\(\int \frac {e^{\frac {e^{2 x}+(-x+x^2) \log (\frac {1}{13} (65+5 x \log (2)))}{\log (\frac {1}{13} (65+5 x \log (2)))}} (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log (\frac {1}{13} (65+5 x \log (2)))+(-13+26 x+(-x+2 x^2) \log (2)) \log ^2(\frac {1}{13} (65+5 x \log (2))))}{(13+x \log (2)) \log ^2(\frac {1}{13} (65+5 x \log (2)))} \, dx\) [2239]

Optimal result

Integrand size = 133, antiderivative size = 27 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{-x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \] Output:

exp(x^2-x+exp(2*x)/ln(5/13*x*ln(2)+5))
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{-x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \] Input:

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]
 

Output:

E^(-x + x^2 + E^(2*x)/Log[5 + (5*x*Log[2])/13])
 

Rubi [F]

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

\(\displaystyle -\int \exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 \log (2) x+65)\right )+e^{2 x}}{\log \left (\frac {5}{13} \log (2) x+5\right )}\right )dx+2 \int \exp \left (\frac {\left (x^2-x\right ) \log \left (\frac {1}{13} (5 \log (2) x+65)\right )+e^{2 x}}{\log \left (\frac {5}{13} \log (2) x+5\right )}\right ) xdx-\log (2) \int \frac {e^{x^2+x+\frac {e^{2 x}}{\log \left (\frac {5}{13} \log (2) x+5\right )}}}{(\log (2) x+13) \log ^2\left (\frac {5}{13} \log (2) x+5\right )}dx+2 \int \frac {e^{x^2+x+\frac {e^{2 x}}{\log \left (\frac {5}{13} \log (2) x+5\right )}}}{\log \left (\frac {5}{13} \log (2) x+5\right )}dx\)

Input:

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]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 9.66 (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	\({\mathrm e}^{\frac {\ln \left (\frac {5 x \ln \left (2\right )}{13}+5\right ) x^{2}-\ln \left (\frac {5 x \ln \left (2\right )}{13}+5\right ) x +{\mathrm e}^{2 x}}{\ln \left (\frac {5 x \ln \left (2\right )}{13}+5\right )}}\)	\(41\)

Input:

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)
 

Output:

exp(((x^2-x)*ln(5/13*x*ln(2)+5)+exp(2*x))/ln(5/13*x*ln(2)+5))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{\left (\frac {{\left (x^{2} - x\right )} \log \left (\frac {5}{13} \, x \log \left (2\right ) + 5\right ) + e^{\left (2 \, x\right )}}{\log \left (\frac {5}{13} \, x \log \left (2\right ) + 5\right )}\right )} \] Input:

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="fricas")
 

Output:

e^(((x^2 - x)*log(5/13*x*log(2) + 5) + e^(2*x))/log(5/13*x*log(2) + 5))
 

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{\frac {\left (x^{2} - x\right ) \log {\left (\frac {5 x \log {\left (2 \right )}}{13} + 5 \right )} + e^{2 x}}{\log {\left (\frac {5 x \log {\left (2 \right )}}{13} + 5 \right )}}} \] Input:

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)
 

Output:

exp(((x**2 - x)*log(5*x*log(2)/13 + 5) + exp(2*x))/log(5*x*log(2)/13 + 5))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:

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="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=e^{\left (x^{2} - x + \frac {e^{\left (2 \, x\right )}}{\log \left (\frac {5}{13} \, x \log \left (2\right ) + 5\right )}\right )} \] Input:

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="giac")
 

Output:

e^(x^2 - x + e^(2*x)/log(5/13*x*log(2) + 5))
 

Mupad [B] (verification not implemented)

Time = 3.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx={\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}}{\ln \left (\frac {5\,x\,\ln \left (2\right )}{13}+5\right )}} \] Input:

int(-(exp((exp(2*x) - log((5*x*log(2))/13 + 5)*(x - x^2))/log((5*x*log(2)) 
/13 + 5))*(exp(2*x)*log(2) + log((5*x*log(2))/13 + 5)^2*(log(2)*(x - 2*x^2 
) - 26*x + 13) - log((5*x*log(2))/13 + 5)*exp(2*x)*(2*x*log(2) + 26)))/(lo 
g((5*x*log(2))/13 + 5)^2*(x*log(2) + 13)),x)
 

Output:

exp(-x)*exp(x^2)*exp(exp(2*x)/log((5*x*log(2))/13 + 5))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^{\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}} \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx=\frac {e^{\frac {e^{2 x}+\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (2\right ) x}{13}+5\right ) x^{2}}{\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (2\right ) x}{13}+5\right )}}}{e^{x}} \] Input:

int((((2*x^2-x)*log(2)+26*x-13)*log(5/13*x*log(2)+5)^2+(2*x*log(2)+26)*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 
)
 

Output:

e**((e**(2*x) + log((5*log(2)*x + 65)/13)*x**2)/log((5*log(2)*x + 65)/13)) 
/e**x