3.5.6 \(\int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log (\frac {5 \log (x)}{e^3})}{(4 x^2-x^3) \log (x)+(-3 x+x^2) \log (x) \log (\frac {5 \log (x)}{e^3})} \, dx\)

Optimal. Leaf size=24 \[ \log \left (\frac {x-(-3+x) \left (x-\log \left (\frac {5 \log (x)}{e^3}\right )\right )}{x^2}\right ) \]

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Rubi [F]  time = 3.91, 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 {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3 + x - 4*x*Log[x] + (6 - x)*Log[x]*Log[(5*Log[x])/E^3])/((4*x^2 - x^3)*Log[x] + (-3*x + x^2)*Log[x]*Log
[(5*Log[x])/E^3]),x]

[Out]

-1/3*x - 2*Log[x] + 8*Defer[Int][(9 - x^2 + x*(1 + Log[5]) + x*Log[Log[x]] - 3*Log[5*Log[x]])^(-1), x] - (2*(2
- Log[5])*Defer[Int][x/(9 - x^2 + x*(1 + Log[5]) + x*Log[Log[x]] - 3*Log[5*Log[x]]), x])/3 + Defer[Int][x^2/(
9 - x^2 + x*(1 + Log[5]) + x*Log[Log[x]] - 3*Log[5*Log[x]]), x]/3 + Defer[Int][1/(Log[x]*(9 - x^2 + x*(1 + Log
[5]) + x*Log[Log[x]] - 3*Log[5*Log[x]])), x] + (2*Defer[Int][(x*Log[Log[x]])/(9 - x^2 + x*(1 + Log[5]) + x*Log
[Log[x]] - 3*Log[5*Log[x]]), x])/3 + 4*Defer[Int][(-9 + x^2 - x*(1 + Log[5]) - x*Log[Log[x]] + 3*Log[5*Log[x]]
)^(-1), x] + ((1 + Log[5])*Defer[Int][x/(-9 + x^2 - x*(1 + Log[5]) - x*Log[Log[x]] + 3*Log[5*Log[x]]), x])/3 +
(2*Defer[Int][x^2/(-9 + x^2 - x*(1 + Log[5]) - x*Log[Log[x]] + 3*Log[5*Log[x]]), x])/3 + 3*Defer[Int][1/(x*Lo
g[x]*(-9 + x^2 - x*(1 + Log[5]) - x*Log[Log[x]] + 3*Log[5*Log[x]])), x] + Defer[Int][(x*Log[Log[x]])/(-9 + x^2
- x*(1 + Log[5]) - x*Log[Log[x]] + 3*Log[5*Log[x]]), x]/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{x \log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ &=\int \left (\frac {-3+x-4 x \log (x)}{x \log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )}+\frac {(6-x) \left (-3 \left (1-\frac {\log (5)}{3}\right )+\log (\log (x))\right )}{x \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )}\right ) \, dx\\ &=\int \frac {-3+x-4 x \log (x)}{x \log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx+\int \frac {(6-x) \left (-3 \left (1-\frac {\log (5)}{3}\right )+\log (\log (x))\right )}{x \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ &=\int \left (\frac {6 \left (-3 \left (1-\frac {\log (5)}{3}\right )+\log (\log (x))\right )}{x \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )}+\frac {3-\log (5 \log (x))}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))}\right ) \, dx+\int \left (\frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )}+\frac {4}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))}+\frac {3}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )}\right ) \, dx\\ &=3 \int \frac {1}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )} \, dx+4 \int \frac {1}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+6 \int \frac {-3 \left (1-\frac {\log (5)}{3}\right )+\log (\log (x))}{x \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx+\int \frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx+\int \frac {3-\log (5 \log (x))}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx\\ &=3 \int \frac {1}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )} \, dx+4 \int \frac {1}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+6 \int \frac {-3+\log (5 \log (x))}{x \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx+\int \left (\frac {1}{3}+\frac {x (-1+x-\log (5)-\log (\log (x)))}{3 \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )}\right ) \, dx+\int \frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ &=\frac {x}{3}+\frac {1}{3} \int \frac {x (-1+x-\log (5)-\log (\log (x)))}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+3 \int \frac {1}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )} \, dx+4 \int \frac {1}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+6 \int \left (-\frac {1}{3 x}+\frac {1-x+\log (5 \log (x))}{3 \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )}\right ) \, dx+\int \frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ &=\frac {x}{3}-2 \log (x)+\frac {1}{3} \int \left (\frac {x^2}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))}+\frac {x (1+\log (5))}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))}+\frac {x \log (\log (x))}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))}\right ) \, dx+2 \int \frac {1-x+\log (5 \log (x))}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+3 \int \frac {1}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )} \, dx+4 \int \frac {1}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\int \frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ &=\frac {x}{3}-2 \log (x)+\frac {1}{3} \int \frac {x^2}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+\frac {1}{3} \int \frac {x \log (\log (x))}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+2 \int \left (-\frac {1}{3}+\frac {12-x^2-2 x \left (1-\frac {\log (5)}{2}\right )+x \log (\log (x))}{3 \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )}\right ) \, dx+3 \int \frac {1}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )} \, dx+4 \int \frac {1}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\frac {1}{3} (1+\log (5)) \int \frac {x}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\int \frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ &=-\frac {x}{3}-2 \log (x)+\frac {1}{3} \int \frac {x^2}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+\frac {1}{3} \int \frac {x \log (\log (x))}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\frac {2}{3} \int \frac {12-x^2-2 x \left (1-\frac {\log (5)}{2}\right )+x \log (\log (x))}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+3 \int \frac {1}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )} \, dx+4 \int \frac {1}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\frac {1}{3} (1+\log (5)) \int \frac {x}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\int \frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ &=-\frac {x}{3}-2 \log (x)+\frac {1}{3} \int \frac {x^2}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+\frac {1}{3} \int \frac {x \log (\log (x))}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\frac {2}{3} \int \left (\frac {12}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))}+\frac {x (-2+\log (5))}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))}+\frac {x \log (\log (x))}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))}+\frac {x^2}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))}\right ) \, dx+3 \int \frac {1}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )} \, dx+4 \int \frac {1}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\frac {1}{3} (1+\log (5)) \int \frac {x}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\int \frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ &=-\frac {x}{3}-2 \log (x)+\frac {1}{3} \int \frac {x^2}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+\frac {1}{3} \int \frac {x \log (\log (x))}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\frac {2}{3} \int \frac {x \log (\log (x))}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+\frac {2}{3} \int \frac {x^2}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+3 \int \frac {1}{x \log (x) \left (-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))\right )} \, dx+4 \int \frac {1}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+8 \int \frac {1}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx-\frac {1}{3} (2 (2-\log (5))) \int \frac {x}{9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))} \, dx+\frac {1}{3} (1+\log (5)) \int \frac {x}{-9+x^2-x (1+\log (5))-x \log (\log (x))+3 \log (5 \log (x))} \, dx+\int \frac {1}{\log (x) \left (9-x^2+x (1+\log (5))+x \log (\log (x))-3 \log (5 \log (x))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.91, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-3+x-4 x \log (x)+(6-x) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )}{\left (4 x^2-x^3\right ) \log (x)+\left (-3 x+x^2\right ) \log (x) \log \left (\frac {5 \log (x)}{e^3}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(-3 + x - 4*x*Log[x] + (6 - x)*Log[x]*Log[(5*Log[x])/E^3])/((4*x^2 - x^3)*Log[x] + (-3*x + x^2)*Log[
x]*Log[(5*Log[x])/E^3]),x]

[Out]

Integrate[(-3 + x - 4*x*Log[x] + (6 - x)*Log[x]*Log[(5*Log[x])/E^3])/((4*x^2 - x^3)*Log[x] + (-3*x + x^2)*Log[
x]*Log[(5*Log[x])/E^3]), x]

