\(\int \frac {(-8 x^2+2 x^3-2 x^4) \log (5) \log (\frac {1}{4} (-4 x+x^2-x^3))+(16-8 x+12 x^2+(-4 x^2+2 x^3-3 x^4) \log (5)) \log (-4+x^2 \log (5))+(8 x^2-2 x^3+2 x^4) \log (5) \log (\frac {1}{4} (-4 x+x^2-x^3)) \log (\log (\frac {1}{4} (-4 x+x^2-x^3)))}{(-16 x+4 x^2-4 x^3+(4 x^3-x^4+x^5) \log (5)) \log (\frac {1}{4} (-4 x+x^2-x^3)) \log ^2(-4+x^2 \log (5))} \, dx\) [700]

Optimal result

Integrand size = 194, antiderivative size = 34 \[ \int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx=\frac {1-\log \left (\log \left (-x+\frac {1}{4} x \left (x-x^2\right )\right )\right )}{\log \left (-4+x^2 \log (5)\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx=\int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx \]

[In]

Int[((-8*x^2 + 2*x^3 - 2*x^4)*Log[5]*Log[(-4*x + x^2 - x^3)/4] + (16 - 8*x + 12*x^2 + (-4*x^2 + 2*x^3 - 3*x^4)
*Log[5])*Log[-4 + x^2*Log[5]] + (8*x^2 - 2*x^3 + 2*x^4)*Log[5]*Log[(-4*x + x^2 - x^3)/4]*Log[Log[(-4*x + x^2 -
x^3)/4]])/((-16*x + 4*x^2 - 4*x^3 + (4*x^3 - x^4 + x^5)*Log[5])*Log[(-4*x + x^2 - x^3)/4]*Log[-4 + x^2*Log[5]
]^2),x]

[Out]

