Integrand size = 67, antiderivative size = 27 \[ \int \frac {e^x \left (-105+25 x+25 x^2-25 x^3\right )+e^x \left (105+55 x+25 x^2\right ) \log (x)-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx=\frac {e^x}{-x+\frac {-\frac {21}{5}+x}{1-\frac {x}{\log (x)}}} \] Output:
exp(x)/((x-21/5)/(1-x/ln(x))-x)
Time = 0.40 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (-105+25 x+25 x^2-25 x^3\right )+e^x \left (105+55 x+25 x^2\right ) \log (x)-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx=-\frac {5 e^x (x-\log (x))}{5 x^2-21 \log (x)} \] Input:
Integrate[(E^x*(-105 + 25*x + 25*x^2 - 25*x^3) + E^x*(105 + 55*x + 25*x^2) *Log[x] - 105*E^x*Log[x]^2)/(25*x^4 - 210*x^2*Log[x] + 441*Log[x]^2),x]
Output:
(-5*E^x*(x - Log[x]))/(5*x^2 - 21*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.
\(\displaystyle \int \frac {e^x \left (25 x^2+55 x+105\right ) \log (x)+e^x \left (-25 x^3+25 x^2+25 x-105\right )-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 e^x \left (-5 x^3+5 x^2+5 x^2 \log (x)+5 x-21 \log ^2(x)+11 x \log (x)+21 \log (x)-21\right )}{\left (5 x^2-21 \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int -\frac {e^x \left (5 x^3-5 \log (x) x^2-5 x^2-11 \log (x) x-5 x+21 \log ^2(x)-21 \log (x)+21\right )}{\left (5 x^2-21 \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {e^x \left (5 x^3-5 \log (x) x^2-5 x^2-11 \log (x) x-5 x+21 \log ^2(x)-21 \log (x)+21\right )}{\left (5 x^2-21 \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (\frac {e^x \left (-5 x^2+11 x+21\right )}{21 \left (5 x^2-21 \log (x)\right )}+\frac {e^x}{21}+\frac {e^x \left (50 x^3-210 x^2-105 x+441\right )}{21 \left (5 x^2-21 \log (x)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left (21 \int \frac {e^x}{\left (5 x^2-21 \log (x)\right )^2}dx-5 \int \frac {e^x x}{\left (5 x^2-21 \log (x)\right )^2}dx-10 \int \frac {e^x x^2}{\left (5 x^2-21 \log (x)\right )^2}dx+\int \frac {e^x}{5 x^2-21 \log (x)}dx+\frac {11}{21} \int \frac {e^x x}{5 x^2-21 \log (x)}dx-\frac {5}{21} \int \frac {e^x x^2}{5 x^2-21 \log (x)}dx+\frac {50}{21} \int \frac {e^x x^3}{\left (5 x^2-21 \log (x)\right )^2}dx+\frac {e^x}{21}\right )\) |
Input:
Int[(E^x*(-105 + 25*x + 25*x^2 - 25*x^3) + E^x*(105 + 55*x + 25*x^2)*Log[x ] - 105*E^x*Log[x]^2)/(25*x^4 - 210*x^2*Log[x] + 441*Log[x]^2),x]
Output:
$Aborted
Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {-105 \,{\mathrm e}^{x} x +105 \,{\mathrm e}^{x} \ln \left (x \right )}{105 x^{2}-441 \ln \left (x \right )}\) | \(27\) |
risch | \(-\frac {5 \,{\mathrm e}^{x}}{21}+\frac {5 \left (5 x -21\right ) x \,{\mathrm e}^{x}}{21 \left (5 x^{2}-21 \ln \left (x \right )\right )}\) | \(28\) |
Input:
int((-105*exp(x)*ln(x)^2+(25*x^2+55*x+105)*exp(x)*ln(x)+(-25*x^3+25*x^2+25 *x-105)*exp(x))/(441*ln(x)^2-210*x^2*ln(x)+25*x^4),x,method=_RETURNVERBOSE )
Output:
1/21*(-105*exp(x)*x+105*exp(x)*ln(x))/(5*x^2-21*ln(x))
Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-105+25 x+25 x^2-25 x^3\right )+e^x \left (105+55 x+25 x^2\right ) \log (x)-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx=-\frac {5 \, {\left (x e^{x} - e^{x} \log \left (x\right )\right )}}{5 \, x^{2} - 21 \, \log \left (x\right )} \] Input:
integrate((-105*exp(x)*log(x)^2+(25*x^2+55*x+105)*exp(x)*log(x)+(-25*x^3+2 5*x^2+25*x-105)*exp(x))/(441*log(x)^2-210*x^2*log(x)+25*x^4),x, algorithm= "fricas")
Output:
-5*(x*e^x - e^x*log(x))/(5*x^2 - 21*log(x))
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {e^x \left (-105+25 x+25 x^2-25 x^3\right )+e^x \left (105+55 x+25 x^2\right ) \log (x)-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx=\frac {\left (- 5 x + 5 \log {\left (x \right )}\right ) e^{x}}{5 x^{2} - 21 \log {\left (x \right )}} \] Input:
integrate((-105*exp(x)*ln(x)**2+(25*x**2+55*x+105)*exp(x)*ln(x)+(-25*x**3+ 25*x**2+25*x-105)*exp(x))/(441*ln(x)**2-210*x**2*ln(x)+25*x**4),x)
Output:
(-5*x + 5*log(x))*exp(x)/(5*x**2 - 21*log(x))
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (-105+25 x+25 x^2-25 x^3\right )+e^x \left (105+55 x+25 x^2\right ) \log (x)-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx=-\frac {5 \, {\left (x - \log \left (x\right )\right )} e^{x}}{5 \, x^{2} - 21 \, \log \left (x\right )} \] Input:
integrate((-105*exp(x)*log(x)^2+(25*x^2+55*x+105)*exp(x)*log(x)+(-25*x^3+2 5*x^2+25*x-105)*exp(x))/(441*log(x)^2-210*x^2*log(x)+25*x^4),x, algorithm= "maxima")
Output:
-5*(x - log(x))*e^x/(5*x^2 - 21*log(x))
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-105+25 x+25 x^2-25 x^3\right )+e^x \left (105+55 x+25 x^2\right ) \log (x)-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx=-\frac {5 \, {\left (x e^{x} - e^{x} \log \left (x\right )\right )}}{5 \, x^{2} - 21 \, \log \left (x\right )} \] Input:
integrate((-105*exp(x)*log(x)^2+(25*x^2+55*x+105)*exp(x)*log(x)+(-25*x^3+2 5*x^2+25*x-105)*exp(x))/(441*log(x)^2-210*x^2*log(x)+25*x^4),x, algorithm= "giac")
Output:
-5*(x*e^x - e^x*log(x))/(5*x^2 - 21*log(x))
Time = 4.47 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (-105+25 x+25 x^2-25 x^3\right )+e^x \left (105+55 x+25 x^2\right ) \log (x)-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx=\frac {5\,{\mathrm {e}}^x\,\left (x-\ln \left (x\right )\right )}{21\,\ln \left (x\right )-5\,x^2} \] Input:
int((exp(x)*(25*x + 25*x^2 - 25*x^3 - 105) - 105*exp(x)*log(x)^2 + exp(x)* log(x)*(55*x + 25*x^2 + 105))/(441*log(x)^2 - 210*x^2*log(x) + 25*x^4),x)
Output:
(5*exp(x)*(x - log(x)))/(21*log(x) - 5*x^2)
Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (-105+25 x+25 x^2-25 x^3\right )+e^x \left (105+55 x+25 x^2\right ) \log (x)-105 e^x \log ^2(x)}{25 x^4-210 x^2 \log (x)+441 \log ^2(x)} \, dx=\frac {5 e^{x} \left (x -\mathrm {log}\left (x \right )\right )}{21 \,\mathrm {log}\left (x \right )-5 x^{2}} \] Input:
int((-105*exp(x)*log(x)^2+(25*x^2+55*x+105)*exp(x)*log(x)+(-25*x^3+25*x^2+ 25*x-105)*exp(x))/(441*log(x)^2-210*x^2*log(x)+25*x^4),x)
Output:
(5*e**x*( - log(x) + x))/(21*log(x) - 5*x**2)