Integrand size = 90, antiderivative size = 32 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 x+\frac {1}{5} \left (\log (3 x)-\log \left (\log \left (\frac {-e^{x/5}+x}{x}\right )\right )\right ) \]
Time = 0.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=\frac {1}{25} \left (125 x+5 \log (x)-5 \log \left (\log \left (1-\frac {e^{x/5}}{x}\right )\right )\right ) \]
Integrate[(E^(x/5)*(5 - x) + (-5*x - 125*x^2 + E^(x/5)*(5 + 125*x))*Log[(- E^(x/5) + x)/x])/((25*E^(x/5)*x - 25*x^2)*Log[(-E^(x/5) + x)/x]),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.
\(\displaystyle \int \frac {\left (-125 x^2-5 x+e^{x/5} (125 x+5)\right ) \log \left (\frac {x-e^{x/5}}{x}\right )+e^{x/5} (5-x)}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {x-e^{x/5}}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-125 x^2-5 x+e^{x/5} (125 x+5)\right ) \log \left (\frac {x-e^{x/5}}{x}\right )+e^{x/5} (5-x)}{25 \left (e^{x/5}-x\right ) x \log \left (1-\frac {e^{x/5}}{x}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \frac {e^{x/5} (5-x)-5 \left (25 x^2+x-e^{x/5} (25 x+1)\right ) \log \left (-\frac {e^{x/5}-x}{x}\right )}{\left (e^{x/5}-x\right ) x \log \left (1-\frac {e^{x/5}}{x}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {1}{25} \int \frac {-\frac {e^{x/5} (x-5)}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}+125 x+5}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{25} \int \left (\frac {x-5}{\left (x-e^{x/5}\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}+\frac {125 \log \left (1-\frac {e^{x/5}}{x}\right ) x-x+5 \log \left (1-\frac {e^{x/5}}{x}\right )+5}{x \log \left (1-\frac {e^{x/5}}{x}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (-\int \frac {1}{\log \left (1-\frac {e^{x/5}}{x}\right )}dx+5 \int \frac {1}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}dx+5 \int \frac {1}{x \log \left (1-\frac {e^{x/5}}{x}\right )}dx-\int \frac {x}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}dx+125 x+5 \log (x)\right )\) |
Int[(E^(x/5)*(5 - x) + (-5*x - 125*x^2 + E^(x/5)*(5 + 125*x))*Log[(-E^(x/5 ) + x)/x])/((25*E^(x/5)*x - 25*x^2)*Log[(-E^(x/5) + x)/x]),x]
3.20.22.3.1 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]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
norman | \(5 x +\frac {\ln \left (x \right )}{5}-\frac {\ln \left (\ln \left (\frac {-{\mathrm e}^{\frac {x}{5}}+x}{x}\right )\right )}{5}\) | \(25\) |
parallelrisch | \(\frac {\ln \left (x \right )}{5}-\frac {\ln \left (\ln \left (-\frac {{\mathrm e}^{\frac {x}{5}}-x}{x}\right )\right )}{5}+5 x\) | \(26\) |
risch | \(5 x +\frac {\ln \left (x \right )}{5}-\frac {\ln \left (\ln \left (-{\mathrm e}^{\frac {x}{5}}+x \right )+\frac {i \left (-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )}{x}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )}{x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )}{x}\right )}^{2}-\pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )}{x}\right )}^{3}+2 i \ln \left (x \right )\right )}{2}\right )}{5}\) | \(145\) |
int((((125*x+5)*exp(1/5*x)-125*x^2-5*x)*ln((-exp(1/5*x)+x)/x)+(5-x)*exp(1/ 5*x))/(25*x*exp(1/5*x)-25*x^2)/ln((-exp(1/5*x)+x)/x),x,method=_RETURNVERBO SE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 \, x + \frac {1}{5} \, \log \left (x\right ) - \frac {1}{5} \, \log \left (\log \left (\frac {x - e^{\left (\frac {1}{5} \, x\right )}}{x}\right )\right ) \]
integrate((((125*x+5)*exp(1/5*x)-125*x^2-5*x)*log((-exp(1/5*x)+x)/x)+(5-x) *exp(1/5*x))/(25*x*exp(1/5*x)-25*x^2)/log((-exp(1/5*x)+x)/x),x, algorithm= \
Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 x + \frac {\log {\left (x \right )}}{5} - \frac {\log {\left (\log {\left (\frac {x - e^{\frac {x}{5}}}{x} \right )} \right )}}{5} \]
integrate((((125*x+5)*exp(1/5*x)-125*x**2-5*x)*ln((-exp(1/5*x)+x)/x)+(5-x) *exp(1/5*x))/(25*x*exp(1/5*x)-25*x**2)/ln((-exp(1/5*x)+x)/x),x)
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 \, x + \frac {1}{5} \, \log \left (x\right ) - \frac {1}{5} \, \log \left (\log \left (x - e^{\left (\frac {1}{5} \, x\right )}\right ) - \log \left (x\right )\right ) \]
integrate((((125*x+5)*exp(1/5*x)-125*x^2-5*x)*log((-exp(1/5*x)+x)/x)+(5-x) *exp(1/5*x))/(25*x*exp(1/5*x)-25*x^2)/log((-exp(1/5*x)+x)/x),x, algorithm= \
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 \, x + \frac {1}{5} \, \log \left (x\right ) - \frac {1}{5} \, \log \left (\log \left (\frac {x - e^{\left (\frac {1}{5} \, x\right )}}{x}\right )\right ) \]
integrate((((125*x+5)*exp(1/5*x)-125*x^2-5*x)*log((-exp(1/5*x)+x)/x)+(5-x) *exp(1/5*x))/(25*x*exp(1/5*x)-25*x^2)/log((-exp(1/5*x)+x)/x),x, algorithm= \
Time = 11.67 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5\,x-\frac {\ln \left (\ln \left (\frac {x-{\left ({\mathrm {e}}^x\right )}^{1/5}}{x}\right )\right )}{5}+\frac {\ln \left (x\right )}{5} \]