Integrand size = 133, antiderivative size = 27 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{-x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \]
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{-x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \]
Integrate[(E^((E^(2*x) + (-x + x^2)*Log[(65 + 5*x*Log[2])/13])/Log[(65 + 5 *x*Log[2])/13])*(-(E^(2*x)*Log[2]) + E^(2*x)*(26 + 2*x*Log[2])*Log[(65 + 5 *x*Log[2])/13] + (-13 + 26*x + (-x + 2*x^2)*Log[2])*Log[(65 + 5*x*Log[2])/ 13]^2))/((13 + x*Log[2])*Log[(65 + 5*x*Log[2])/13]^2),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.
\(\displaystyle \int \frac {\left (\left (\left (2 x^2-x\right ) \log (2)+26 x-13\right ) \log ^2\left (\frac {1}{13} (5 x \log (2)+65)\right )+e^{2 x} (2 x \log (2)+26) \log \left (\frac {1}{13} (5 x \log (2)+65)\right )-e^{2 x} \log (2)\right ) \exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 x \log (2)+65)\right )+e^{2 x}}{\log \left (\frac {1}{13} (5 x \log (2)+65)\right )}\right )}{(x \log (2)+13) \log ^2\left (\frac {1}{13} (5 x \log (2)+65)\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (\left (\left (2 x^2-x\right ) \log (2)+26 x-13\right ) \log ^2\left (\frac {1}{13} (5 x \log (2)+65)\right )+e^{2 x} (2 x \log (2)+26) \log \left (\frac {1}{13} (5 x \log (2)+65)\right )-e^{2 x} \log (2)\right ) \exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 x \log (2)+65)\right )+e^{2 x}}{\log \left (\frac {5}{13} x \log (2)+5\right )}\right )}{(x \log (2)+13) \log ^2\left (\frac {5}{13} x \log (2)+5\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (2 x \log (2) \log \left (\frac {5}{13} x \log (2)+5\right )+26 \log \left (\frac {5}{13} x \log (2)+5\right )-\log (2)\right ) \exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 x \log (2)+65)\right )+e^{2 x}}{\log \left (\frac {5}{13} x \log (2)+5\right )}+2 x\right )}{(x \log (2)+13) \log ^2\left (\frac {5}{13} x \log (2)+5\right )}+2 x \exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 x \log (2)+65)\right )+e^{2 x}}{\log \left (\frac {5}{13} x \log (2)+5\right )}\right )-\exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 x \log (2)+65)\right )+e^{2 x}}{\log \left (\frac {5}{13} x \log (2)+5\right )}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 \log (2) x+65)\right )+e^{2 x}}{\log \left (\frac {5}{13} \log (2) x+5\right )}\right )dx+2 \int \exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 \log (2) x+65)\right )+e^{2 x}}{\log \left (\frac {5}{13} \log (2) x+5\right )}\right ) xdx-\log (2) \int \frac {e^{x^2+x+\frac {e^{2 x}}{\log \left (\frac {5}{13} \log (2) x+5\right )}}}{(\log (2) x+13) \log ^2\left (\frac {5}{13} \log (2) x+5\right )}dx+2 \int \frac {e^{x^2+x+\frac {e^{2 x}}{\log \left (\frac {5}{13} \log (2) x+5\right )}}}{\log \left (\frac {5}{13} \log (2) x+5\right )}dx\) |
