Integrand size = 107, antiderivative size = 23 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log \left (-1+\frac {e^{-2 x}}{x}-(5+x)^2\right )} \]
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \]
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]
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 |
\(\displaystyle -2 \int \frac {1}{\log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+52 i \int \frac {1}{(-2 x-(10-2 i)) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+(10+50 i) \int \frac {1}{(2 x+(10-2 i)) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+(10+2 i) \int \frac {1}{(2 x+(10+2 i)) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+\int \frac {1}{\log \left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx-3 \int \frac {1}{\left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+52 i \int \frac {1}{(-2 x-(10-2 i)) \left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx-2 \int \frac {x}{\left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+(10+50 i) \int \frac {1}{(2 x+(10-2 i)) \left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx+(10+2 i) \int \frac {1}{(2 x+(10+2 i)) \left (e^{2 x} x^3+10 e^{2 x} x^2+26 e^{2 x} x-1\right ) \log ^2\left (-x^2-10 x-26+\frac {e^{-2 x}}{x}\right )}dx\) |
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]
3.10.27.3.1 Defintions of rubi rules used
Time = 2.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35
method | result | size |
parallelrisch | \(\frac {x}{\ln \left ({\mathrm e}^{\ln \left (\frac {{\mathrm e}^{-1}}{x}\right )-2 x +1}-x^{2}-10 x -26\right )}\) | \(31\) |
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)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log \left (-x^{2} - 10 \, x + e^{\left (-2 \, x + \log \left (\frac {e^{\left (-1\right )}}{x}\right ) + 1\right )} - 26\right )} \]
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=\
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log {\left (- x^{2} - 10 x - 26 + \frac {e^{1 - 2 x}}{e x} \right )}} \]
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)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=-\frac {x}{2 \, x - \log \left (-{\left (x^{3} + 10 \, x^{2} + 26 \, x\right )} e^{\left (2 \, x\right )} + 1\right ) + \log \left (x\right )} \]
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=\
Time = 0.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\frac {x}{\log \left (-x^{3} - 10 \, x^{2} - 26 \, x + e^{\left (-2 \, x\right )}\right ) - \log \left (x\right )} \]
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=\
Timed out. \[ \int \frac {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx=\int -\frac {10\,x+{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}\,\left (2\,x+1\right )+2\,x^2-\ln \left ({\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}-10\,x-x^2-26\right )\,\left (10\,x-{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}+x^2+26\right )}{{\ln \left ({\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}-10\,x-x^2-26\right )}^2\,\left (10\,x-{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}+x^2+26\right )} \,d x \]
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)