\(\int \frac {e^{-1+x} (15625-15625 x)+(-31250 e^{-1+x}-31250 x) \log (\frac {x}{e^{-1+x}+x})}{e^{-1+x} x^3+x^4} \, dx\) [51]

Optimal result

Integrand size = 52, antiderivative size = 17 \[ \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx=\frac {15625 \log \left (\frac {x}{e^{-1+x}+x}\right )}{x^2} \]

[Out]

15625*ln(x/(exp(-1+x)+x))/x^2

Rubi [F]

\[ \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx=\int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx \]

[In]

Int[(E^(-1 + x)*(15625 - 15625*x) + (-31250*E^(-1 + x) - 31250*x)*Log[x/(E^(-1 + x) + x)])/(E^(-1 + x)*x^3 + x
^4),x]

[Out]

(15625*(1 + x)^2)/(2*x^2) + (15625*Log[x/(E^x + E*x)])/x^2 - 15625*Defer[Int][E^x/(x^3*(E^x + E*x)), x] - 1562
5*E*Defer[Int][1/(x^2*(E^x + E*x)), x] + 15625*Defer[Int][E^x/(x^2*(E^x + E*x)), x] + 15625*E*Defer[Int][1/(x*
(E^x + E*x)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e \left (e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )\right )}{x^3 \left (e^x+e x\right )} \, dx \\ & = e \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{x^3 \left (e^x+e x\right )} \, dx \\ & = e \int \left (\frac {15625 (-1+x)}{x^2 \left (e^x+e x\right )}-\frac {15625 \left (1+x+2 \log \left (\frac {x}{e^x+e x}\right )\right )}{e x^3}\right ) \, dx \\ & = -\left (15625 \int \frac {1+x+2 \log \left (\frac {x}{e^x+e x}\right )}{x^3} \, dx\right )+(15625 e) \int \frac {-1+x}{x^2 \left (e^x+e x\right )} \, dx \\ & = -\left (15625 \int \left (\frac {1+x}{x^3}+\frac {2 \log \left (\frac {x}{e^x+e x}\right )}{x^3}\right ) \, dx\right )+(15625 e) \int \left (-\frac {1}{x^2 \left (e^x+e x\right )}+\frac {1}{x \left (e^x+e x\right )}\right ) \, dx \\ & = -\left (15625 \int \frac {1+x}{x^3} \, dx\right )-31250 \int \frac {\log \left (\frac {x}{e^x+e x}\right )}{x^3} \, dx-(15625 e) \int \frac {1}{x^2 \left (e^x+e x\right )} \, dx+(15625 e) \int \frac {1}{x \left (e^x+e x\right )} \, dx \\ & = \frac {15625 (1+x)^2}{2 x^2}+\frac {15625 \log \left (\frac {x}{e^x+e x}\right )}{x^2}-15625 \int \frac {e^x (1-x)}{x^3 \left (e^x+e x\right )} \, dx-(15625 e) \int \frac {1}{x^2 \left (e^x+e x\right )} \, dx+(15625 e) \int \frac {1}{x \left (e^x+e x\right )} \, dx \\ & = \frac {15625 (1+x)^2}{2 x^2}+\frac {15625 \log \left (\frac {x}{e^x+e x}\right )}{x^2}-15625 \int \left (\frac {e^x}{x^3 \left (e^x+e x\right )}-\frac {e^x}{x^2 \left (e^x+e x\right )}\right ) \, dx-(15625 e) \int \frac {1}{x^2 \left (e^x+e x\right )} \, dx+(15625 e) \int \frac {1}{x \left (e^x+e x\right )} \, dx \\ & = \frac {15625 (1+x)^2}{2 x^2}+\frac {15625 \log \left (\frac {x}{e^x+e x}\right )}{x^2}-15625 \int \frac {e^x}{x^3 \left (e^x+e x\right )} \, dx+15625 \int \frac {e^x}{x^2 \left (e^x+e x\right )} \, dx-(15625 e) \int \frac {1}{x^2 \left (e^x+e x\right )} \, dx+(15625 e) \int \frac {1}{x \left (e^x+e x\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx=-15625 \left (-\frac {1}{x^2}-\frac {\log \left (\frac {x}{e^x+e x}\right )}{x^2}\right ) \]