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fricas [A]  time = 0.85, size = 36, normalized size = 1.50 \begin {gather*} \log \left (x - 3\right ) - 2 \, \log \relax (x) + \log \left (-\frac {x^{2} - {\left (x - 3\right )} \log \left (5 \, e^{\left (-3\right )} \log \relax (x)\right ) - 4 \, x}{x - 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+6)*log(x)*log(5*log(x)/exp(3))-4*x*log(x)+x-3)/((x^2-3*x)*log(x)*log(5*log(x)/exp(3))+(-x^3+4*x
^2)*log(x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.44, size = 28, normalized size = 1.17 \begin {gather*} \log \left (-x^{2} + x \log \left (5 \, \log \relax (x)\right ) + x - 3 \, \log \left (5 \, \log \relax (x)\right ) + 9\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+6)*log(x)*log(5*log(x)/exp(3))-4*x*log(x)+x-3)/((x^2-3*x)*log(x)*log(5*log(x)/exp(3))+(-x^3+4*x
^2)*log(x)),x, algorithm="giac")

[Out]

log(-x^2 + x*log(5*log(x)) + x - 3*log(5*log(x)) + 9) - 2*log(x)

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maple [A]  time = 0.08, size = 30, normalized size = 1.25

method	result	size
risch	\(\ln \left (x -3\right )-2 \ln \relax (x )+\ln \left (\ln \left (5 \ln \relax (x ) {\mathrm e}^{-3}\right )-\frac {\left (x -4\right ) x}{x -3}\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x+6)*ln(x)*ln(5*ln(x)/exp(3))-4*x*ln(x)+x-3)/((x^2-3*x)*ln(x)*ln(5*ln(x)/exp(3))+(-x^3+4*x^2)*ln(x)),x,
method=_RETURNVERBOSE)

[Out]

ln(x-3)-2*ln(x)+ln(ln(5*ln(x)*exp(-3))-(x-4)*x/(x-3))

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maxima [A]  time = 0.66, size = 41, normalized size = 1.71 \begin {gather*} \log \left (x - 3\right ) - 2 \, \log \relax (x) + \log \left (-\frac {x^{2} - x {\left (\log \relax (5) + 1\right )} - {\left (x - 3\right )} \log \left (\log \relax (x)\right ) + 3 \, \log \relax (5) - 9}{x - 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+6)*log(x)*log(5*log(x)/exp(3))-4*x*log(x)+x-3)/((x^2-3*x)*log(x)*log(5*log(x)/exp(3))+(-x^3+4*x
^2)*log(x)),x, algorithm="maxima")

[Out]

log(x - 3) - 2*log(x) + log(-(x^2 - x*(log(5) + 1) - (x - 3)*log(log(x)) + 3*log(5) - 9)/(x - 3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {4\,x\,\ln \relax (x)-x+\ln \left (5\,{\mathrm {e}}^{-3}\,\ln \relax (x)\right )\,\ln \relax (x)\,\left (x-6\right )+3}{\ln \relax (x)\,\left (4\,x^2-x^3\right )-\ln \left (5\,{\mathrm {e}}^{-3}\,\ln \relax (x)\right )\,\ln \relax (x)\,\left (3\,x-x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x*log(x) - x + log(5*exp(-3)*log(x))*log(x)*(x - 6) + 3)/(log(x)*(4*x^2 - x^3) - log(5*exp(-3)*log(x))
*log(x)*(3*x - x^2)),x)

[Out]

int(-(4*x*log(x) - x + log(5*exp(-3)*log(x))*log(x)*(x - 6) + 3)/(log(x)*(4*x^2 - x^3) - log(5*exp(-3)*log(x))
*log(x)*(3*x - x^2)), x)

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sympy [A]  time = 0.72, size = 31, normalized size = 1.29 \begin {gather*} - 2 \log {\relax (x )} + \log {\left (x - 3 \right )} + \log {\left (\log {\left (\frac {5 \log {\relax (x )}}{e^{3}} \right )} + \frac {- x^{2} + 4 x}{x - 3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+6)*ln(x)*ln(5*ln(x)/exp(3))-4*x*ln(x)+x-3)/((x**2-3*x)*ln(x)*ln(5*ln(x)/exp(3))+(-x**3+4*x**2)*
ln(x)),x)

[Out]

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

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