\(\int \frac {e^{x/5} (5-x)+(-5 x-125 x^2+e^{x/5} (5+125 x)) \log (\frac {-e^{x/5}+x}{x})}{(25 e^{x/5} x-25 x^2) \log (\frac {-e^{x/5}+x}{x})} \, dx\) [4922]

Optimal result

Integrand size = 90, antiderivative size = 32 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 x+\frac {1}{5} \left (\log (3 x)-\log \left (\log \left (\frac {-e^{x/5}+x}{x}\right )\right )\right ) \]

[Out]

5*x-1/5*ln(ln((-exp(1/5*x)+x)/x))+1/5*ln(3*x)

Rubi [F]

\[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=\int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx \]

[In]

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]

[Out]

5*x + Log[x]/5 - Defer[Int][Log[1 - E^(x/5)/x]^(-1), x]/25 + Defer[Int][1/((E^(x/5) - x)*Log[1 - E^(x/5)/x]),
x]/5 + Defer[Int][1/(x*Log[1 - E^(x/5)/x]), x]/5 - Defer[Int][x/((E^(x/5) - x)*Log[1 - E^(x/5)/x]), x]/25

Rubi steps \begin{align*} \text {integral}& = \int \frac {5+125 x-\frac {e^{x/5} (-5+x)}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}}{25 x} \, dx \\ & = \frac {1}{25} \int \frac {5+125 x-\frac {e^{x/5} (-5+x)}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}}{x} \, dx \\ & = \frac {1}{25} \int \left (\frac {-5+x}{\left (-e^{x/5}+x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}+\frac {5-x+5 \log \left (1-\frac {e^{x/5}}{x}\right )+125 x \log \left (1-\frac {e^{x/5}}{x}\right )}{x \log \left (1-\frac {e^{x/5}}{x}\right )}\right ) \, dx \\ & = \frac {1}{25} \int \frac {-5+x}{\left (-e^{x/5}+x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx+\frac {1}{25} \int \frac {5-x+5 \log \left (1-\frac {e^{x/5}}{x}\right )+125 x \log \left (1-\frac {e^{x/5}}{x}\right )}{x \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx \\ & = \frac {1}{25} \int \left (\frac {5}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}-\frac {x}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )}\right ) \, dx+\frac {1}{25} \int \frac {5-x+5 (1+25 x) \log \left (1-\frac {e^{x/5}}{x}\right )}{x \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx \\ & = \frac {1}{25} \int \left (\frac {5 (1+25 x)}{x}+\frac {5-x}{x \log \left (1-\frac {e^{x/5}}{x}\right )}\right ) \, dx-\frac {1}{25} \int \frac {x}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx+\frac {1}{5} \int \frac {1}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx \\ & = \frac {1}{25} \int \frac {5-x}{x \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx-\frac {1}{25} \int \frac {x}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx+\frac {1}{5} \int \frac {1+25 x}{x} \, dx+\frac {1}{5} \int \frac {1}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx \\ & = \frac {1}{25} \int \left (-\frac {1}{\log \left (1-\frac {e^{x/5}}{x}\right )}+\frac {5}{x \log \left (1-\frac {e^{x/5}}{x}\right )}\right ) \, dx-\frac {1}{25} \int \frac {x}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx+\frac {1}{5} \int \left (25+\frac {1}{x}\right ) \, dx+\frac {1}{5} \int \frac {1}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx \\ & = 5 x+\frac {\log (x)}{5}-\frac {1}{25} \int \frac {1}{\log \left (1-\frac {e^{x/5}}{x}\right )} \, dx-\frac {1}{25} \int \frac {x}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx+\frac {1}{5} \int \frac {1}{\left (e^{x/5}-x\right ) \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx+\frac {1}{5} \int \frac {1}{x \log \left (1-\frac {e^{x/5}}{x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=\frac {1}{25} \left (125 x+5 \log (x)-5 \log \left (\log \left (1-\frac {e^{x/5}}{x}\right )\right )\right ) \]

[In]

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]

[Out]

(125*x + 5*Log[x] - 5*Log[Log[1 - E^(x/5)/x]])/25

Maple [A] (verified)

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	\(5 x +\frac {\ln \left (x \right )}{5}-\frac {\ln \left (\ln \left (-{\mathrm e}^{\frac {x}{5}}+x \right )+\frac {i \left (-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )}{x}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )}{x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )}{x}\right )}^{2}-\pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{\frac {x}{5}}+x \right )}{x}\right )}^{3}+2 i \ln \left (x \right )\right )}{2}\right )}{5}\)	\(145\)

[In]

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=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 \, x + \frac {1}{5} \, \log \left (x\right ) - \frac {1}{5} \, \log \left (\log \left (\frac {x - e^{\left (\frac {1}{5} \, x\right )}}{x}\right )\right ) \]

[In]

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="fricas")

[Out]

5*x + 1/5*log(x) - 1/5*log(log((x - e^(1/5*x))/x))

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 x + \frac {\log {\left (x \right )}}{5} - \frac {\log {\left (\log {\left (\frac {x - e^{\frac {x}{5}}}{x} \right )} \right )}}{5} \]

[In]

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)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 \, x + \frac {1}{5} \, \log \left (x\right ) - \frac {1}{5} \, \log \left (\log \left (x - e^{\left (\frac {1}{5} \, x\right )}\right ) - \log \left (x\right )\right ) \]

[In]

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="maxima")

[Out]

5*x + 1/5*log(x) - 1/5*log(log(x - e^(1/5*x)) - log(x))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5 \, x + \frac {1}{5} \, \log \left (x\right ) - \frac {1}{5} \, \log \left (\log \left (\frac {x - e^{\left (\frac {1}{5} \, x\right )}}{x}\right )\right ) \]

[In]

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="giac")

[Out]

5*x + 1/5*log(x) - 1/5*log(log((x - e^(1/5*x))/x))

Mupad [B] (verification not implemented)

Time = 11.67 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{x/5} (5-x)+\left (-5 x-125 x^2+e^{x/5} (5+125 x)\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )}{\left (25 e^{x/5} x-25 x^2\right ) \log \left (\frac {-e^{x/5}+x}{x}\right )} \, dx=5\,x-\frac {\ln \left (\ln \left (\frac {x-{\left ({\mathrm {e}}^x\right )}^{1/5}}{x}\right )\right )}{5}+\frac {\ln \left (x\right )}{5} \]

[In]

int(-(exp(x/5)*(x - 5) + log((x - exp(x/5))/x)*(5*x - exp(x/5)*(125*x + 5) + 125*x^2))/(log((x - exp(x/5))/x)*
(25*x*exp(x/5) - 25*x^2)),x)

[Out]

5*x - log(log((x - exp(x)^(1/5))/x))/5 + log(x)/5