\(\int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+(2-8 x+2 x^2-8 x^3) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx\) [8901]

Optimal result

Integrand size = 91, antiderivative size = 17 \[ \int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx=\frac {x}{-1-x^2+\log (1-4 x)} \]

[Out]

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

Rubi [F]

\[ \int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx=\int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx \]

[In]

Int[(1 - 8*x - x^2 + 4*x^3 + (-1 + 4*x)*Log[1 - 4*x])/(-1 + 4*x - 2*x^2 + 8*x^3 - x^4 + 4*x^5 + (2 - 8*x + 2*x
^2 - 8*x^3)*Log[1 - 4*x] + (-1 + 4*x)*Log[1 - 4*x]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+8 x+x^2-4 x^3-(-1+4 x) \log (1-4 x)}{(1-4 x) \left (1+x^2-\log (1-4 x)\right )^2} \, dx \\ & = \int \left (\frac {2 x \left (-2-x+4 x^2\right )}{(-1+4 x) \left (1+x^2-\log (1-4 x)\right )^2}+\frac {1}{-1-x^2+\log (1-4 x)}\right ) \, dx \\ & = 2 \int \frac {x \left (-2-x+4 x^2\right )}{(-1+4 x) \left (1+x^2-\log (1-4 x)\right )^2} \, dx+\int \frac {1}{-1-x^2+\log (1-4 x)} \, dx \\ & = 2 \int \left (-\frac {1}{2 \left (1+x^2-\log (1-4 x)\right )^2}+\frac {x^2}{\left (1+x^2-\log (1-4 x)\right )^2}-\frac {1}{2 (-1+4 x) \left (1+x^2-\log (1-4 x)\right )^2}\right ) \, dx+\int \frac {1}{-1-x^2+\log (1-4 x)} \, dx \\ & = 2 \int \frac {x^2}{\left (1+x^2-\log (1-4 x)\right )^2} \, dx-\int \frac {1}{\left (1+x^2-\log (1-4 x)\right )^2} \, dx-\int \frac {1}{(-1+4 x) \left (1+x^2-\log (1-4 x)\right )^2} \, dx+\int \frac {1}{-1-x^2+\log (1-4 x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx=\frac {x}{-1-x^2+\log (1-4 x)} \]

[In]

Integrate[(1 - 8*x - x^2 + 4*x^3 + (-1 + 4*x)*Log[1 - 4*x])/(-1 + 4*x - 2*x^2 + 8*x^3 - x^4 + 4*x^5 + (2 - 8*x
+ 2*x^2 - 8*x^3)*Log[1 - 4*x] + (-1 + 4*x)*Log[1 - 4*x]^2),x]

[Out]

x/(-1 - x^2 + Log[1 - 4*x])

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12

method	result	size
norman	\(-\frac {x}{x^{2}-\ln \left (-4 x +1\right )+1}\)	\(19\)
risch	\(-\frac {x}{x^{2}-\ln \left (-4 x +1\right )+1}\)	\(19\)
parallelrisch	\(-\frac {x}{x^{2}-\ln \left (-4 x +1\right )+1}\)	\(19\)
derivativedivides	\(\frac {16 x}{-\left (-4 x +1\right )^{2}+16 \ln \left (-4 x +1\right )-8 x -15}\)	\(28\)
default	\(\frac {16 x}{-\left (-4 x +1\right )^{2}+16 \ln \left (-4 x +1\right )-8 x -15}\)	\(28\)

[In]

int(((-1+4*x)*ln(-4*x+1)+4*x^3-x^2-8*x+1)/((-1+4*x)*ln(-4*x+1)^2+(-8*x^3+2*x^2-8*x+2)*ln(-4*x+1)+4*x^5-x^4+8*x
^3-2*x^2+4*x-1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx=-\frac {x}{x^{2} - \log \left (-4 \, x + 1\right ) + 1} \]

[In]

integrate(((-1+4*x)*log(-4*x+1)+4*x^3-x^2-8*x+1)/((-1+4*x)*log(-4*x+1)^2+(-8*x^3+2*x^2-8*x+2)*log(-4*x+1)+4*x^
5-x^4+8*x^3-2*x^2+4*x-1),x, algorithm="fricas")

[Out]

-x/(x^2 - log(-4*x + 1) + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx=\frac {x}{- x^{2} + \log {\left (1 - 4 x \right )} - 1} \]

[In]

integrate(((-1+4*x)*ln(-4*x+1)+4*x**3-x**2-8*x+1)/((-1+4*x)*ln(-4*x+1)**2+(-8*x**3+2*x**2-8*x+2)*ln(-4*x+1)+4*
x**5-x**4+8*x**3-2*x**2+4*x-1),x)

[Out]

x/(-x**2 + log(1 - 4*x) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx=-\frac {x}{x^{2} - \log \left (-4 \, x + 1\right ) + 1} \]

[In]

integrate(((-1+4*x)*log(-4*x+1)+4*x^3-x^2-8*x+1)/((-1+4*x)*log(-4*x+1)^2+(-8*x^3+2*x^2-8*x+2)*log(-4*x+1)+4*x^
5-x^4+8*x^3-2*x^2+4*x-1),x, algorithm="maxima")

[Out]

-x/(x^2 - log(-4*x + 1) + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx=-\frac {x}{x^{2} - \log \left (-4 \, x + 1\right ) + 1} \]

[In]

integrate(((-1+4*x)*log(-4*x+1)+4*x^3-x^2-8*x+1)/((-1+4*x)*log(-4*x+1)^2+(-8*x^3+2*x^2-8*x+2)*log(-4*x+1)+4*x^
5-x^4+8*x^3-2*x^2+4*x-1),x, algorithm="giac")

[Out]

-x/(x^2 - log(-4*x + 1) + 1)

Mupad [B] (verification not implemented)

Time = 13.88 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {1-8 x-x^2+4 x^3+(-1+4 x) \log (1-4 x)}{-1+4 x-2 x^2+8 x^3-x^4+4 x^5+\left (2-8 x+2 x^2-8 x^3\right ) \log (1-4 x)+(-1+4 x) \log ^2(1-4 x)} \, dx=-\frac {x}{x^2-\ln \left (1-4\,x\right )+1} \]

[In]

int((log(1 - 4*x)*(4*x - 1) - 8*x - x^2 + 4*x^3 + 1)/(4*x + log(1 - 4*x)^2*(4*x - 1) - log(1 - 4*x)*(8*x - 2*x
^2 + 8*x^3 - 2) - 2*x^2 + 8*x^3 - x^4 + 4*x^5 - 1),x)

[Out]

-x/(x^2 - log(1 - 4*x) + 1)