\(\int \frac {32 x^2+24 x^3-8 x^4+(-8 x+7 x^2+3 x^3-2 x^4) \log (1-2 x+x^2) \log (\log ^4(1-2 x+x^2))}{(-5-5 x+5 x^2+5 x^3) \log (1-2 x+x^2)} \, dx\) [9740]

Optimal result

Integrand size = 86, antiderivative size = 28 \[ \int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx=\frac {(4-x) x^2 \log \left (\log ^4\left ((1-x)^2\right )\right )}{5 (1+x)} \]

[Out]

1/5*x^2/(1+x)*(-x+4)*ln(ln((1-x)^2)^4)

Rubi [F]

\[ \int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx=\int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx \]

[In]

Int[(32*x^2 + 24*x^3 - 8*x^4 + (-8*x + 7*x^2 + 3*x^3 - 2*x^4)*Log[1 - 2*x + x^2]*Log[Log[1 - 2*x + x^2]^4])/((
-5 - 5*x + 5*x^2 + 5*x^3)*Log[1 - 2*x + x^2]),x]

[Out]

(8*(1 - x)*ExpIntegralEi[Log[(-1 + x)^2]/2])/(5*Sqrt[(-1 + x)^2]) + (6*Log[Log[(-1 + x)^2]])/5 - (1 - x)*Log[L
og[(-1 + x)^2]^4] - (4*LogIntegral[(-1 + x)^2])/5 - 4*Defer[Int][1/((1 + x)*Log[(-1 + x)^2]), x] - (2*Defer[In
t][x*Log[Log[(-1 + x)^2]^4], x])/5 - Defer[Int][Log[Log[(-1 + x)^2]^4]/(1 + x)^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {32 x^2}{5 (-1+x) (1+x)^2 \log \left ((-1+x)^2\right )}+\frac {24 x^3}{5 (-1+x) (1+x)^2 \log \left ((-1+x)^2\right )}-\frac {8 x^4}{5 (-1+x) (1+x)^2 \log \left ((-1+x)^2\right )}-\frac {x \left (-8-x+2 x^2\right ) \log \left (\log ^4\left ((-1+x)^2\right )\right )}{5 (1+x)^2}\right ) \, dx \\ & = -\left (\frac {1}{5} \int \frac {x \left (-8-x+2 x^2\right ) \log \left (\log ^4\left ((-1+x)^2\right )\right )}{(1+x)^2} \, dx\right )-\frac {8}{5} \int \frac {x^4}{(-1+x) (1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \int \frac {x^3}{(-1+x) (1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {32}{5} \int \frac {x^2}{(-1+x) (1+x)^2 \log \left ((-1+x)^2\right )} \, dx \\ & = -\left (\frac {1}{5} \int \left (-5 \log \left (\log ^4\left ((-1+x)^2\right )\right )+2 x \log \left (\log ^4\left ((-1+x)^2\right )\right )+\frac {5 \log \left (\log ^4\left ((-1+x)^2\right )\right )}{(1+x)^2}\right ) \, dx\right )-\frac {8}{5} \int \left (-\frac {1}{\log \left ((-1+x)^2\right )}+\frac {1}{4 (-1+x) \log \left ((-1+x)^2\right )}+\frac {x}{\log \left ((-1+x)^2\right )}-\frac {1}{2 (1+x)^2 \log \left ((-1+x)^2\right )}+\frac {7}{4 (1+x) \log \left ((-1+x)^2\right )}\right ) \, dx+\frac {24}{5} \int \left (\frac {1}{\log \left ((-1+x)^2\right )}+\frac {1}{4 (-1+x) \log \left ((-1+x)^2\right )}+\frac {1}{2 (1+x)^2 \log \left ((-1+x)^2\right )}-\frac {5}{4 (1+x) \log \left ((-1+x)^2\right )}\right ) \, dx+\frac {32}{5} \int \left (\frac {1}{4 (-1+x) \log \left ((-1+x)^2\right )}-\frac {1}{2 (1+x)^2 \log \left ((-1+x)^2\right )}+\frac {3}{4 (1+x) \log \left ((-1+x)^2\right )}\right ) \, dx \\ & = -\left (\frac {2}{5} \int \frac {1}{(-1+x) \log \left ((-1+x)^2\right )} \, dx\right )-\frac {2}{5} \int x \log \left (\log ^4\left ((-1+x)^2\right )\right ) \, dx+\frac {4}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {6}{5} \int \frac {1}{(-1+x) \log \left ((-1+x)^2\right )} \, dx+\frac {8}{5} \int \frac {1}{\log \left ((-1+x)^2\right )} \, dx+\frac {8}{5} \int \frac {1}{(-1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {8}{5} \int \frac {x}{\log \left ((-1+x)^2\right )} \, dx+\frac {12}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx-\frac {14}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {16}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{\log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-6 \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx+\int \log \left (\log ^4\left ((-1+x)^2\right )\right ) \, dx-\int \frac {\log \left (\log ^4\left ((-1+x)^2\right )\right )}{(1+x)^2} \, dx \\ & = -\left (\frac {2}{5} \int x \log \left (\log ^4\left ((-1+x)^2\right )\right ) \, dx\right )-\frac {2}{5} \text {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,-1+x\right )+\frac {4}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {6}{5} \text {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,-1+x\right )-\frac {8}{5} \int \left (\frac {1}{\log \left ((-1+x)^2\right )}+\frac {-1+x}{\log \left ((-1+x)^2\right )}\right ) \, dx+\frac {8}{5} \text {Subst}\left (\int \frac {1}{\log \left (x^2\right )} \, dx,x,-1+x\right )+\frac {8}{5} \text {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,-1+x\right )+\frac {12}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx-\frac {14}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {16}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \text {Subst}\left (\int \frac {1}{\log \left (x^2\right )} \, dx,x,-1+x\right )-6 \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\int \frac {\log \left (\log ^4\left ((-1+x)^2\right )\right )}{(1+x)^2} \, dx+\text {Subst}\left (\int \log \left (\log ^4\left (x^2\right )\right ) \, dx,x,-1+x\right ) \\ & = (-1+x) \log \left (\log ^4\left ((-1+x)^2\right )\right )-\frac {1}{5} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )-\frac {2}{5} \int x \log \left (\log ^4\left ((-1+x)^2\right )\right ) \, dx+\frac {3}{5} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )+\frac {4}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {4}{5} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )-\frac {8}{5} \int \frac {1}{\log \left ((-1+x)^2\right )} \, dx-\frac {8}{5} \int \frac {-1+x}{\log \left ((-1+x)^2\right )} \, dx+\frac {12}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx-\frac {14}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {16}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-6 \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-8 \text {Subst}\left (\int \frac {1}{\log \left (x^2\right )} \, dx,x,-1+x\right )+\frac {(4 (-1+x)) \text {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )}{5 \sqrt {(-1+x)^2}}+\frac {(12 (-1+x)) \text {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )}{5 \sqrt {(-1+x)^2}}-\int \frac {\log \left (\log ^4\left ((-1+x)^2\right )\right )}{(1+x)^2} \, dx \\ & = -\frac {16 (1-x) \operatorname {ExpIntegralEi}\left (\frac {1}{2} \log \left ((-1+x)^2\right )\right )}{5 \sqrt {(-1+x)^2}}+\frac {6}{5} \log \left (\log \left ((-1+x)^2\right )\right )+(-1+x) \log \left (\log ^4\left ((-1+x)^2\right )\right )-\frac {2}{5} \int x \log \left (\log ^4\left ((-1+x)^2\right )\right ) \, dx+\frac {4}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx-\frac {8}{5} \text {Subst}\left (\int \frac {1}{\log \left (x^2\right )} \, dx,x,-1+x\right )-\frac {8}{5} \text {Subst}\left (\int \frac {x}{\log \left (x^2\right )} \, dx,x,-1+x\right )+\frac {12}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx-\frac {14}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {16}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-6 \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {(4 (-1+x)) \text {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )}{\sqrt {(-1+x)^2}}-\int \frac {\log \left (\log ^4\left ((-1+x)^2\right )\right )}{(1+x)^2} \, dx \\ & = \frac {4 (1-x) \operatorname {ExpIntegralEi}\left (\frac {1}{2} \log \left ((-1+x)^2\right )\right )}{5 \sqrt {(-1+x)^2}}+\frac {6}{5} \log \left (\log \left ((-1+x)^2\right )\right )+(-1+x) \log \left (\log ^4\left ((-1+x)^2\right )\right )-\frac {2}{5} \int x \log \left (\log ^4\left ((-1+x)^2\right )\right ) \, dx+\frac {4}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx-\frac {4}{5} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,(-1+x)^2\right )+\frac {12}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx-\frac {14}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {16}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-6 \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {(4 (-1+x)) \text {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left ((-1+x)^2\right )\right )}{5 \sqrt {(-1+x)^2}}-\int \frac {\log \left (\log ^4\left ((-1+x)^2\right )\right )}{(1+x)^2} \, dx \\ & = \frac {8 (1-x) \operatorname {ExpIntegralEi}\left (\frac {1}{2} \log \left ((-1+x)^2\right )\right )}{5 \sqrt {(-1+x)^2}}+\frac {6}{5} \log \left (\log \left ((-1+x)^2\right )\right )+(-1+x) \log \left (\log ^4\left ((-1+x)^2\right )\right )-\frac {4}{5} \operatorname {LogIntegral}\left ((-1+x)^2\right )-\frac {2}{5} \int x \log \left (\log ^4\left ((-1+x)^2\right )\right ) \, dx+\frac {4}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {12}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx-\frac {14}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\frac {16}{5} \int \frac {1}{(1+x)^2 \log \left ((-1+x)^2\right )} \, dx+\frac {24}{5} \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-6 \int \frac {1}{(1+x) \log \left ((-1+x)^2\right )} \, dx-\int \frac {\log \left (\log ^4\left ((-1+x)^2\right )\right )}{(1+x)^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx=\frac {1}{5} \left (-20 \log \left (\log \left ((-1+x)^2\right )\right )-\frac {\left (-5-5 x-4 x^2+x^3\right ) \log \left (\log ^4\left ((-1+x)^2\right )\right )}{1+x}\right ) \]

