\(\int \frac {1-20 x+12 x^3+(19-12 x^2) \log (x)+(x-\log (x)) \log (-\frac {e^{4+e^4}}{-x+\log (x)})}{-x+\log (x)} \, dx\) [7494]

Optimal result

Integrand size = 54, antiderivative size = 28 \[ \int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx=x \left (19-4 x^2-\log \left (\frac {e^{4+e^4}}{x-\log (x)}\right )\right ) \]

[Out]

x*(19-4*x^2-ln(exp(4+exp(4))/(x-ln(x))))

Rubi [F]

\[ \int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx=\int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx \]

[In]

Int[(1 - 20*x + 12*x^3 + (19 - 12*x^2)*Log[x] + (x - Log[x])*Log[-(E^(4 + E^4)/(-x + Log[x]))])/(-x + Log[x]),
x]

[Out]

19*x - (4 + E^4)*x - 4*x^3 + Defer[Int][x/(x - Log[x]), x] + Defer[Int][(-x + Log[x])^(-1), x] - Defer[Int][Lo
g[(x - Log[x])^(-1)], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-4 \left (1+\frac {e^4}{4}\right )+\frac {-1+20 x-12 x^3-19 \log (x)+12 x^2 \log (x)}{x-\log (x)}-\log \left (\frac {1}{x-\log (x)}\right )\right ) \, dx \\ & = -\left (\left (4+e^4\right ) x\right )+\int \frac {-1+20 x-12 x^3-19 \log (x)+12 x^2 \log (x)}{x-\log (x)} \, dx-\int \log \left (\frac {1}{x-\log (x)}\right ) \, dx \\ & = -\left (\left (4+e^4\right ) x\right )+\int \left (19-12 x^2+\frac {-1+x}{x-\log (x)}\right ) \, dx-\int \log \left (\frac {1}{x-\log (x)}\right ) \, dx \\ & = 19 x-\left (4+e^4\right ) x-4 x^3+\int \frac {-1+x}{x-\log (x)} \, dx-\int \log \left (\frac {1}{x-\log (x)}\right ) \, dx \\ & = 19 x-\left (4+e^4\right ) x-4 x^3+\int \left (\frac {x}{x-\log (x)}+\frac {1}{-x+\log (x)}\right ) \, dx-\int \log \left (\frac {1}{x-\log (x)}\right ) \, dx \\ & = 19 x-\left (4+e^4\right ) x-4 x^3+\int \frac {x}{x-\log (x)} \, dx+\int \frac {1}{-x+\log (x)} \, dx-\int \log \left (\frac {1}{x-\log (x)}\right ) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx=-x \left (-15+e^4+4 x^2+\log \left (\frac {1}{x-\log (x)}\right )\right ) \]

[In]

Integrate[(1 - 20*x + 12*x^3 + (19 - 12*x^2)*Log[x] + (x - Log[x])*Log[-(E^(4 + E^4)/(-x + Log[x]))])/(-x + Lo
g[x]),x]

[Out]

-(x*(-15 + E^4 + 4*x^2 + Log[(x - Log[x])^(-1)]))

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
risch	\(x \ln \left (x -\ln \left (x \right )\right )-\frac {x \left (-30+8 x^{2}+2 \,{\mathrm e}^{4}\right )}{2}\)	\(25\)
default	\(-4 x^{3}-x \,{\mathrm e}^{4}+15 x -x \ln \left (\frac {1}{x -\ln \left (x \right )}\right )\)	\(27\)
parallelrisch	\(-4 x^{3}-\ln \left (-\frac {{\mathrm e}^{4+{\mathrm e}^{4}}}{\ln \left (x \right )-x}\right ) x +19 x\)	\(29\)

[In]

int(((x-ln(x))*ln(-exp(4+exp(4))/(ln(x)-x))+(-12*x^2+19)*ln(x)+12*x^3-20*x+1)/(ln(x)-x),x,method=_RETURNVERBOS
E)

[Out]

x*ln(x-ln(x))-1/2*x*(-30+8*x^2+2*exp(4))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx=-4 \, x^{3} - x \log \left (\frac {e^{\left (e^{4} + 4\right )}}{x - \log \left (x\right )}\right ) + 19 \, x \]

[In]

integrate(((x-log(x))*log(-exp(4+exp(4))/(log(x)-x))+(-12*x^2+19)*log(x)+12*x^3-20*x+1)/(log(x)-x),x, algorith
m="fricas")

[Out]

-4*x^3 - x*log(e^(e^4 + 4)/(x - log(x))) + 19*x

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx=- 4 x^{3} - x \log {\left (- \frac {e^{4 + e^{4}}}{- x + \log {\left (x \right )}} \right )} + 19 x \]

[In]

integrate(((x-ln(x))*ln(-exp(4+exp(4))/(ln(x)-x))+(-12*x**2+19)*ln(x)+12*x**3-20*x+1)/(ln(x)-x),x)

[Out]

-4*x**3 - x*log(-exp(4 + exp(4))/(-x + log(x))) + 19*x

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx=-4 \, x^{3} - x {\left (e^{4} - 15\right )} + x \log \left (x - \log \left (x\right )\right ) \]

[In]

integrate(((x-log(x))*log(-exp(4+exp(4))/(log(x)-x))+(-12*x^2+19)*log(x)+12*x^3-20*x+1)/(log(x)-x),x, algorith
m="maxima")

[Out]

-4*x^3 - x*(e^4 - 15) + x*log(x - log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx=-4 \, x^{3} - x e^{4} + x \log \left (x - \log \left (x\right )\right ) + 15 \, x \]

[In]

integrate(((x-log(x))*log(-exp(4+exp(4))/(log(x)-x))+(-12*x^2+19)*log(x)+12*x^3-20*x+1)/(log(x)-x),x, algorith
m="giac")

[Out]

-4*x^3 - x*e^4 + x*log(x - log(x)) + 15*x

Mupad [B] (verification not implemented)

Time = 13.61 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1-20 x+12 x^3+\left (19-12 x^2\right ) \log (x)+(x-\log (x)) \log \left (-\frac {e^{4+e^4}}{-x+\log (x)}\right )}{-x+\log (x)} \, dx=15\,x-x\,{\mathrm {e}}^4-x\,\ln \left (\frac {1}{x-\ln \left (x\right )}\right )-4\,x^3 \]

[In]

int(-(12*x^3 - 20*x - log(x)*(12*x^2 - 19) + log(exp(exp(4) + 4)/(x - log(x)))*(x - log(x)) + 1)/(x - log(x)),
x)

[Out]

15*x - x*exp(4) - x*log(1/(x - log(x))) - 4*x^3