Integrand size = 80, antiderivative size = 34 \[ \int \frac {e^5 \left (-x-3 x^3\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (-5+x-45 x^2+9 x^3\right )+e^5 \left (30 x-6 x^2\right ) \log (5)\right ) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx=-4+\frac {e^5 x \left (\frac {\frac {x}{3}+x^3}{x}-x \log (5)\right )}{\log (5-x)} \] Output:
((x^3+1/3*x)/x-x*ln(5))/ln(5-x)*exp(5)*x-4
Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {e^5 \left (-x-3 x^3\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (-5+x-45 x^2+9 x^3\right )+e^5 \left (30 x-6 x^2\right ) \log (5)\right ) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx=\frac {e^5 x \left (1+3 x^2-3 x \log (5)\right )}{3 \log (5-x)} \] Input:
Integrate[(E^5*(-x - 3*x^3) + 3*E^5*x^2*Log[5] + (E^5*(-5 + x - 45*x^2 + 9 *x^3) + E^5*(30*x - 6*x^2)*Log[5])*Log[5 - x])/((-15 + 3*x)*Log[5 - x]^2), x]
Output:
(E^5*x*(1 + 3*x^2 - 3*x*Log[5]))/(3*Log[5 - x])
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.99 (sec) , antiderivative size = 214, normalized size of antiderivative = 6.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {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.
| \(\displaystyle \int \frac {e^5 \left (-3 x^3-x\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (30 x-6 x^2\right ) \log (5)+e^5 \left (9 x^3-45 x^2+x-5\right )\right ) \log (5-x)}{(3 x-15) \log ^2(5-x)} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^5 \left (9 x^2-6 x \log (5)+1\right )}{3 \log (5-x)}-\frac {e^5 x \left (3 x^2-3 x \log (5)+1\right )}{3 (x-5) \log ^2(5-x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -20 e^5 \operatorname {ExpIntegralEi}(2 \log (5-x))+2 e^5 (15-\log (5)) \operatorname {ExpIntegralEi}(2 \log (5-x))-2 e^5 (5-\log (5)) \operatorname {ExpIntegralEi}(2 \log (5-x))+25 e^5 \operatorname {LogIntegral}(5-x)+5 e^5 (5-\log (5)) \operatorname {LogIntegral}(5-x)-\frac {2}{3} e^5 (113-15 \log (5)) \operatorname {LogIntegral}(5-x)+\frac {1}{3} e^5 (76-15 \log (5)) \operatorname {LogIntegral}(5-x)-\frac {e^5 (5-x) x^2}{\log (5-x)}-\frac {e^5 (5-x) x (5-\log (5))}{\log (5-x)}-\frac {e^5 (5-x) (76-15 \log (5))}{3 \log (5-x)}+\frac {5 e^5 (76-15 \log (5))}{3 \log (5-x)}\) |
Input:
Int[(E^5*(-x - 3*x^3) + 3*E^5*x^2*Log[5] + (E^5*(-5 + x - 45*x^2 + 9*x^3) + E^5*(30*x - 6*x^2)*Log[5])*Log[5 - x])/((-15 + 3*x)*Log[5 - x]^2),x]
Output:
-20*E^5*ExpIntegralEi[2*Log[5 - x]] - 2*E^5*ExpIntegralEi[2*Log[5 - x]]*(5 - Log[5]) + 2*E^5*ExpIntegralEi[2*Log[5 - x]]*(15 - Log[5]) - (E^5*(5 - x )*x^2)/Log[5 - x] + (5*E^5*(76 - 15*Log[5]))/(3*Log[5 - x]) - (E^5*(5 - x) *(76 - 15*Log[5]))/(3*Log[5 - x]) - (E^5*(5 - x)*x*(5 - Log[5]))/Log[5 - x ] + 25*E^5*LogIntegral[5 - x] + (E^5*(76 - 15*Log[5])*LogIntegral[5 - x])/ 3 - (2*E^5*(113 - 15*Log[5])*LogIntegral[5 - x])/3 + 5*E^5*(5 - Log[5])*Lo gIntegral[5 - x]
Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76
| method | result | size |
| risch | \(-\frac {x \,{\mathrm e}^{5} \left (3 x \ln \left (5\right )-3 x^{2}-1\right )}{3 \ln \left (5-x \right )}\) | \(26\) |
| norman | \(\frac {x^{3} {\mathrm e}^{5}+\frac {x \,{\mathrm e}^{5}}{3}-x^{2} {\mathrm e}^{5} \ln \left (5\right )}{\ln \left (5-x \right )}\) | \(31\) |
| parallelrisch | \(-\frac {3 x^{2} {\mathrm e}^{5} \ln \left (5\right )-3 x^{3} {\mathrm e}^{5}-x \,{\mathrm e}^{5}}{3 \ln \left (5-x \right )}\) | \(33\) |
