\(\int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+(-26+\frac {e^{-2 x}}{x}-10 x-x^2) \log (-26+\frac {e^{-2 x}}{x}-10 x-x^2)}{(-26+\frac {e^{-2 x}}{x}-10 x-x^2) \log ^2(-26+\frac {e^{-2 x}}{x}-10 x-x^2)} \, dx\) [927]

Optimal result

Integrand size = 107, antiderivative size = 23 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log \left (-1+\frac {e^{-2 x}}{x}-(5+x)^2\right )} \] Output:

x/ln(exp(ln(exp(-1)/x)-2*x+1)-(5+x)^2-1)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \] Input:

Integrate[(10*x + 2*x^2 + (1 + 2*x)/(E^(2*x)*x) + (-26 + 1/(E^(2*x)*x) - 1 
0*x - x^2)*Log[-26 + 1/(E^(2*x)*x) - 10*x - x^2])/((-26 + 1/(E^(2*x)*x) - 
10*x - x^2)*Log[-26 + 1/(E^(2*x)*x) - 10*x - x^2]^2),x]
 

Output:

x/Log[-26 + 1/(E^(2*x)*x) - 10*x - x^2]
 

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.

\(\displaystyle \int \frac {2 x^2+\left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right ) \log \left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right )+10 x+\frac {e^{-2 x} (2 x+1)}{x}}{\left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right ) \log ^2\left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {-2 x^2+x^2 \log \left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right )+10 x \log \left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right )+26 \log \left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right )-10 x}{\left (x^2+10 x+26\right ) \log ^2\left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right )}-\frac {2 x^3+23 x^2+72 x+26}{\left (x^2+10 x+26\right ) \left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x+\frac {e^{-2 x}}{x}-26\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \int \frac {1}{\log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+52 i \int \frac {1}{(-2 x-(10-2 i)) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+(10+50 i) \int \frac {1}{(2 x+(10-2 i)) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+(10+2 i) \int \frac {1}{(2 x+(10+2 i)) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+\int \frac {1}{\log \left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx-3 \int \frac {1}{\left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+52 i \int \frac {1}{(-2 x-(10-2 i)) \left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx-2 \int \frac {x}{\left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+(10+50 i) \int \frac {1}{(2 x+(10-2 i)) \left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+(10+2 i) \int \frac {1}{(2 x+(10+2 i)) \left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx\)

Input:

Int[(10*x + 2*x^2 + (1 + 2*x)/(E^(2*x)*x) + (-26 + 1/(E^(2*x)*x) - 10*x - 
x^2)*Log[-26 + 1/(E^(2*x)*x) - 10*x - x^2])/((-26 + 1/(E^(2*x)*x) - 10*x - 
 x^2)*Log[-26 + 1/(E^(2*x)*x) - 10*x - x^2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35

method	result	size
parallelrisch	\(\frac {x}{\ln \left ({\mathrm e}^{\ln \left (\frac {{\mathrm e}^{-1}}{x}\right )-2 x +1}-x^{2}-10 x -26\right )}\)	\(31\)

Input:

int(((exp(ln(1/x/exp(1))+1-2*x)-x^2-10*x-26)*ln(exp(ln(1/x/exp(1))+1-2*x)- 
x^2-10*x-26)+(1+2*x)*exp(ln(1/x/exp(1))+1-2*x)+2*x^2+10*x)/(exp(ln(1/x/exp 
(1))+1-2*x)-x^2-10*x-26)/ln(exp(ln(1/x/exp(1))+1-2*x)-x^2-10*x-26)^2,x,met 
hod=_RETURNVERBOSE)
 

Output:

x/ln(exp(ln(1/x/exp(1))+1-2*x)-x^2-10*x-26)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log \left (-x^{2} - 10 \, x + e^{\left (-2 \, x + \log \left (\frac {e^{\left (-1\right )}}{x}\right ) + 1\right )} - 26\right )} \] Input:

integrate(((exp(log(1/x/exp(1))+1-2*x)-x^2-10*x-26)*log(exp(log(1/x/exp(1) 
)+1-2*x)-x^2-10*x-26)+(1+2*x)*exp(log(1/x/exp(1))+1-2*x)+2*x^2+10*x)/(exp( 
log(1/x/exp(1))+1-2*x)-x^2-10*x-26)/log(exp(log(1/x/exp(1))+1-2*x)-x^2-10* 
x-26)^2,x, algorithm="fricas")
 

Output:

x/log(-x^2 - 10*x + e^(-2*x + log(e^(-1)/x) + 1) - 26)
 