Log[-4 + x^2*Log[5]]^(-1) + ((2*I)*Defer[Int][1/((1 + I*Sqrt[15] - 2*x)*Log[-1/4*(x*(4 - x + x^2))]*Log[-4 + x
^2*Log[5]]), x])/Sqrt[15] - Defer[Int][1/(x*Log[-1/4*(x*(4 - x + x^2))]*Log[-4 + x^2*Log[5]]), x] - ((15 - I*S
qrt[15])*Log[25]*Defer[Int][1/((-1 - I*Sqrt[15] + 2*x)*Log[-1/4*(x*(4 - x + x^2))]*Log[-4 + x^2*Log[5]]), x])/
(15*Log[5]) + ((2*I)*Defer[Int][1/((-1 + I*Sqrt[15] + 2*x)*Log[-1/4*(x*(4 - x + x^2))]*Log[-4 + x^2*Log[5]]),
x])/Sqrt[15] - ((15 + I*Sqrt[15])*Log[25]*Defer[Int][1/((-1 + I*Sqrt[15] + 2*x)*Log[-1/4*(x*(4 - x + x^2))]*Lo
g[-4 + x^2*Log[5]]), x])/(15*Log[5]) - Sqrt[Log[5]]*Defer[Int][Log[Log[-1/4*(x*(4 - x + x^2))]]/((2 - x*Sqrt[L
og[5]])*Log[-4 + x^2*Log[5]]^2), x] + Sqrt[Log[5]]*Defer[Int][Log[Log[-1/4*(x*(4 - x + x^2))]]/((2 + x*Sqrt[Lo
g[5]])*Log[-4 + x^2*Log[5]]^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-\frac {\left (16-8 x-4 x^2 (-3+\log (5))-3 x^4 \log (5)+x^3 \log (25)\right ) \log \left (-4+x^2 \log (5)\right )}{\left (4-x+x^2\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )}-2 x^2 \log (5) \left (-1+\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )\right )}{x \left (4-x^2 \log (5)\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx \\ & = \int \left (\frac {8 x^2 \log (5) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )-2 x^3 \log (5) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )+2 x^4 \log (5) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )-16 \log \left (-4+x^2 \log (5)\right )+8 x \log \left (-4+x^2 \log (5)\right )-12 x^2 \left (1-\frac {\log (5)}{3}\right ) \log \left (-4+x^2 \log (5)\right )+3 x^4 \log (5) \log \left (-4+x^2 \log (5)\right )-x^3 \log (25) \log \left (-4+x^2 \log (5)\right )}{x \left (4-x+x^2\right ) \left (4-x^2 \log (5)\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )}+\frac {2 x \log (5) \log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (-4+x^2 \log (5)\right ) \log ^2\left (-4+x^2 \log (5)\right )}\right ) \, dx \\ & = (2 \log (5)) \int \frac {x \log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (-4+x^2 \log (5)\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\int \frac {8 x^2 \log (5) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )-2 x^3 \log (5) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )+2 x^4 \log (5) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )-16 \log \left (-4+x^2 \log (5)\right )+8 x \log \left (-4+x^2 \log (5)\right )-12 x^2 \left (1-\frac {\log (5)}{3}\right ) \log \left (-4+x^2 \log (5)\right )+3 x^4 \log (5) \log \left (-4+x^2 \log (5)\right )-x^3 \log (25) \log \left (-4+x^2 \log (5)\right )}{x \left (4-x+x^2\right ) \left (4-x^2 \log (5)\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx \\ & = (2 \log (5)) \int \left (-\frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{2 \left (2-x \sqrt {\log (5)}\right ) \sqrt {\log (5)} \log ^2\left (-4+x^2 \log (5)\right )}+\frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{2 \left (2+x \sqrt {\log (5)}\right ) \sqrt {\log (5)} \log ^2\left (-4+x^2 \log (5)\right )}\right ) \, dx+\int \frac {2 x^2 \log (5)-\frac {\left (16-8 x-4 x^2 (-3+\log (5))-3 x^4 \log (5)+x^3 \log (25)\right ) \log \left (-4+x^2 \log (5)\right )}{\left (4-x+x^2\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )}}{x \left (4-x^2 \log (5)\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx \\ & = -\left (\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx\right )+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\int \left (-\frac {2 x \log (5)}{\left (-4+x^2 \log (5)\right ) \log ^2\left (-4+x^2 \log (5)\right )}+\frac {-16+8 x-4 x^2 (3-\log (5))+3 x^4 \log (5)-x^3 \log (25)}{x \left (4-x+x^2\right ) \left (4-x^2 \log (5)\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}\right ) \, dx \\ & = -\left (\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx\right )+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx-(2 \log (5)) \int \frac {x}{\left (-4+x^2 \log (5)\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\int \frac {-16+8 x-4 x^2 (3-\log (5))+3 x^4 \log (5)-x^3 \log (25)}{x \left (4-x+x^2\right ) \left (4-x^2 \log (5)\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx \\ & = -\left (\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx\right )+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx-\log (5) \text {Subst}\left (\int \frac {1}{(-4+x \log (5)) \log ^2(-4+x \log (5))} \, dx,x,x^2\right )+\int \frac {-4-3 x^2+\frac {x \log (25)}{\log (5)}}{x \left (4-x+x^2\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx \\ & = -\left (\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx\right )+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\int \left (-\frac {1}{x \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}+\frac {\log (5)-x \log (25)}{\left (4-x+x^2\right ) \log (5) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}\right ) \, dx-\text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-4+x^2 \log (5)\right ) \\ & = \frac {\int \frac {\log (5)-x \log (25)}{\left (4-x+x^2\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx}{\log (5)}-\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx-\int \frac {1}{x \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx-\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (-4+x^2 \log (5)\right )\right ) \\ & = \frac {1}{\log \left (-4+x^2 \log (5)\right )}+\frac {\int \left (\frac {\log (5)}{\left (4-x+x^2\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}-\frac {x \log (25)}{\left (4-x+x^2\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}\right ) \, dx}{\log (5)}-\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx-\int \frac {1}{x \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx \\ & = \frac {1}{\log \left (-4+x^2 \log (5)\right )}-\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx-\frac {\log (25) \int \frac {x}{\left (4-x+x^2\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx}{\log (5)}-\int \frac {1}{x \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx+\int \frac {1}{\left (4-x+x^2\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx \\ & = \frac {1}{\log \left (-4+x^2 \log (5)\right )}-\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx-\frac {\log (25) \int \left (\frac {1-\frac {i}{\sqrt {15}}}{\left (-1-i \sqrt {15}+2 x\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}+\frac {1+\frac {i}{\sqrt {15}}}{\left (-1+i \sqrt {15}+2 x\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}\right ) \, dx}{\log (5)}+\int \left (\frac {2 i}{\sqrt {15} \left (1+i \sqrt {15}-2 x\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}+\frac {2 i}{\sqrt {15} \left (-1+i \sqrt {15}+2 x\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )}\right ) \, dx-\int \frac {1}{x \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx \\ & = \frac {1}{\log \left (-4+x^2 \log (5)\right )}+\frac {(2 i) \int \frac {1}{\left (1+i \sqrt {15}-2 x\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx}{\sqrt {15}}+\frac {(2 i) \int \frac {1}{\left (-1+i \sqrt {15}+2 x\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx}{\sqrt {15}}-\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2-x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx+\sqrt {\log (5)} \int \frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\left (2+x \sqrt {\log (5)}\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx-\frac {\left (\left (15-i \sqrt {15}\right ) \log (25)\right ) \int \frac {1}{\left (-1-i \sqrt {15}+2 x\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx}{15 \log (5)}-\frac {\left (\left (15+i \sqrt {15}\right ) \log (25)\right ) \int \frac {1}{\left (-1+i \sqrt {15}+2 x\right ) \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx}{15 \log (5)}-\int \frac {1}{x \log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right ) \log \left (-4+x^2 \log (5)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx=\frac {1}{\log \left (-4+x^2 \log (5)\right )}-\frac {\log \left (\log \left (-\frac {1}{4} x \left (4-x+x^2\right )\right )\right )}{\log \left (-4+x^2 \log (5)\right )} \]

