\(\int \frac {-5+9 x-8 x^2+(5 x-4 x^2) \log (\frac {1}{4} (5 x-4 x^2))}{(-5 x+4 x^2) \log (x)+(5 x^2-4 x^3) \log (\frac {1}{4} (5 x-4 x^2))} \, dx\) [2353]

Optimal result

Integrand size = 76, antiderivative size = 19 \[ \int \frac {-5+9 x-8 x^2+\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{\left (-5 x+4 x^2\right ) \log (x)+\left (5 x^2-4 x^3\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )} \, dx=\log \left (\log (x)-x \log \left (\frac {5 x}{4}-x^2\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6873, 6816} \[ \int \frac {-5+9 x-8 x^2+\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{\left (-5 x+4 x^2\right ) \log (x)+\left (5 x^2-4 x^3\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )} \, dx=\log \left (\log (x)-x \log \left (\frac {1}{4} (5-4 x) x\right )\right ) \]

[In]

Int[(-5 + 9*x - 8*x^2 + (5*x - 4*x^2)*Log[(5*x - 4*x^2)/4])/((-5*x + 4*x^2)*Log[x] + (5*x^2 - 4*x^3)*Log[(5*x
- 4*x^2)/4]),x]

[Out]

Log[Log[x] - x*Log[((5 - 4*x)*x)/4]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5-9 x+8 x^2-\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{(5-4 x) x \left (\log (x)-x \log \left (\frac {1}{4} (5-4 x) x\right )\right )} \, dx \\ & = \log \left (\log (x)-x \log \left (\frac {1}{4} (5-4 x) x\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-5+9 x-8 x^2+\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{\left (-5 x+4 x^2\right ) \log (x)+\left (5 x^2-4 x^3\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )} \, dx=\log \left (\log (x)-x \log \left (\frac {1}{4} (5-4 x) x\right )\right ) \]

[In]

Integrate[(-5 + 9*x - 8*x^2 + (5*x - 4*x^2)*Log[(5*x - 4*x^2)/4])/((-5*x + 4*x^2)*Log[x] + (5*x^2 - 4*x^3)*Log
[(5*x - 4*x^2)/4]),x]

[Out]

Log[Log[x] - x*Log[((5 - 4*x)*x)/4]]

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\ln \left (\ln \left (-x^{2}+\frac {5}{4} x \right ) x -\ln \left (x \right )\right )\)	\(19\)
default	\(\ln \left (2 x \ln \left (2\right )-x \ln \left (-4 x^{2}+5 x \right )+\ln \left (x \right )\right )\)	\(23\)
risch	\(\ln \left (x \right )+\ln \left (\ln \left (x -\frac {5}{4}\right )+\frac {i \left (\pi x \,\operatorname {csgn}\left (i \left (x -\frac {5}{4}\right )\right ) \operatorname {csgn}\left (i x \left (x -\frac {5}{4}\right )\right )^{2}-\pi x \,\operatorname {csgn}\left (i \left (x -\frac {5}{4}\right )\right ) \operatorname {csgn}\left (i x \left (x -\frac {5}{4}\right )\right ) \operatorname {csgn}\left (i x \right )+\pi x \operatorname {csgn}\left (i x \left (x -\frac {5}{4}\right )\right )^{3}+\pi x \operatorname {csgn}\left (i x \left (x -\frac {5}{4}\right )\right )^{2} \operatorname {csgn}\left (i x \right )-2 \pi x \operatorname {csgn}\left (i x \left (x -\frac {5}{4}\right )\right )^{2}+4 i \ln \left (2\right ) x -2 i x \ln \left (x \right )+2 \pi x +2 i \ln \left (x \right )\right )}{2 x}\right )\)	\(127\)

[In]

int(((-4*x^2+5*x)*ln(-x^2+5/4*x)-8*x^2+9*x-5)/((4*x^2-5*x)*ln(x)+(-4*x^3+5*x^2)*ln(-x^2+5/4*x)),x,method=_RETU
RNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-5+9 x-8 x^2+\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{\left (-5 x+4 x^2\right ) \log (x)+\left (5 x^2-4 x^3\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )} \, dx=\log \left (-x \log \left (-x^{2} + \frac {5}{4} \, x\right ) + \log \left (x\right )\right ) \]

[In]

integrate(((-4*x^2+5*x)*log(-x^2+5/4*x)-8*x^2+9*x-5)/((4*x^2-5*x)*log(x)+(-4*x^3+5*x^2)*log(-x^2+5/4*x)),x, al
gorithm="fricas")

[Out]

log(-x*log(-x^2 + 5/4*x) + log(x))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-5+9 x-8 x^2+\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{\left (-5 x+4 x^2\right ) \log (x)+\left (5 x^2-4 x^3\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (- x^{2} + \frac {5 x}{4} \right )} - \frac {\log {\left (x \right )}}{x} \right )} \]

[In]

integrate(((-4*x**2+5*x)*ln(-x**2+5/4*x)-8*x**2+9*x-5)/((4*x**2-5*x)*ln(x)+(-4*x**3+5*x**2)*ln(-x**2+5/4*x)),x
)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {-5+9 x-8 x^2+\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{\left (-5 x+4 x^2\right ) \log (x)+\left (5 x^2-4 x^3\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )} \, dx=\log \left (x\right ) + \log \left (-\frac {2 \, x \log \left (2\right ) - {\left (x - 1\right )} \log \left (x\right ) - x \log \left (-4 \, x + 5\right )}{x}\right ) \]

[In]

integrate(((-4*x^2+5*x)*log(-x^2+5/4*x)-8*x^2+9*x-5)/((4*x^2-5*x)*log(x)+(-4*x^3+5*x^2)*log(-x^2+5/4*x)),x, al
gorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-5+9 x-8 x^2+\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{\left (-5 x+4 x^2\right ) \log (x)+\left (5 x^2-4 x^3\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )} \, dx=\log \left (-2 \, x \log \left (2\right ) + x \log \left (x\right ) + x \log \left (-4 \, x + 5\right ) - \log \left (x\right )\right ) \]

[In]

integrate(((-4*x^2+5*x)*log(-x^2+5/4*x)-8*x^2+9*x-5)/((4*x^2-5*x)*log(x)+(-4*x^3+5*x^2)*log(-x^2+5/4*x)),x, al
gorithm="giac")

[Out]

log(-2*x*log(2) + x*log(x) + x*log(-4*x + 5) - log(x))

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {-5+9 x-8 x^2+\left (5 x-4 x^2\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )}{\left (-5 x+4 x^2\right ) \log (x)+\left (5 x^2-4 x^3\right ) \log \left (\frac {1}{4} \left (5 x-4 x^2\right )\right )} \, dx=\ln \left (\ln \left (\frac {5\,x}{4}-x^2\right )-\frac {\ln \left (x\right )}{x}\right )+\ln \left (x\right ) \]

[In]

int((9*x + log((5*x)/4 - x^2)*(5*x - 4*x^2) - 8*x^2 - 5)/(log((5*x)/4 - x^2)*(5*x^2 - 4*x^3) - log(x)*(5*x - 4
*x^2)),x)

[Out]

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