\(\int \frac {-1+x+e^x x}{-e^x x-x^2+x \log (\frac {103}{5})+x \log (x)} \, dx\) [4107]

Optimal result

Integrand size = 33, antiderivative size = 21 \[ \int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx=\log (5)+\log \left (\frac {1}{-e^x-x+\log \left (\frac {103}{5}\right )+\log (x)}\right ) \]

[Out]

ln(5)+ln(1/(ln(103/5)-exp(x)-x+ln(x)))

Rubi [F]

\[ \int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx=\int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx \]

[In]

Int[(-1 + x + E^x*x)/(-(E^x*x) - x^2 + x*Log[103/5] + x*Log[x]),x]

[Out]

-x + Defer[Int][1/(x*(E^x + x - Log[(103*x)/5])), x] + Defer[Int][x/(E^x + x - Log[(103*x)/5]), x] - 5*(1 + Lo
g[103/5])*Defer[Subst][Defer[Int][(E^(5*x) + 5*x - Log[103*x])^(-1), x], x, x/5] - 5*Defer[Subst][Defer[Int][L
og[5*x]/(E^(5*x) + 5*x - Log[103*x]), x], x, x/5]

Rubi steps \begin{align*} \text {integral}& = \int \left (-1+\frac {1+x^2-x \left (1+\log \left (\frac {103}{5}\right )\right )-x \log (x)}{x \left (e^x+x-\log \left (\frac {103 x}{5}\right )\right )}\right ) \, dx \\ & = -x+\int \frac {1+x^2-x \left (1+\log \left (\frac {103}{5}\right )\right )-x \log (x)}{x \left (e^x+x-\log \left (\frac {103 x}{5}\right )\right )} \, dx \\ & = -x+\int \left (\frac {1}{x \left (e^x+x-\log \left (\frac {103 x}{5}\right )\right )}+\frac {x}{e^x+x-\log \left (\frac {103 x}{5}\right )}-\frac {1+\log \left (\frac {103}{5}\right )}{e^x+x-\log \left (\frac {103 x}{5}\right )}-\frac {\log (x)}{e^x+x-\log \left (\frac {103 x}{5}\right )}\right ) \, dx \\ & = -x+\left (-1-\log \left (\frac {103}{5}\right )\right ) \int \frac {1}{e^x+x-\log \left (\frac {103 x}{5}\right )} \, dx+\int \frac {1}{x \left (e^x+x-\log \left (\frac {103 x}{5}\right )\right )} \, dx+\int \frac {x}{e^x+x-\log \left (\frac {103 x}{5}\right )} \, dx-\int \frac {\log (x)}{e^x+x-\log \left (\frac {103 x}{5}\right )} \, dx \\ & = -x-5 \text {Subst}\left (\int \frac {\log (5 x)}{e^{5 x}+5 x-\log (103 x)} \, dx,x,\frac {x}{5}\right )-\left (5 \left (1+\log \left (\frac {103}{5}\right )\right )\right ) \text {Subst}\left (\int \frac {1}{e^{5 x}+5 x-\log (103 x)} \, dx,x,\frac {x}{5}\right )+\int \frac {1}{x \left (e^x+x-\log \left (\frac {103 x}{5}\right )\right )} \, dx+\int \frac {x}{e^x+x-\log \left (\frac {103 x}{5}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx=-\log \left (e^x+x-\log \left (\frac {103 x}{5}\right )\right ) \]

[In]

Integrate[(-1 + x + E^x*x)/(-(E^x*x) - x^2 + x*Log[103/5] + x*Log[x]),x]

[Out]

-Log[E^x + x - Log[(103*x)/5]]

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76

method	result	size
norman	\(-\ln \left (\ln \left (\frac {103}{5}\right )-{\mathrm e}^{x}-x +\ln \left (x \right )\right )\)	\(16\)
parallelrisch	\(-\ln \left (-\ln \left (\frac {103}{5}\right )+{\mathrm e}^{x}+x -\ln \left (x \right )\right )\)	\(16\)
risch	\(-\ln \left (\ln \left (x \right )-{\mathrm e}^{x}+\ln \left (103\right )-\ln \left (5\right )-x \right )\)	\(20\)

[In]

int((exp(x)*x+x-1)/(x*ln(x)-exp(x)*x+x*ln(103/5)-x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(103/5)-exp(x)-x+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx=-\log \left (-x - e^{x} + \log \left (\frac {103}{5}\right ) + \log \left (x\right )\right ) \]

[In]

integrate((exp(x)*x+x-1)/(x*log(x)-exp(x)*x+x*log(103/5)-x^2),x, algorithm="fricas")

[Out]

-log(-x - e^x + log(103/5) + log(x))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx=- \log {\left (x + e^{x} - \log {\left (x \right )} - \log {\left (103 \right )} + \log {\left (5 \right )} \right )} \]

[In]

integrate((exp(x)*x+x-1)/(x*ln(x)-exp(x)*x+x*ln(103/5)-x**2),x)

[Out]

-log(x + exp(x) - log(x) - log(103) + log(5))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx=-\log \left (x + e^{x} - \log \left (103\right ) + \log \left (5\right ) - \log \left (x\right )\right ) \]

[In]

integrate((exp(x)*x+x-1)/(x*log(x)-exp(x)*x+x*log(103/5)-x^2),x, algorithm="maxima")

[Out]

-log(x + e^x - log(103) + log(5) - log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx=-\log \left (-x - e^{x} + \log \left (103\right ) - \log \left (5\right ) + \log \left (x\right )\right ) \]

[In]

integrate((exp(x)*x+x-1)/(x*log(x)-exp(x)*x+x*log(103/5)-x^2),x, algorithm="giac")

[Out]

-log(-x - e^x + log(103) - log(5) + log(x))

Mupad [B] (verification not implemented)

Time = 9.96 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {-1+x+e^x x}{-e^x x-x^2+x \log \left (\frac {103}{5}\right )+x \log (x)} \, dx=-\ln \left (x-\ln \left (\frac {103\,x}{5}\right )+{\mathrm {e}}^x\right ) \]

[In]

int((x + x*exp(x) - 1)/(x*log(103/5) - x*exp(x) + x*log(x) - x^2),x)

[Out]

-log(x - log((103*x)/5) + exp(x))