| parts | \(-\frac {{\mathrm e}^{5} \left (-6 \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )-9 \,\operatorname {expIntegral}_{1}\left (-3 \ln \left (5-x \right )\right )+30 \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )+90 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )-226 \,\operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )\right )}{3}+10 \,{\mathrm e}^{5} \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )+\frac {{\mathrm e}^{5} x}{3 \ln \left (5-x \right )}+30 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )-\frac {226 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )}{3}-\frac {{\mathrm e}^{5} \ln \left (5\right ) x^{2}}{\ln \left (5-x \right )}-2 \,{\mathrm e}^{5} \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )+\frac {{\mathrm e}^{5} x^{3}}{\ln \left (5-x \right )}-3 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-3 \ln \left (5-x \right )\right )\) | \(189\) |
| derivativedivides | \(2 \,{\mathrm e}^{5} \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )+3 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-3 \ln \left (5-x \right )\right )-10 \,{\mathrm e}^{5} \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )+{\mathrm e}^{5} \ln \left (5\right ) \left (-\frac {\left (5-x \right )^{2}}{\ln \left (5-x \right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )\right )-30 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )+{\mathrm e}^{5} \left (-\frac {\left (5-x \right )^{3}}{\ln \left (5-x \right )}-3 \,\operatorname {expIntegral}_{1}\left (-3 \ln \left (5-x \right )\right )\right )-10 \,{\mathrm e}^{5} \ln \left (5\right ) \left (-\frac {5-x}{\ln \left (5-x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )\right )+\frac {226 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )}{3}-15 \,{\mathrm e}^{5} \left (-\frac {\left (5-x \right )^{2}}{\ln \left (5-x \right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )\right )-\frac {25 \,{\mathrm e}^{5} \ln \left (5\right )}{\ln \left (5-x \right )}+\frac {226 \,{\mathrm e}^{5} \left (-\frac {5-x}{\ln \left (5-x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )\right )}{3}+\frac {380 \,{\mathrm e}^{5}}{3 \ln \left (5-x \right )}\) | \(270\) |
| default | \(2 \,{\mathrm e}^{5} \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )+3 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-3 \ln \left (5-x \right )\right )-10 \,{\mathrm e}^{5} \ln \left (5\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )+{\mathrm e}^{5} \ln \left (5\right ) \left (-\frac {\left (5-x \right )^{2}}{\ln \left (5-x \right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )\right )-30 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )+{\mathrm e}^{5} \left (-\frac {\left (5-x \right )^{3}}{\ln \left (5-x \right )}-3 \,\operatorname {expIntegral}_{1}\left (-3 \ln \left (5-x \right )\right )\right )-10 \,{\mathrm e}^{5} \ln \left (5\right ) \left (-\frac {5-x}{\ln \left (5-x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )\right )+\frac {226 \,{\mathrm e}^{5} \operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )}{3}-15 \,{\mathrm e}^{5} \left (-\frac {\left (5-x \right )^{2}}{\ln \left (5-x \right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (5-x \right )\right )\right )-\frac {25 \,{\mathrm e}^{5} \ln \left (5\right )}{\ln \left (5-x \right )}+\frac {226 \,{\mathrm e}^{5} \left (-\frac {5-x}{\ln \left (5-x \right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (5-x \right )\right )\right )}{3}+\frac {380 \,{\mathrm e}^{5}}{3 \ln \left (5-x \right )}\) | \(270\) |
Input:
int((((-6*x^2+30*x)*exp(5)*ln(5)+(9*x^3-45*x^2+x-5)*exp(5))*ln(5-x)+3*x^2* exp(5)*ln(5)+(-3*x^3-x)*exp(5))/(3*x-15)/ln(5-x)^2,x,method=_RETURNVERBOSE )
Output:
-1/3*x*exp(5)*(3*x*ln(5)-3*x^2-1)/ln(5-x)
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {e^5 \left (-x-3 x^3\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (-5+x-45 x^2+9 x^3\right )+e^5 \left (30 x-6 x^2\right ) \log (5)\right ) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx=-\frac {3 \, x^{2} e^{5} \log \left (5\right ) - {\left (3 \, x^{3} + x\right )} e^{5}}{3 \, \log \left (-x + 5\right )} \] Input:
integrate((((-6*x^2+30*x)*exp(5)*log(5)+(9*x^3-45*x^2+x-5)*exp(5))*log(5-x )+3*x^2*exp(5)*log(5)+(-3*x^3-x)*exp(5))/(3*x-15)/log(5-x)^2,x, algorithm= "fricas")
