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

Optimal. Leaf size=19 \[ \log \left (\log (x)-x \log \left (\frac {5 x}{4}-x^2\right )\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6741, 6684} \begin {gather*} \log \left (\log (x)-x \log \left (\frac {1}{4} (5-4 x) x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 6684

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

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 18, normalized size = 0.95 \begin {gather*} \log \left (\log (x)-x \log \left (\frac {1}{4} (5-4 x) x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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]]

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fricas [A]  time = 0.45, size = 17, normalized size = 0.89 \begin {gather*} \log \left (-x \log \left (-x^{2} + \frac {5}{4} \, x\right ) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.22, size = 23, normalized size = 1.21 \begin {gather*} \log \left (-2 \, x \log \relax (2) + x \log \relax (x) + x \log \left (-4 \, x + 5\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 0.11, size = 127, normalized size = 6.68

method	result	size
risch	\(\ln \relax (x )+\ln \left (\ln \left (x -\frac {5}{4}\right )+\frac {i \left (\pi x \,\mathrm {csgn}\left (i \left (x -\frac {5}{4}\right )\right ) \mathrm {csgn}\left (i x \left (x -\frac {5}{4}\right )\right )^{2}-\pi x \,\mathrm {csgn}\left (i \left (x -\frac {5}{4}\right )\right ) \mathrm {csgn}\left (i x \left (x -\frac {5}{4}\right )\right ) \mathrm {csgn}\left (i x \right )+\pi x \mathrm {csgn}\left (i x \left (x -\frac {5}{4}\right )\right )^{3}+\pi x \mathrm {csgn}\left (i x \left (x -\frac {5}{4}\right )\right )^{2} \mathrm {csgn}\left (i x \right )-2 \pi x \mathrm {csgn}\left (i x \left (x -\frac {5}{4}\right )\right )^{2}+4 i x \ln \relax (2)-2 i x \ln \relax (x )+2 \pi x +2 i \ln \relax (x )\right )}{2 x}\right )\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[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(x)+ln(ln(x-5/4)+1/2*I/x*(Pi*x*csgn(I*(x-5/4))*csgn(I*x*(x-5/4))^2-Pi*x*csgn(I*(x-5/4))*csgn(I*x*(x-5/4))*cs
gn(I*x)+Pi*x*csgn(I*x*(x-5/4))^3+Pi*x*csgn(I*x*(x-5/4))^2*csgn(I*x)-2*Pi*x*csgn(I*x*(x-5/4))^2+4*I*x*ln(2)-2*I
*x*ln(x)+2*Pi*x+2*I*ln(x)))

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maxima [A]  time = 0.73, size = 31, normalized size = 1.63 \begin {gather*} \log \relax (x) + \log \left (-\frac {2 \, x \log \relax (2) - {\left (x - 1\right )} \log \relax (x) - x \log \left (-4 \, x + 5\right )}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.76, size = 22, normalized size = 1.16 \begin {gather*} \ln \left (\ln \left (\frac {5\,x}{4}-x^2\right )-\frac {\ln \relax (x)}{x}\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.45, size = 19, normalized size = 1.00 \begin {gather*} \log {\relax (x )} + \log {\left (\log {\left (- x^{2} + \frac {5 x}{4} \right )} - \frac {\log {\relax (x )}}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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