\(\int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx\) [9456]

Optimal result

Integrand size = 42, antiderivative size = 17 \[ \int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx=1+\frac {4 x}{2-\log (4)+x \log (x)} \]

[Out]

1+4*x/(2+x*ln(x)-2*ln(2))

Rubi [F]

\[ \int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx=\int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx \]

[In]

Int[(8 - 4*x - 4*Log[4])/(4 - 4*Log[4] + Log[4]^2 + (4*x - 2*x*Log[4])*Log[x] + x^2*Log[x]^2),x]

[Out]

8*(1 - Log[2])*Defer[Int][(2*(1 - Log[2]) + x*Log[x])^(-2), x] - 4*Defer[Int][x/(2*(1 - Log[2]) + x*Log[x])^2,
x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 (2-x-\log (4))}{(2 (1-\log (2))+x \log (x))^2} \, dx \\ & = 4 \int \frac {2-x-\log (4)}{(2 (1-\log (2))+x \log (x))^2} \, dx \\ & = 4 \int \left (-\frac {x}{(2 (1-\log (2))+x \log (x))^2}+\frac {2 (1-\log (2))}{(2 (1-\log (2))+x \log (x))^2}\right ) \, dx \\ & = -\left (4 \int \frac {x}{(2 (1-\log (2))+x \log (x))^2} \, dx\right )+(8 (1-\log (2))) \int \frac {1}{(2 (1-\log (2))+x \log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx=\frac {4 x}{2-\log (4)+x \log (x)} \]

[In]

Integrate[(8 - 4*x - 4*Log[4])/(4 - 4*Log[4] + Log[4]^2 + (4*x - 2*x*Log[4])*Log[x] + x^2*Log[x]^2),x]

[Out]

(4*x)/(2 - Log[4] + x*Log[x])

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {4 x}{-x \ln \left (x \right )+2 \ln \left (2\right )-2}\)	\(17\)
norman	\(-\frac {4 x}{-x \ln \left (x \right )+2 \ln \left (2\right )-2}\)	\(17\)
risch	\(-\frac {4 x}{-x \ln \left (x \right )+2 \ln \left (2\right )-2}\)	\(17\)
parallelrisch	\(-\frac {4 x}{-x \ln \left (x \right )+2 \ln \left (2\right )-2}\)	\(17\)

[In]

int((-8*ln(2)-4*x+8)/(x^2*ln(x)^2+(-4*x*ln(2)+4*x)*ln(x)+4*ln(2)^2-8*ln(2)+4),x,method=_RETURNVERBOSE)

[Out]

-4*x/(-x*ln(x)+2*ln(2)-2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx=\frac {4 \, x}{x \log \left (x\right ) - 2 \, \log \left (2\right ) + 2} \]

[In]

integrate((-8*log(2)-4*x+8)/(x^2*log(x)^2+(-4*x*log(2)+4*x)*log(x)+4*log(2)^2-8*log(2)+4),x, algorithm="fricas
")

[Out]

4*x/(x*log(x) - 2*log(2) + 2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx=\frac {4 x}{x \log {\left (x \right )} - 2 \log {\left (2 \right )} + 2} \]

[In]

integrate((-8*ln(2)-4*x+8)/(x**2*ln(x)**2+(-4*x*ln(2)+4*x)*ln(x)+4*ln(2)**2-8*ln(2)+4),x)

[Out]

4*x/(x*log(x) - 2*log(2) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx=\frac {4 \, x}{x \log \left (x\right ) - 2 \, \log \left (2\right ) + 2} \]

[In]

integrate((-8*log(2)-4*x+8)/(x^2*log(x)^2+(-4*x*log(2)+4*x)*log(x)+4*log(2)^2-8*log(2)+4),x, algorithm="maxima
")

[Out]

4*x/(x*log(x) - 2*log(2) + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx=\frac {4 \, x}{x \log \left (x\right ) - 2 \, \log \left (2\right ) + 2} \]

[In]

integrate((-8*log(2)-4*x+8)/(x^2*log(x)^2+(-4*x*log(2)+4*x)*log(x)+4*log(2)^2-8*log(2)+4),x, algorithm="giac")

[Out]

4*x/(x*log(x) - 2*log(2) + 2)

Mupad [B] (verification not implemented)

Time = 14.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx=\frac {4\,x}{x\,\ln \left (x\right )-\ln \left (4\right )+2} \]

[In]

int(-(4*x + 8*log(2) - 8)/(x^2*log(x)^2 - 8*log(2) + log(x)*(4*x - 4*x*log(2)) + 4*log(2)^2 + 4),x)

[Out]

(4*x)/(x*log(x) - log(4) + 2)