Int[(E^((E^(2*x) + (-x + x^2)*Log[(65 + 5*x*Log[2])/13])/Log[(65 + 5*x*Log [2])/13])*(-(E^(2*x)*Log[2]) + E^(2*x)*(26 + 2*x*Log[2])*Log[(65 + 5*x*Log [2])/13] + (-13 + 26*x + (-x + 2*x^2)*Log[2])*Log[(65 + 5*x*Log[2])/13]^2) )/((13 + x*Log[2])*Log[(65 + 5*x*Log[2])/13]^2),x]
3.23.39.3.1 Defintions of rubi rules used
Time = 10.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (x^{2}-x \right ) \ln \left (\frac {5 x \ln \left (2\right )}{13}+5\right )+{\mathrm e}^{2 x}}{\ln \left (\frac {5 x \ln \left (2\right )}{13}+5\right )}}\) | \(34\) |
risch | \({\mathrm e}^{\frac {\ln \left (\frac {5 x \ln \left (2\right )}{13}+5\right ) x^{2}-\ln \left (\frac {5 x \ln \left (2\right )}{13}+5\right ) x +{\mathrm e}^{2 x}}{\ln \left (\frac {5 x \ln \left (2\right )}{13}+5\right )}}\) | \(41\) |
int((((2*x^2-x)*ln(2)+26*x-13)*ln(5/13*x*ln(2)+5)^2+(2*x*ln(2)+26)*exp(2*x )*ln(5/13*x*ln(2)+5)-ln(2)*exp(2*x))*exp(((x^2-x)*ln(5/13*x*ln(2)+5)+exp(2 *x))/ln(5/13*x*ln(2)+5))/(x*ln(2)+13)/ln(5/13*x*ln(2)+5)^2,x,method=_RETUR NVERBOSE)
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{\left (\frac {{\left (x^{2} - x\right )} \log \left (\frac {5}{13} \, x \log \left (2\right ) + 5\right ) + e^{\left (2 \, x\right )}}{\log \left (\frac {5}{13} \, x \log \left (2\right ) + 5\right )}\right )} \]
integrate((((2*x^2-x)*log(2)+26*x-13)*log(5/13*x*log(2)+5)^2+(2*x*log(2)+2 6)*exp(2*x)*log(5/13*x*log(2)+5)-log(2)*exp(2*x))*exp(((x^2-x)*log(5/13*x* log(2)+5)+exp(2*x))/log(5/13*x*log(2)+5))/(x*log(2)+13)/log(5/13*x*log(2)+ 5)^2,x, algorithm=\
Time = 0.40 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{\frac {\left (x^{2} - x\right ) \log {\left (\frac {5 x \log {\left (2 \right )}}{13} + 5 \right )} + e^{2 x}}{\log {\left (\frac {5 x \log {\left (2 \right )}}{13} + 5 \right )}}} \]
integrate((((2*x**2-x)*ln(2)+26*x-13)*ln(5/13*x*ln(2)+5)**2+(2*x*ln(2)+26) *exp(2*x)*ln(5/13*x*ln(2)+5)-ln(2)*exp(2*x))*exp(((x**2-x)*ln(5/13*x*ln(2) +5)+exp(2*x))/ln(5/13*x*ln(2)+5))/(x*ln(2)+13)/ln(5/13*x*ln(2)+5)**2,x)
Exception generated. \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=\text {Exception raised: RuntimeError} \]
integrate((((2*x^2-x)*log(2)+26*x-13)*log(5/13*x*log(2)+5)^2+(2*x*log(2)+2 6)*exp(2*x)*log(5/13*x*log(2)+5)-log(2)*exp(2*x))*exp(((x^2-x)*log(5/13*x* log(2)+5)+exp(2*x))/log(5/13*x*log(2)+5))/(x*log(2)+13)/log(5/13*x*log(2)+ 5)^2,x, algorithm=\
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
Time = 0.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{\left (x^{2} - x + \frac {e^{\left (2 \, x\right )}}{\log \left (\frac {5}{13} \, x \log \left (2\right ) + 5\right )}\right )} \]
integrate((((2*x^2-x)*log(2)+26*x-13)*log(5/13*x*log(2)+5)^2+(2*x*log(2)+2 6)*exp(2*x)*log(5/13*x*log(2)+5)-log(2)*exp(2*x))*exp(((x^2-x)*log(5/13*x* log(2)+5)+exp(2*x))/log(5/13*x*log(2)+5))/(x*log(2)+13)/log(5/13*x*log(2)+ 5)^2,x, algorithm=\
Time = 12.78 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx={\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}}{\ln \left (\frac {5\,x\,\ln \left (2\right )}{13}+5\right )}} \]