Integrand size = 89, antiderivative size = 34 \[ \int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx=\frac {8 \left (-\frac {1}{4 x}-x+\log \left (e^{\frac {e^x}{5}}-x\right )\right )}{x \log (2)} \]
Time = 0.49 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx=-\frac {\frac {10}{x^2}-\frac {40 \log \left (e^{\frac {e^x}{5}}-x\right )}{x}}{5 \log (2)} \]
Integrate[(-20*x - 40*x^2 + E^(E^x/5)*(20 + 8*E^x*x^2) + (-40*E^(E^x/5)*x + 40*x^2)*Log[E^(E^x/5) - x])/(5*E^(E^x/5)*x^3*Log[2] - 5*x^4*Log[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.
\(\displaystyle \int \frac {-40 x^2+e^{\frac {e^x}{5}} \left (8 e^x x^2+20\right )+\left (40 x^2-40 e^{\frac {e^x}{5}} x\right ) \log \left (e^{\frac {e^x}{5}}-x\right )-20 x}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-40 x^2+e^{\frac {e^x}{5}} \left (8 e^x x^2+20\right )+\left (40 x^2-40 e^{\frac {e^x}{5}} x\right ) \log \left (e^{\frac {e^x}{5}}-x\right )-20 x}{5 \left (e^{\frac {e^x}{5}}-x\right ) x^3 \log (2)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {4 \left (10 x^2+5 x-e^{\frac {e^x}{5}} \left (2 e^x x^2+5\right )+10 \left (e^{\frac {e^x}{5}} x-x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )\right )}{\left (e^{\frac {e^x}{5}}-x\right ) x^3}dx}{5 \log (2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 \int \frac {10 x^2+5 x-e^{\frac {e^x}{5}} \left (2 e^x x^2+5\right )+10 \left (e^{\frac {e^x}{5}} x-x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{\left (e^{\frac {e^x}{5}}-x\right ) x^3}dx}{5 \log (2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {4 \int \left (\frac {5 \left (-2 \log \left (e^{\frac {e^x}{5}}-x\right ) x^2+2 x^2+2 e^{\frac {e^x}{5}} \log \left (e^{\frac {e^x}{5}}-x\right ) x+x-e^{\frac {e^x}{5}}\right )}{\left (e^{\frac {e^x}{5}}-x\right ) x^3}+\frac {2 e^{\frac {1}{5} \left (5 x+e^x\right )}}{x \left (x-e^{\frac {e^x}{5}}\right )}\right )dx}{5 \log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \left (2 \int \frac {e^{\frac {1}{5} \left (5 x+e^x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x}dx+2 \int \frac {e^{\frac {1}{5} \left (5 x+e^x\right )}}{x \left (x-e^{\frac {e^x}{5}}\right )}dx+\frac {5}{2 x^2}-\frac {10 \log \left (e^{\frac {e^x}{5}}-x\right )}{x}\right )}{5 \log (2)}\) |
Int[(-20*x - 40*x^2 + E^(E^x/5)*(20 + 8*E^x*x^2) + (-40*E^(E^x/5)*x + 40*x ^2)*Log[E^(E^x/5) - x])/(5*E^(E^x/5)*x^3*Log[2] - 5*x^4*Log[2]),x]
3.22.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.88 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74
method | result | size |
parallelrisch | \(\frac {-10+40 \ln \left ({\mathrm e}^{\frac {{\mathrm e}^{x}}{5}}-x \right ) x}{5 \ln \left (2\right ) x^{2}}\) | \(25\) |
risch | \(\frac {8 \ln \left ({\mathrm e}^{\frac {{\mathrm e}^{x}}{5}}-x \right )}{x \ln \left (2\right )}-\frac {2}{\ln \left (2\right ) x^{2}}\) | \(30\) |
int(((-40*x*exp(1/5*exp(x))+40*x^2)*ln(exp(1/5*exp(x))-x)+(8*exp(x)*x^2+20 )*exp(1/5*exp(x))-40*x^2-20*x)/(5*x^3*ln(2)*exp(1/5*exp(x))-5*x^4*ln(2)),x ,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx=\frac {2 \, {\left (4 \, x \log \left (-x + e^{\left (\frac {1}{5} \, e^{x}\right )}\right ) - 1\right )}}{x^{2} \log \left (2\right )} \]
integrate(((-40*x*exp(1/5*exp(x))+40*x^2)*log(exp(1/5*exp(x))-x)+(8*exp(x) *x^2+20)*exp(1/5*exp(x))-40*x^2-20*x)/(5*x^3*log(2)*exp(1/5*exp(x))-5*x^4* log(2)),x, algorithm=\
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx=\frac {8 \log {\left (- x + e^{\frac {e^{x}}{5}} \right )}}{x \log {\left (2 \right )}} - \frac {2}{x^{2} \log {\left (2 \right )}} \]
integrate(((-40*x*exp(1/5*exp(x))+40*x**2)*ln(exp(1/5*exp(x))-x)+(8*exp(x) *x**2+20)*exp(1/5*exp(x))-40*x**2-20*x)/(5*x**3*ln(2)*exp(1/5*exp(x))-5*x* *4*ln(2)),x)
Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx=\frac {2 \, {\left (4 \, x \log \left (-x + e^{\left (\frac {1}{5} \, e^{x}\right )}\right ) - 1\right )}}{x^{2} \log \left (2\right )} \]
integrate(((-40*x*exp(1/5*exp(x))+40*x^2)*log(exp(1/5*exp(x))-x)+(8*exp(x) *x^2+20)*exp(1/5*exp(x))-40*x^2-20*x)/(5*x^3*log(2)*exp(1/5*exp(x))-5*x^4* log(2)),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx=\frac {2 \, {\left (4 \, x \log \left (-{\left (x e^{x} - e^{\left (x + \frac {1}{5} \, e^{x}\right )}\right )} e^{\left (-x\right )}\right ) - 1\right )}}{x^{2} \log \left (2\right )} \]
integrate(((-40*x*exp(1/5*exp(x))+40*x^2)*log(exp(1/5*exp(x))-x)+(8*exp(x) *x^2+20)*exp(1/5*exp(x))-40*x^2-20*x)/(5*x^3*log(2)*exp(1/5*exp(x))-5*x^4* log(2)),x, algorithm=\
Time = 14.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68 \[ \int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx=\frac {8\,x\,\ln \left ({\mathrm {e}}^{\frac {{\mathrm {e}}^x}{5}}-x\right )-2}{x^2\,\ln \left (2\right )} \]