3.58.99 \(\int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{(4 x^2-15 x^3-4 x^4+(-5 x+20 x^2) \log (x)) \log ^2(\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)})} \, dx\) [5799]

Optimal. Leaf size=30 \[ \frac {1}{\log \left (\frac {1-4 x}{5 \left (1-x+\frac {5 (-x+\log (x))}{x}\right )}\right )} \]

[Out]

1/ln((1-4*x)/(25/x*(ln(x)-x)-5*x+5))

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Rubi [F]
time = 3.57, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{\left (4 x^2-15 x^3-4 x^4+\left (-5 x+20 x^2\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5 + 20*x + 17*x^2 + (5 - 40*x)*Log[x])/((4*x^2 - 15*x^3 - 4*x^4 + (-5*x + 20*x^2)*Log[x])*Log[(x - 4*x^2
)/(-20*x - 5*x^2 + 25*Log[x])]^2),x]

[Out]

(-17*Defer[Int][1/((4*x + x^2 - 5*Log[x])*Log[(x*(-1 + 4*x))/(5*(x*(4 + x) - 5*Log[x]))]^2), x])/4 - 5*Defer[I
nt][1/(x*(4*x + x^2 - 5*Log[x])*Log[(x*(-1 + 4*x))/(5*(x*(4 + x) - 5*Log[x]))]^2), x] - (17*Defer[Int][1/((-1
+ 4*x)*(4*x + x^2 - 5*Log[x])*Log[(x*(-1 + 4*x))/(5*(x*(4 + x) - 5*Log[x]))]^2), x])/4 + 5*Defer[Int][Log[x]/(
x*(4*x + x^2 - 5*Log[x])*Log[(x*(-1 + 4*x))/(5*(x*(4 + x) - 5*Log[x]))]^2), x] + 20*Defer[Int][Log[x]/((-1 + 4
*x)*(4*x + x^2 - 5*Log[x])*Log[(x*(-1 + 4*x))/(5*(x*(4 + x) - 5*Log[x]))]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{(1-4 x) x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (-\frac {(1-4 x) x}{5 \left (4 x+x^2-5 \log (x)\right )}\right )} \, dx\\ &=\int \left (\frac {-5+20 x+17 x^2+5 \log (x)-40 x \log (x)}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}-\frac {4 \left (-5+20 x+17 x^2+5 \log (x)-40 x \log (x)\right )}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}\right ) \, dx\\ &=-\left (4 \int \frac {-5+20 x+17 x^2+5 \log (x)-40 x \log (x)}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx\right )+\int \frac {-5+20 x+17 x^2+5 \log (x)-40 x \log (x)}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx\\ &=-\left (4 \int \left (-\frac {5}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {20 x}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {17 x^2}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {5 \log (x)}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}-\frac {40 x \log (x)}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}\right ) \, dx\right )+\int \left (\frac {20}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}-\frac {5}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {17 x}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}-\frac {40 \log (x)}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {5 \log (x)}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}\right ) \, dx\\ &=-\left (5 \int \frac {1}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx\right )+5 \int \frac {\log (x)}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+17 \int \frac {x}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+20 \int \frac {1}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+20 \int \frac {1}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-20 \int \frac {\log (x)}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-40 \int \frac {\log (x)}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-68 \int \frac {x^2}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-80 \int \frac {x}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+160 \int \frac {x \log (x)}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx\\ &=-\left (5 \int \frac {1}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx\right )+5 \int \frac {\log (x)}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+17 \int \frac {x}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+20 \int \frac {1}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+20 \int \frac {1}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-20 \int \frac {\log (x)}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-40 \int \frac {\log (x)}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-68 \int \left (\frac {1}{16 \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {x}{4 \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {1}{16 (-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}\right ) \, dx-80 \int \left (\frac {1}{4 \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {1}{4 (-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}\right ) \, dx+160 \int \left (\frac {\log (x)}{4 \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}+\frac {\log (x)}{4 (-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )}\right ) \, dx\\ &=-\left (\frac {17}{4} \int \frac {1}{\left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx\right )-\frac {17}{4} \int \frac {1}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-5 \int \frac {1}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+5 \int \frac {\log (x)}{x \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx-20 \int \frac {\log (x)}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx+40 \int \frac {\log (x)}{(-1+4 x) \left (4 x+x^2-5 \log (x)\right ) \log ^2\left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.09, size = 25, normalized size = 0.83 \begin {gather*} \frac {1}{\log \left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 20*x + 17*x^2 + (5 - 40*x)*Log[x])/((4*x^2 - 15*x^3 - 4*x^4 + (-5*x + 20*x^2)*Log[x])*Log[(x -
4*x^2)/(-20*x - 5*x^2 + 25*Log[x])]^2),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 8.66, size = 368, normalized size = 12.27

