3.4.65 \(\int \frac {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+(-45 x^2+30 x^3) \log ((3 x-2 x^2) \log (\log (2)))} \, dx\)

Optimal. Leaf size=23 \[ \log \left (\frac {1}{5}-\frac {4}{3 x}+\log ((3-2 x) x \log (\log (2)))\right ) \]

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Rubi [F]  time = 0.64, 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 {-60-5 x+60 x^2}{60 x-49 x^2+6 x^3+\left (-45 x^2+30 x^3\right ) \log \left (\left (3 x-2 x^2\right ) \log (\log (2))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-60 - 5*x + 60*x^2)/(60*x - 49*x^2 + 6*x^3 + (-45*x^2 + 30*x^3)*Log[(3*x - 2*x^2)*Log[Log[2]]]),x]

[Out]

30*Defer[Int][(-20 + 3*x + 15*x*Log[(3 - 2*x)*x*Log[Log[2]]])^(-1), x] + 20*Defer[Int][1/(x*(-20 + 3*x + 15*x*
Log[(3 - 2*x)*x*Log[Log[2]]])), x] + 45*Defer[Int][1/((-3 + 2*x)*(-20 + 3*x + 15*x*Log[(3 - 2*x)*x*Log[Log[2]]
])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-60-5 x+60 x^2}{(3-2 x) x (20-3 x-15 x \log ((3-2 x) x \log (\log (2))))} \, dx\\ &=\int \left (\frac {30}{-20+3 x+15 x \log ((3-2 x) x \log (\log (2)))}+\frac {20}{x (-20+3 x+15 x \log ((3-2 x) x \log (\log (2))))}+\frac {45}{(-3+2 x) (-20+3 x+15 x \log ((3-2 x) x \log (\log (2))))}\right ) \, dx\\ &=20 \int \frac {1}{x (-20+3 x+15 x \log ((3-2 x) x \log (\log (2))))} \, dx+30 \int \frac {1}{-20+3 x+15 x \log ((3-2 x) x \log (\log (2)))} \, dx+45 \int \frac {1}{(-3+2 x) (-20+3 x+15 x \log ((3-2 x) x \log (\log (2))))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 25, normalized size = 1.09 \begin {gather*} -\log (x)+\log (20-3 x-15 x \log ((3-2 x) x \log (\log (2)))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60 - 5*x + 60*x^2)/(60*x - 49*x^2 + 6*x^3 + (-45*x^2 + 30*x^3)*Log[(3*x - 2*x^2)*Log[Log[2]]]),x]

[Out]

-Log[x] + Log[20 - 3*x - 15*x*Log[(3 - 2*x)*x*Log[Log[2]]]]

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fricas [A]  time = 1.30, size = 28, normalized size = 1.22 \begin {gather*} \log \left (\frac {15 \, x \log \left (-{\left (2 \, x^{2} - 3 \, x\right )} \log \left (\log \relax (2)\right )\right ) + 3 \, x - 20}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((60*x^2-5*x-60)/((30*x^3-45*x^2)*log((-2*x^2+3*x)*log(log(2)))+6*x^3-49*x^2+60*x),x, algorithm="fric
as")

[Out]

log((15*x*log(-(2*x^2 - 3*x)*log(log(2))) + 3*x - 20)/x)

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giac [A]  time = 0.31, size = 30, normalized size = 1.30 \begin {gather*} \log \left (15 \, x \log \left (-2 \, x^{2} \log \left (\log \relax (2)\right ) + 3 \, x \log \left (\log \relax (2)\right )\right ) + 3 \, x - 20\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((60*x^2-5*x-60)/((30*x^3-45*x^2)*log((-2*x^2+3*x)*log(log(2)))+6*x^3-49*x^2+60*x),x, algorithm="giac
")

[Out]

log(15*x*log(-2*x^2*log(log(2)) + 3*x*log(log(2))) + 3*x - 20) - log(x)

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maple [A]  time = 0.05, size = 27, normalized size = 1.17

method	result	size
risch	\(\ln \left (\ln \left (\left (-2 x^{2}+3 x \right ) \ln \left (\ln \relax (2)\right )\right )+\frac {3 x -20}{15 x}\right )\)	\(27\)
norman	\(-\ln \relax (x )+\ln \left (15 x \ln \left (\left (-2 x^{2}+3 x \right ) \ln \left (\ln \relax (2)\right )\right )+3 x -20\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*x^2-5*x-60)/((30*x^3-45*x^2)*ln((-2*x^2+3*x)*ln(ln(2)))+6*x^3-49*x^2+60*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln((-2*x^2+3*x)*ln(ln(2)))+1/15/x*(3*x-20))

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maxima [A]  time = 0.95, size = 33, normalized size = 1.43 \begin {gather*} \log \left (\frac {3 \, x {\left (5 \, \log \left (\log \left (\log \relax (2)\right )\right ) + 1\right )} + 15 \, x \log \relax (x) + 15 \, x \log \left (-2 \, x + 3\right ) - 20}{15 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((60*x^2-5*x-60)/((30*x^3-45*x^2)*log((-2*x^2+3*x)*log(log(2)))+6*x^3-49*x^2+60*x),x, algorithm="maxi
ma")

[Out]

log(1/15*(3*x*(5*log(log(log(2))) + 1) + 15*x*log(x) + 15*x*log(-2*x + 3) - 20)/x)

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mupad [B]  time = 2.94, size = 29, normalized size = 1.26 \begin {gather*} \ln \left (3\,x+15\,x\,\left (\ln \left (x\,\left (2\,x-3\right )\right )+\ln \left (-\ln \left (\ln \relax (2)\right )\right )\right )-20\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x - 60*x^2 + 60)/(60*x - log(log(log(2))*(3*x - 2*x^2))*(45*x^2 - 30*x^3) - 49*x^2 + 6*x^3),x)

[Out]

log(3*x + 15*x*(log(x*(2*x - 3)) + log(-log(log(2)))) - 20) - log(x)

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sympy [A]  time = 0.28, size = 24, normalized size = 1.04 \begin {gather*} \log {\left (\log {\left (\left (- 2 x^{2} + 3 x\right ) \log {\left (\log {\relax (2 )} \right )} \right )} + \frac {3 x - 20}{15 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((60*x**2-5*x-60)/((30*x**3-45*x**2)*ln((-2*x**2+3*x)*ln(ln(2)))+6*x**3-49*x**2+60*x),x)

[Out]

log(log((-2*x**2 + 3*x)*log(log(2))) + (3*x - 20)/(15*x))

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