[In]

Integrate[(32*x^2 + 24*x^3 - 8*x^4 + (-8*x + 7*x^2 + 3*x^3 - 2*x^4)*Log[1 - 2*x + x^2]*Log[Log[1 - 2*x + x^2]^
4])/((-5 - 5*x + 5*x^2 + 5*x^3)*Log[1 - 2*x + x^2]),x]

[Out]

(-20*Log[Log[(-1 + x)^2]] - ((-5 - 5*x - 4*x^2 + x^3)*Log[Log[(-1 + x)^2]^4])/(1 + x))/5

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(96\) vs. \(2(26)=52\).

Time = 2.65 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.46

method	result	size
parallelrisch	\(\frac {-x^{3} \ln \left (\ln \left (x^{2}-2 x +1\right )^{4}\right )+4 \ln \left (\ln \left (x^{2}-2 x +1\right )^{4}\right ) x^{2}-16 \ln \left (\ln \left (x^{2}-2 x +1\right )\right ) x +4 \ln \left (\ln \left (x^{2}-2 x +1\right )^{4}\right ) x -16 \ln \left (\ln \left (x^{2}-2 x +1\right )\right )+4 \ln \left (\ln \left (x^{2}-2 x +1\right )^{4}\right )}{5 x +5}\)	\(97\)

[In]

int(((-2*x^4+3*x^3+7*x^2-8*x)*ln(x^2-2*x+1)*ln(ln(x^2-2*x+1)^4)-8*x^4+24*x^3+32*x^2)/(5*x^3+5*x^2-5*x-5)/ln(x^
2-2*x+1),x,method=_RETURNVERBOSE)

[Out]