method	result	size
risch	\(\frac {2 i}{\pi \,\mathrm {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \mathrm {csgn}\left (\frac {i}{-x^{2}+5 \ln \left (x \right )-4 x}\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )-\pi \,\mathrm {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{2}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right ) \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{-x^{2}+5 \ln \left (x \right )-4 x}\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{3}+\pi \,\mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right ) \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{3}-2 i \ln \left (x^{2}-5 \ln \left (x \right )+4 x \right )-2 i \ln \left (5\right )+2 i \ln \left (x \right )+4 i \ln \left (2\right )+2 i \ln \left (x -\frac {1}{4}\right )}\)	\(362\)
default	\(\frac {2 i}{2 \pi \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \mathrm {csgn}\left (\frac {i}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )-\pi \,\mathrm {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )^{2}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right ) \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right ) \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )^{3}-2 \pi +2 i \ln \left (x \right )-4 i \ln \left (5\right )-2 i \ln \left (-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}\right )+4 i \ln \left (2\right )+2 i \ln \left (x -\frac {1}{4}\right )}\)	\(368\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-40*x+5)*ln(x)+17*x^2+20*x-5)/((20*x^2-5*x)*ln(x)-4*x^4-15*x^3+4*x^2)/ln((-4*x^2+x)/(25*ln(x)-5*x^2-20*x
))^2,x,method=_RETURNVERBOSE)

[Out]

2*I/(2*Pi*csgn(I*x/(-1/5*x^2+ln(x)-4/5*x)*(x-1/4))^2+Pi*csgn(I*(x-1/4))*csgn(I/(-1/5*x^2+ln(x)-4/5*x))*csgn(I*
(x-1/4)/(-1/5*x^2+ln(x)-4/5*x))-Pi*csgn(I*(x-1/4))*csgn(I*(x-1/4)/(-1/5*x^2+ln(x)-4/5*x))^2-Pi*csgn(I/(-1/5*x^
2+ln(x)-4/5*x))*csgn(I*(x-1/4)/(-1/5*x^2+ln(x)-4/5*x))^2+Pi*csgn(I*x)*csgn(I*(x-1/4)/(-1/5*x^2+ln(x)-4/5*x))*c
sgn(I*x/(-1/5*x^2+ln(x)-4/5*x)*(x-1/4))-Pi*csgn(I*x)*csgn(I*x/(-1/5*x^2+ln(x)-4/5*x)*(x-1/4))^2+Pi*csgn(I*(x-1
/4)/(-1/5*x^2+ln(x)-4/5*x))^3-Pi*csgn(I*(x-1/4)/(-1/5*x^2+ln(x)-4/5*x))*csgn(I*x/(-1/5*x^2+ln(x)-4/5*x)*(x-1/4
))^2-Pi*csgn(I*x/(-1/5*x^2+ln(x)-4/5*x)*(x-1/4))^3-2*Pi+2*I*ln(x)-4*I*ln(5)-2*I*ln(-1/5*x^2+ln(x)-4/5*x)+4*I*l
n(2)+2*I*ln(x-1/4))

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Maxima [A]
time = 0.65, size = 31, normalized size = 1.03 \begin {gather*} -\frac {1}{\log \left (5\right ) + \log \left (x^{2} + 4 \, x - 5 \, \log \left (x\right )\right ) - \log \left (4 \, x - 1\right ) - \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x+5)*log(x)+17*x^2+20*x-5)/((20*x^2-5*x)*log(x)-4*x^4-15*x^3+4*x^2)/log((-4*x^2+x)/(25*log(x)-
5*x^2-20*x))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 27, normalized size = 0.90 \begin {gather*} \frac {1}{\log \left (\frac {4 \, x^{2} - x}{5 \, {\left (x^{2} + 4 \, x - 5 \, \log \left (x\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x+5)*log(x)+17*x^2+20*x-5)/((20*x^2-5*x)*log(x)-4*x^4-15*x^3+4*x^2)/log((-4*x^2+x)/(25*log(x)-
5*x^2-20*x))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.20, size = 22, normalized size = 0.73 \begin {gather*} \frac {1}{\log {\left (\frac {- 4 x^{2} + x}{- 5 x^{2} - 20 x + 25 \log {\left (x \right )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x+5)*ln(x)+17*x**2+20*x-5)/((20*x**2-5*x)*ln(x)-4*x**4-15*x**3+4*x**2)/ln((-4*x**2+x)/(25*ln(x
)-5*x**2-20*x))**2,x)

[Out]

1/log((-4*x**2 + x)/(-5*x**2 - 20*x + 25*log(x)))

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Giac [A]
time = 0.44, size = 31, normalized size = 1.03 \begin {gather*} -\frac {1}{\log \left (5 \, x^{2} + 20 \, x - 25 \, \log \left (x\right )\right ) - \log \left (4 \, x - 1\right ) - \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x+5)*log(x)+17*x^2+20*x-5)/((20*x^2-5*x)*log(x)-4*x^4-15*x^3+4*x^2)/log((-4*x^2+x)/(25*log(x)-
5*x^2-20*x))^2,x, algorithm="giac")

[Out]

-1/(log(5*x^2 + 20*x - 25*log(x)) - log(4*x - 1) - log(x))

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Mupad [B]
time = 5.51, size = 27, normalized size = 0.90 \begin {gather*} \frac {1}{\ln \left (-\frac {x-4\,x^2}{20\,x-25\,\ln \left (x\right )+5\,x^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x - log(x)*(40*x - 5) + 17*x^2 - 5)/(log(-(x - 4*x^2)/(20*x - 25*log(x) + 5*x^2))^2*(log(x)*(5*x - 20
*x^2) - 4*x^2 + 15*x^3 + 4*x^4)),x)

[Out]

1/log(-(x - 4*x^2)/(20*x - 25*log(x) + 5*x^2))

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