Output:
-1/3*(3*x^2*e^5*log(5) - (3*x^3 + x)*e^5)/log(-x + 5)
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {e^5 \left (-x-3 x^3\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (-5+x-45 x^2+9 x^3\right )+e^5 \left (30 x-6 x^2\right ) \log (5)\right ) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx=\frac {3 x^{3} e^{5} - 3 x^{2} e^{5} \log {\left (5 \right )} + x e^{5}}{3 \log {\left (5 - x \right )}} \] Input:
integrate((((-6*x**2+30*x)*exp(5)*ln(5)+(9*x**3-45*x**2+x-5)*exp(5))*ln(5- x)+3*x**2*exp(5)*ln(5)+(-3*x**3-x)*exp(5))/(3*x-15)/ln(5-x)**2,x)
Output:
(3*x**3*exp(5) - 3*x**2*exp(5)*log(5) + x*exp(5))/(3*log(5 - x))
Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^5 \left (-x-3 x^3\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (-5+x-45 x^2+9 x^3\right )+e^5 \left (30 x-6 x^2\right ) \log (5)\right ) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx=\frac {3 \, x^{3} e^{5} - 3 \, x^{2} e^{5} \log \left (5\right ) + x e^{5}}{3 \, \log \left (-x + 5\right )} + \frac {5}{3} \, e^{5} \] Input:
integrate((((-6*x^2+30*x)*exp(5)*log(5)+(9*x^3-45*x^2+x-5)*exp(5))*log(5-x )+3*x^2*exp(5)*log(5)+(-3*x^3-x)*exp(5))/(3*x-15)/log(5-x)^2,x, algorithm= "maxima")
Output:
1/3*(3*x^3*e^5 - 3*x^2*e^5*log(5) + x*e^5)/log(-x + 5) + 5/3*e^5
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.94 \[ \int \frac {e^5 \left (-x-3 x^3\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (-5+x-45 x^2+9 x^3\right )+e^5 \left (30 x-6 x^2\right ) \log (5)\right ) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx=\frac {3 \, {\left (x - 5\right )}^{3} e^{5} - 3 \, {\left (x - 5\right )}^{2} e^{5} \log \left (5\right ) + 45 \, {\left (x - 5\right )}^{2} e^{5} - 30 \, {\left (x - 5\right )} e^{5} \log \left (5\right ) + 226 \, {\left (x - 5\right )} e^{5} - 75 \, e^{5} \log \left (5\right ) + 380 \, e^{5}}{3 \, \log \left (-x + 5\right )} \] Input:
integrate((((-6*x^2+30*x)*exp(5)*log(5)+(9*x^3-45*x^2+x-5)*exp(5))*log(5-x )+3*x^2*exp(5)*log(5)+(-3*x^3-x)*exp(5))/(3*x-15)/log(5-x)^2,x, algorithm= "giac")
Output:
1/3*(3*(x - 5)^3*e^5 - 3*(x - 5)^2*e^5*log(5) + 45*(x - 5)^2*e^5 - 30*(x - 5)*e^5*log(5) + 226*(x - 5)*e^5 - 75*e^5*log(5) + 380*e^5)/log(-x + 5)
Time = 1.83 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.41 \[ \int \frac {e^5 \left (-x-3 x^3\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (-5+x-45 x^2+9 x^3\right )+e^5 \left (30 x-6 x^2\right ) \log (5)\right ) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx=\frac {3\,{\mathrm {e}}^5\,x^5-3\,{\mathrm {e}}^5\,\left (\ln \left (5\right )+10\right )\,x^4+2\,{\mathrm {e}}^5\,\left (15\,\ln \left (5\right )+38\right )\,x^3-5\,{\mathrm {e}}^5\,\left (15\,\ln \left (5\right )+2\right )\,x^2+25\,{\mathrm {e}}^5\,x}{75\,\ln \left (5-x\right )-30\,x\,\ln \left (5-x\right )+3\,x^2\,\ln \left (5-x\right )} \] Input:
int((log(5 - x)*(exp(5)*(x - 45*x^2 + 9*x^3 - 5) + exp(5)*log(5)*(30*x - 6 *x^2)) - exp(5)*(x + 3*x^3) + 3*x^2*exp(5)*log(5))/(log(5 - x)^2*(3*x - 15 )),x)
Output:
(25*x*exp(5) + 3*x^5*exp(5) - 3*x^4*exp(5)*(log(5) + 10) - 5*x^2*exp(5)*(1 5*log(5) + 2) + 2*x^3*exp(5)*(15*log(5) + 38))/(75*log(5 - x) - 30*x*log(5 - x) + 3*x^2*log(5 - x))
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {e^5 \left (-x-3 x^3\right )+3 e^5 x^2 \log (5)+\left (e^5 \left (-5+x-45 x^2+9 x^3\right )+e^5 \left (30 x-6 x^2\right ) \log (5)\right ) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx=\frac {e^{5} x \left (-3 \,\mathrm {log}\left (5\right ) x +3 x^{2}+1\right )}{3 \,\mathrm {log}\left (-x +5\right )} \] Input:
int((((-6*x^2+30*x)*exp(5)*log(5)+(9*x^3-45*x^2+x-5)*exp(5))*log(5-x)+3*x^ 2*exp(5)*log(5)+(-3*x^3-x)*exp(5))/(3*x-15)/log(5-x)^2,x)
Output:
(e**5*x*( - 3*log(5)*x + 3*x**2 + 1))/(3*log( - x + 5))