1/5*(-x^3*ln(ln(x^2-2*x+1)^4)+4*ln(ln(x^2-2*x+1)^4)*x^2-16*ln(ln(x^2-2*x+1))*x+4*ln(ln(x^2-2*x+1)^4)*x-16*ln(l
n(x^2-2*x+1))+4*ln(ln(x^2-2*x+1)^4))/(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx=-\frac {{\left (x^{3} - 4 \, x^{2}\right )} \log \left (\log \left (x^{2} - 2 \, x + 1\right )^{4}\right )}{5 \, {\left (x + 1\right )}} \]

[In]

integrate(((-2*x^4+3*x^3+7*x^2-8*x)*log(x^2-2*x+1)*log(log(x^2-2*x+1)^4)-8*x^4+24*x^3+32*x^2)/(5*x^3+5*x^2-5*x
-5)/log(x^2-2*x+1),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx=- 4 \log {\left (\log {\left (x^{2} - 2 x + 1 \right )} \right )} + \frac {\left (- x^{3} + 4 x^{2} + 5 x + 5\right ) \log {\left (\log {\left (x^{2} - 2 x + 1 \right )}^{4} \right )}}{5 x + 5} \]

[In]

integrate(((-2*x**4+3*x**3+7*x**2-8*x)*ln(x**2-2*x+1)*ln(ln(x**2-2*x+1)**4)-8*x**4+24*x**3+32*x**2)/(5*x**3+5*
x**2-5*x-5)/ln(x**2-2*x+1),x)

[Out]

-4*log(log(x**2 - 2*x + 1)) + (-x**3 + 4*x**2 + 5*x + 5)*log(log(x**2 - 2*x + 1)**4)/(5*x + 5)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).

Time = 0.34 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx=-\frac {4 \, {\left (x^{3} \log \left (2\right ) - 4 \, x^{2} \log \left (2\right ) - 5 \, x \log \left (2\right ) + {\left (x^{3} - 4 \, x^{2}\right )} \log \left (\log \left (x - 1\right )\right ) - 5 \, \log \left (2\right )\right )}}{5 \, {\left (x + 1\right )}} \]

[In]

integrate(((-2*x^4+3*x^3+7*x^2-8*x)*log(x^2-2*x+1)*log(log(x^2-2*x+1)^4)-8*x^4+24*x^3+32*x^2)/(5*x^3+5*x^2-5*x
-5)/log(x^2-2*x+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx=-\frac {1}{5} \, {\left (x^{2} - 5 \, x - \frac {5}{x + 1}\right )} \log \left (\log \left (x^{2} - 2 \, x + 1\right )^{4}\right ) - 4 \, \log \left (\log \left (x^{2} - 2 \, x + 1\right )\right ) \]

[In]

integrate(((-2*x^4+3*x^3+7*x^2-8*x)*log(x^2-2*x+1)*log(log(x^2-2*x+1)^4)-8*x^4+24*x^3+32*x^2)/(5*x^3+5*x^2-5*x
-5)/log(x^2-2*x+1),x, algorithm="giac")

[Out]

-1/5*(x^2 - 5*x - 5/(x + 1))*log(log(x^2 - 2*x + 1)^4) - 4*log(log(x^2 - 2*x + 1))

Mupad [B] (verification not implemented)

Time = 15.78 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {32 x^2+24 x^3-8 x^4+\left (-8 x+7 x^2+3 x^3-2 x^4\right ) \log \left (1-2 x+x^2\right ) \log \left (\log ^4\left (1-2 x+x^2\right )\right )}{\left (-5-5 x+5 x^2+5 x^3\right ) \log \left (1-2 x+x^2\right )} \, dx=-\frac {x^2\,\ln \left ({\ln \left (x^2-2\,x+1\right )}^4\right )\,\left (x-4\right )}{5\,\left (x+1\right )} \]

[In]

int(-(32*x^2 + 24*x^3 - 8*x^4 - log(log(x^2 - 2*x + 1)^4)*log(x^2 - 2*x + 1)*(8*x - 7*x^2 - 3*x^3 + 2*x^4))/(l
og(x^2 - 2*x + 1)*(5*x - 5*x^2 - 5*x^3 + 5)),x)

[Out]

-(x^2*log(log(x^2 - 2*x + 1)^4)*(x - 4))/(5*(x + 1))