[In]

Integrate[((-8*x^2 + 2*x^3 - 2*x^4)*Log[5]*Log[(-4*x + x^2 - x^3)/4] + (16 - 8*x + 12*x^2 + (-4*x^2 + 2*x^3 -
3*x^4)*Log[5])*Log[-4 + x^2*Log[5]] + (8*x^2 - 2*x^3 + 2*x^4)*Log[5]*Log[(-4*x + x^2 - x^3)/4]*Log[Log[(-4*x +
x^2 - x^3)/4]])/((-16*x + 4*x^2 - 4*x^3 + (4*x^3 - x^4 + x^5)*Log[5])*Log[(-4*x + x^2 - x^3)/4]*Log[-4 + x^2*
Log[5]]^2),x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.14 (sec) , antiderivative size = 147, normalized size of antiderivative = 4.32

\[-\frac {\ln \left (-2 \ln \left (2\right )+i \pi +\ln \left (x \right )+\ln \left (x^{2}-x +4\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \left (x^{2}-x +4\right )\right ) \left (-\operatorname {csgn}\left (i x \left (x^{2}-x +4\right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (x^{2}-x +4\right )\right )+\operatorname {csgn}\left (i \left (x^{2}-x +4\right )\right )\right )}{2}+i \pi {\operatorname {csgn}\left (i x \left (x^{2}-x +4\right )\right )}^{2} \left (\operatorname {csgn}\left (i x \left (x^{2}-x +4\right )\right )-1\right )\right )}{\ln \left (x^{2} \ln \left (5\right )-4\right )}+\frac {1}{\ln \left (x^{2} \ln \left (5\right )-4\right )}\]

[In]

int(((2*x^4-2*x^3+8*x^2)*ln(5)*ln(-1/4*x^3+1/4*x^2-x)*ln(ln(-1/4*x^3+1/4*x^2-x))+((-3*x^4+2*x^3-4*x^2)*ln(5)+1
2*x^2-8*x+16)*ln(x^2*ln(5)-4)+(-2*x^4+2*x^3-8*x^2)*ln(5)*ln(-1/4*x^3+1/4*x^2-x))/((x^5-x^4+4*x^3)*ln(5)-4*x^3+
4*x^2-16*x)/ln(-1/4*x^3+1/4*x^2-x)/ln(x^2*ln(5)-4)^2,x)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx=-\frac {\log \left (\log \left (-\frac {1}{4} \, x^{3} + \frac {1}{4} \, x^{2} - x\right )\right ) - 1}{\log \left (x^{2} \log \left (5\right ) - 4\right )} \]