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log {\left (- x^{2} - 10 x - 26 + \frac {e^{1 - 2 x}}{e x} \right )}} \] Input:

integrate(((exp(ln(1/x/exp(1))+1-2*x)-x**2-10*x-26)*ln(exp(ln(1/x/exp(1))+ 
1-2*x)-x**2-10*x-26)+(1+2*x)*exp(ln(1/x/exp(1))+1-2*x)+2*x**2+10*x)/(exp(l 
n(1/x/exp(1))+1-2*x)-x**2-10*x-26)/ln(exp(ln(1/x/exp(1))+1-2*x)-x**2-10*x- 
26)**2,x)
 

Output:

x/log(-x**2 - 10*x - 26 + exp(-1)*exp(1 - 2*x)/x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=-\frac {x}{2 \, x - \log \left (-{\left (x^{3} + 10 \, x^{2} + 26 \, x\right )} e^{\left (2 \, x\right )} + 1\right ) + \log \left (x\right )} \] Input:

integrate(((exp(log(1/x/exp(1))+1-2*x)-x^2-10*x-26)*log(exp(log(1/x/exp(1) 
)+1-2*x)-x^2-10*x-26)+(1+2*x)*exp(log(1/x/exp(1))+1-2*x)+2*x^2+10*x)/(exp( 
log(1/x/exp(1))+1-2*x)-x^2-10*x-26)/log(exp(log(1/x/exp(1))+1-2*x)-x^2-10* 
x-26)^2,x, algorithm="maxima")
 

Output:

-x/(2*x - log(-(x^3 + 10*x^2 + 26*x)*e^(2*x) + 1) + log(x))
 

Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log \left (-x^{3} - 10 \, x^{2} - 26 \, x + e^{\left (-2 \, x\right )}\right ) - \log \left (x\right )} \] Input:

integrate(((exp(log(1/x/exp(1))+1-2*x)-x^2-10*x-26)*log(exp(log(1/x/exp(1) 
)+1-2*x)-x^2-10*x-26)+(1+2*x)*exp(log(1/x/exp(1))+1-2*x)+2*x^2+10*x)/(exp( 
log(1/x/exp(1))+1-2*x)-x^2-10*x-26)/log(exp(log(1/x/exp(1))+1-2*x)-x^2-10* 
x-26)^2,x, algorithm="giac")
 

Output:

x/(log(-x^3 - 10*x^2 - 26*x + e^(-2*x)) - log(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\int -\frac {10\,x+{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}\,\left (2\,x+1\right )+2\,x^2-\ln \left ({\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}-10\,x-x^2-26\right )\,\left (10\,x-{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}+x^2+26\right )}{{\ln \left ({\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}-10\,x-x^2-26\right )}^2\,\left (10\,x-{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}+x^2+26\right )} \,d x \] Input:

int(-(10*x + exp(log(exp(-1)/x) - 2*x + 1)*(2*x + 1) + 2*x^2 - log(exp(log 
(exp(-1)/x) - 2*x + 1) - 10*x - x^2 - 26)*(10*x - exp(log(exp(-1)/x) - 2*x 
 + 1) + x^2 + 26))/(log(exp(log(exp(-1)/x) - 2*x + 1) - 10*x - x^2 - 26)^2 
*(10*x - exp(log(exp(-1)/x) - 2*x + 1) + x^2 + 26)),x)
 

Output:

int(-(10*x + exp(log(exp(-1)/x) - 2*x + 1)*(2*x + 1) + 2*x^2 - log(exp(log 
(exp(-1)/x) - 2*x + 1) - 10*x - x^2 - 26)*(10*x - exp(log(exp(-1)/x) - 2*x 
 + 1) + x^2 + 26))/(log(exp(log(exp(-1)/x) - 2*x + 1) - 10*x - x^2 - 26)^2 
*(10*x - exp(log(exp(-1)/x) - 2*x + 1) + x^2 + 26)), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\mathrm {log}\left (\frac {-e^{2 x} x^{3}-10 e^{2 x} x^{2}-26 e^{2 x} x +1}{e^{2 x} x}\right )} \] Input:

int(((exp(log(1/x/exp(1))+1-2*x)-x^2-10*x-26)*log(exp(log(1/x/exp(1))+1-2* 
x)-x^2-10*x-26)+(1+2*x)*exp(log(1/x/exp(1))+1-2*x)+2*x^2+10*x)/(exp(log(1/ 
x/exp(1))+1-2*x)-x^2-10*x-26)/log(exp(log(1/x/exp(1))+1-2*x)-x^2-10*x-26)^ 
2,x)
 

Output:

x/log(( - e**(2*x)*x**3 - 10*e**(2*x)*x**2 - 26*e**(2*x)*x + 1)/(e**(2*x)* 
x))