(x+1) (x+\log (x-5)-5)}\right )dx\) |
Input:
Int[(2*E^(-E^3 + I*Pi + x)*(125 - 50*x + 5*x^2) + 2*E^(-E^3 + I*Pi + x)*(- 25 + 5*x)*Log[-5 + x] + (2*E^(-E^3 + I*Pi + x)*(145 + 90*x - 50*x^2 + 5*x^ 3) + 2*E^(-E^3 + I*Pi + x)*(-25 - 20*x + 5*x^2)*Log[-5 + x])*Log[1 + x])/( -125 - 50*x + 60*x^2 - 14*x^3 + x^4 + (50 + 30*x - 18*x^2 + 2*x^3)*Log[-5 + x] + (-5 - 4*x + x^2)*Log[-5 + x]^2),x]
Output:
$Aborted
Time = 3.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-\frac {10 \,{\mathrm e}^{x -{\mathrm e}^{3}} \ln \left (1+x \right )}{-5+x +\ln \left (-5+x \right )}\) | \(23\) |
parallelrisch | \(-\frac {5 \,{\mathrm e}^{\ln \left (2\right )+x -{\mathrm e}^{3}} \ln \left (1+x \right )}{-5+x +\ln \left (-5+x \right )}\) | \(25\) |
Input:
int(((-(5*x^2-20*x-25)*exp(ln(2)+x-exp(3))*ln(-5+x)-(5*x^3-50*x^2+90*x+145 )*exp(ln(2)+x-exp(3)))*ln(1+x)-(5*x-25)*exp(ln(2)+x-exp(3))*ln(-5+x)-(5*x^ 2-50*x+125)*exp(ln(2)+x-exp(3)))/((x^2-4*x-5)*ln(-5+x)^2+(2*x^3-18*x^2+30* x+50)*ln(-5+x)+x^4-14*x^3+60*x^2-50*x-125),x,method=_RETURNVERBOSE)
Output:
-10*exp(x-exp(3))/(-5+x+ln(-5+x))*ln(1+x)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx=-\frac {5 \, e^{\left (x - e^{3} + \log \left (2\right )\right )} \log \left (x + 1\right )}{x + \log \left (x - 5\right ) - 5} \] Input:
integrate(((-(5*x^2-20*x-25)*exp(log(2)+x-exp(3))*log(-5+x)-(5*x^3-50*x^2+ 90*x+145)*exp(log(2)+x-exp(3)))*log(1+x)-(5*x-25)*exp(log(2)+x-exp(3))*log (-5+x)-(5*x^2-50*x+125)*exp(log(2)+x-exp(3)))/((x^2-4*x-5)*log(-5+x)^2+(2* x^3-18*x^2+30*x+50)*log(-5+x)+x^4-14*x^3+60*x^2-50*x-125),x, algorithm="fr icas")
Output:
-5*e^(x - e^3 + log(2))*log(x + 1)/(x + log(x - 5) - 5)
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx=- \frac {10 e^{x - e^{3}} \log {\left (x + 1 \right )}}{x + \log {\left (x - 5 \right )} - 5} \] Input:
integrate(((-(5*x**2-20*x-25)*exp(ln(2)+x-exp(3))*ln(-5+x)-(5*x**3-50*x**2 +90*x+145)*exp(ln(2)+x-exp(3)))*ln(1+x)-(5*x-25)*exp(ln(2)+x-exp(3))*ln(-5 +x)-(5*x**2-50*x+125)*exp(ln(2)+x-exp(3)))/((x**2-4*x-5)*ln(-5+x)**2+(2*x* *3-18*x**2+30*x+50)*ln(-5+x)+x**4-14*x**3+60*x**2-50*x-125),x)
Output:
-10*exp(x - exp(3))*log(x + 1)/(x + log(x - 5) - 5)
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx=-\frac {10 \, e^{x} \log \left (x + 1\right )}{x e^{\left (e^{3}\right )} + e^{\left (e^{3}\right )} \log \left (x - 5\right ) - 5 \, e^{\left (e^{3}\right )}} \] Input:
integrate(((-(5*x^2-20*x-25)*exp(log(2)+x-exp(3))*log(-5+x)-(5*x^3-50*x^2+ 90*x+145)*exp(log(2)+x-exp(3)))*log(1+x)-(5*x-25)*exp(log(2)+x-exp(3))*log (-5+x)-(5*x^2-50*x+125)*exp(log(2)+x-exp(3)))/((x^2-4*x-5)*log(-5+x)^2+(2* x^3-18*x^2+30*x+50)*log(-5+x)+x^4-14*x^3+60*x^2-50*x-125),x, algorithm="ma xima")
Output:
-10*e^x*log(x + 1)/(x*e^(e^3) + e^(e^3)*log(x - 5) - 5*e^(e^3))
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx=-\frac {10 \, e^{\left (x - e^{3}\right )} \log \left (x + 1\right )}{x + \log \left (x - 5\right ) - 5} \] Input:
integrate(((-(5*x^2-20*x-25)*exp(log(2)+x-exp(3))*log(-5+x)-(5*x^3-50*x^2+ 90*x+145)*exp(log(2)+x-exp(3)))*log(1+x)-(5*x-25)*exp(log(2)+x-exp(3))*log (-5+x)-(5*x^2-50*x+125)*exp(log(2)+x-exp(3)))/((x^2-4*x-5)*log(-5+x)^2+(2* x^3-18*x^2+30*x+50)*log(-5+x)+x^4-14*x^3+60*x^2-50*x-125),x, algorithm="gi ac")
Output:
-10*e^(x - e^3)*log(x + 1)/(x + log(x - 5) - 5)
Time = 0.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx=-\frac {10\,\ln \left (x+1\right )\,{\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^x}{x+\ln \left (x-5\right )-5} \] Input:
int((exp(x - exp(3) + log(2))*(5*x^2 - 50*x + 125) + log(x + 1)*(exp(x - e xp(3) + log(2))*(90*x - 50*x^2 + 5*x^3 + 145) - log(x - 5)*exp(x - exp(3) + log(2))*(20*x - 5*x^2 + 25)) + log(x - 5)*exp(x - exp(3) + log(2))*(5*x - 25))/(50*x - log(x - 5)*(30*x - 18*x^2 + 2*x^3 + 50) + log(x - 5)^2*(4*x - x^2 + 5) - 60*x^2 + 14*x^3 - x^4 + 125),x)
Output:
-(10*log(x + 1)*exp(-exp(3))*exp(x))/(x + log(x - 5) - 5)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx=-\frac {10 e^{x} \mathrm {log}\left (x +1\right )}{e^{e^{3}} \left (\mathrm {log}\left (-5+x \right )+x -5\right )} \] Input:
int(((-(5*x^2-20*x-25)*exp(log(2)+x-exp(3))*log(-5+x)-(5*x^3-50*x^2+90*x+1 45)*exp(log(2)+x-exp(3)))*log(1+x)-(5*x-25)*exp(log(2)+x-exp(3))*log(-5+x) -(5*x^2-50*x+125)*exp(log(2)+x-exp(3)))/((x^2-4*x-5)*log(-5+x)^2+(2*x^3-18 *x^2+30*x+50)*log(-5+x)+x^4-14*x^3+60*x^2-50*x-125),x)
Output:
( - 10*e**x*log(x + 1))/(e**(e**3)*(log(x - 5) + x - 5))