[In]

integrate(((2*x^4-2*x^3+8*x^2)*log(5)*log(-1/4*x^3+1/4*x^2-x)*log(log(-1/4*x^3+1/4*x^2-x))+((-3*x^4+2*x^3-4*x^
2)*log(5)+12*x^2-8*x+16)*log(x^2*log(5)-4)+(-2*x^4+2*x^3-8*x^2)*log(5)*log(-1/4*x^3+1/4*x^2-x))/((x^5-x^4+4*x^
3)*log(5)-4*x^3+4*x^2-16*x)/log(-1/4*x^3+1/4*x^2-x)/log(x^2*log(5)-4)^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((2*x**4-2*x**3+8*x**2)*ln(5)*ln(-1/4*x**3+1/4*x**2-x)*ln(ln(-1/4*x**3+1/4*x**2-x))+((-3*x**4+2*x**3
-4*x**2)*ln(5)+12*x**2-8*x+16)*ln(x**2*ln(5)-4)+(-2*x**4+2*x**3-8*x**2)*ln(5)*ln(-1/4*x**3+1/4*x**2-x))/((x**5
-x**4+4*x**3)*ln(5)-4*x**3+4*x**2-16*x)/ln(-1/4*x**3+1/4*x**2-x)/ln(x**2*ln(5)-4)**2,x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx=-\frac {\log \left (-2 \, \log \left (2\right ) + \log \left (-x^{2} + x - 4\right ) + \log \left (x\right )\right ) - 1}{\log \left (x^{2} \log \left (5\right ) - 4\right )} \]

[In]

integrate(((2*x^4-2*x^3+8*x^2)*log(5)*log(-1/4*x^3+1/4*x^2-x)*log(log(-1/4*x^3+1/4*x^2-x))+((-3*x^4+2*x^3-4*x^
2)*log(5)+12*x^2-8*x+16)*log(x^2*log(5)-4)+(-2*x^4+2*x^3-8*x^2)*log(5)*log(-1/4*x^3+1/4*x^2-x))/((x^5-x^4+4*x^
3)*log(5)-4*x^3+4*x^2-16*x)/log(-1/4*x^3+1/4*x^2-x)/log(x^2*log(5)-4)^2,x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((2*x^4-2*x^3+8*x^2)*log(5)*log(-1/4*x^3+1/4*x^2-x)*log(log(-1/4*x^3+1/4*x^2-x))+((-3*x^4+2*x^3-4*x^
2)*log(5)+12*x^2-8*x+16)*log(x^2*log(5)-4)+(-2*x^4+2*x^3-8*x^2)*log(5)*log(-1/4*x^3+1/4*x^2-x))/((x^5-x^4+4*x^
3)*log(5)-4*x^3+4*x^2-16*x)/log(-1/4*x^3+1/4*x^2-x)/log(x^2*log(5)-4)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 17.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {\left (-8 x^2+2 x^3-2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )+\left (16-8 x+12 x^2+\left (-4 x^2+2 x^3-3 x^4\right ) \log (5)\right ) \log \left (-4+x^2 \log (5)\right )+\left (8 x^2-2 x^3+2 x^4\right ) \log (5) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right )\right )}{\left (-16 x+4 x^2-4 x^3+\left (4 x^3-x^4+x^5\right ) \log (5)\right ) \log \left (\frac {1}{4} \left (-4 x+x^2-x^3\right )\right ) \log ^2\left (-4+x^2 \log (5)\right )} \, dx=-\frac {\ln \left (\ln \left (-\frac {x^3}{4}+\frac {x^2}{4}-x\right )\right )-1}{\ln \left (x^2\,\ln \left (5\right )-4\right )} \]

[In]

int((log(x^2*log(5) - 4)*(8*x + log(5)*(4*x^2 - 2*x^3 + 3*x^4) - 12*x^2 - 16) + log(x^2/4 - x - x^3/4)*log(5)*
(8*x^2 - 2*x^3 + 2*x^4) - log(x^2/4 - x - x^3/4)*log(5)*log(log(x^2/4 - x - x^3/4))*(8*x^2 - 2*x^3 + 2*x^4))/(
log(x^2/4 - x - x^3/4)*log(x^2*log(5) - 4)^2*(16*x - log(5)*(4*x^3 - x^4 + x^5) - 4*x^2 + 4*x^3)),x)

[Out]

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