[In]

Integrate[(E^(-1 + x)*(15625 - 15625*x) + (-31250*E^(-1 + x) - 31250*x)*Log[x/(E^(-1 + x) + x)])/(E^(-1 + x)*x
^3 + x^4),x]

[Out]

-15625*(-x^(-2) - Log[x/(E^x + E*x)]/x^2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {15625 \ln \left (\frac {x}{{\mathrm e}^{-1+x}+x}\right )}{x^{2}}\)	\(17\)
parallelrisch	\(\frac {15625 \ln \left (\frac {x}{{\mathrm e}^{-1+x}+x}\right )}{x^{2}}\)	\(17\)
risch	\(-\frac {15625 \ln \left ({\mathrm e}^{-1+x}+x \right )}{x^{2}}+\frac {-\frac {15625 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{-1+x}+x}\right ) \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{-1+x}+x}\right )}{2}+\frac {15625 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{-1+x}+x}\right )^{2}}{2}+\frac {15625 i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{-1+x}+x}\right ) \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{-1+x}+x}\right )^{2}}{2}-\frac {15625 i \pi \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{-1+x}+x}\right )^{3}}{2}+15625 \ln \left (x \right )}{x^{2}}\)	\(132\)

[In]

int(((-31250*exp(-1+x)-31250*x)*ln(x/(exp(-1+x)+x))+(-15625*x+15625)*exp(-1+x))/(x^3*exp(-1+x)+x^4),x,method=_
RETURNVERBOSE)

[Out]

15625*ln(x/(exp(-1+x)+x))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx=\frac {15625 \, \log \left (\frac {x}{x + e^{\left (x - 1\right )}}\right )}{x^{2}} \]

[In]

integrate(((-31250*exp(-1+x)-31250*x)*log(x/(exp(-1+x)+x))+(-15625*x+15625)*exp(-1+x))/(x^3*exp(-1+x)+x^4),x,
algorithm="fricas")

[Out]

15625*log(x/(x + e^(x - 1)))/x^2

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx=\frac {15625 \log {\left (\frac {x}{x + e^{x - 1}} \right )}}{x^{2}} \]

[In]

integrate(((-31250*exp(-1+x)-31250*x)*ln(x/(exp(-1+x)+x))+(-15625*x+15625)*exp(-1+x))/(x**3*exp(-1+x)+x**4),x)

[Out]

15625*log(x/(x + exp(x - 1)))/x**2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx=-\frac {15625 \, {\left (\log \left (x e + e^{x}\right ) - \log \left (x\right ) - 1\right )}}{x^{2}} \]

[In]

integrate(((-31250*exp(-1+x)-31250*x)*log(x/(exp(-1+x)+x))+(-15625*x+15625)*exp(-1+x))/(x^3*exp(-1+x)+x^4),x,
algorithm="maxima")

[Out]

-15625*(log(x*e + e^x) - log(x) - 1)/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx=\frac {15625 \, {\left (\log \left (\frac {x}{x e + e^{x}}\right ) + 1\right )}}{x^{2}} \]

[In]

integrate(((-31250*exp(-1+x)-31250*x)*log(x/(exp(-1+x)+x))+(-15625*x+15625)*exp(-1+x))/(x^3*exp(-1+x)+x^4),x,
algorithm="giac")

[Out]

15625*(log(x/(x*e + e^x)) + 1)/x^2

Mupad [B] (verification not implemented)

Time = 8.98 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-1+x} (15625-15625 x)+\left (-31250 e^{-1+x}-31250 x\right ) \log \left (\frac {x}{e^{-1+x}+x}\right )}{e^{-1+x} x^3+x^4} \, dx=\frac {15625\,\left (\ln \left (\frac {x}{{\mathrm {e}}^x+x\,\mathrm {e}}\right )+1\right )}{x^2} \]

[In]

int(-(exp(x - 1)*(15625*x - 15625) + log(x/(x + exp(x - 1)))*(31250*x + 31250*exp(x - 1)))/(x^3*exp(x - 1) + x
^4),x)

[Out]

(15625*(log(x/(exp(x) + x*exp(1))) + 1))/x^2