\(\int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx\) [6046]

Optimal result

Integrand size = 29, antiderivative size = 14 \[ \int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=4-\frac {x}{5-x+\log (x)} \]

[Out]

4-x/(ln(x)-x+5)

Rubi [F]

\[ \int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=\int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx \]

[In]

Int[(-4 - Log[x])/(25 - 10*x + x^2 + (10 - 2*x)*Log[x] + Log[x]^2),x]

[Out]

Defer[Int][(-5 + x - Log[x])^(-2), x] - Defer[Int][x/(-5 + x - Log[x])^2, x] + Defer[Int][(-5 + x - Log[x])^(-
1), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4-\log (x)}{(5-x+\log (x))^2} \, dx \\ & = \int \left (\frac {1-x}{(-5+x-\log (x))^2}+\frac {1}{-5+x-\log (x)}\right ) \, dx \\ & = \int \frac {1-x}{(-5+x-\log (x))^2} \, dx+\int \frac {1}{-5+x-\log (x)} \, dx \\ & = \int \left (\frac {1}{(-5+x-\log (x))^2}-\frac {x}{(-5+x-\log (x))^2}\right ) \, dx+\int \frac {1}{-5+x-\log (x)} \, dx \\ & = \int \frac {1}{(-5+x-\log (x))^2} \, dx-\int \frac {x}{(-5+x-\log (x))^2} \, dx+\int \frac {1}{-5+x-\log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=-\frac {x}{5-x+\log (x)} \]

[In]

Integrate[(-4 - Log[x])/(25 - 10*x + x^2 + (10 - 2*x)*Log[x] + Log[x]^2),x]

[Out]

-(x/(5 - x + Log[x]))

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {x}{-\ln \left (x \right )+x -5}\)	\(12\)
parallelrisch	\(\frac {x}{-\ln \left (x \right )+x -5}\)	\(12\)
default	\(-\frac {x}{\ln \left (x \right )-x +5}\)	\(13\)
norman	\(\frac {5+\ln \left (x \right )}{-\ln \left (x \right )+x -5}\)	\(15\)

[In]

int((-ln(x)-4)/(ln(x)^2+(-2*x+10)*ln(x)+x^2-10*x+25),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=\frac {x}{x - \log \left (x\right ) - 5} \]

[In]

integrate((-log(x)-4)/(log(x)^2+(-2*x+10)*log(x)+x^2-10*x+25),x, algorithm="fricas")

[Out]

x/(x - log(x) - 5)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=- \frac {x}{- x + \log {\left (x \right )} + 5} \]

[In]

integrate((-ln(x)-4)/(ln(x)**2+(-2*x+10)*ln(x)+x**2-10*x+25),x)

[Out]

-x/(-x + log(x) + 5)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=\frac {x}{x - \log \left (x\right ) - 5} \]

[In]

integrate((-log(x)-4)/(log(x)^2+(-2*x+10)*log(x)+x^2-10*x+25),x, algorithm="maxima")

[Out]

x/(x - log(x) - 5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=\frac {x}{x - \log \left (x\right ) - 5} \]

[In]

integrate((-log(x)-4)/(log(x)^2+(-2*x+10)*log(x)+x^2-10*x+25),x, algorithm="giac")

[Out]

x/(x - log(x) - 5)

Mupad [B] (verification not implemented)

Time = 10.69 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=-\frac {x}{\ln \left (x\right )-x+5} \]

[In]

int(-(log(x) + 4)/(log(x)^2 - 10*x - log(x)*(2*x - 10) + x^2 + 25),x)

[Out]

-x/(log(x) - x + 5)