3.88.80 \(\int \frac {-3 x-30 x^2+9 x^3+(-10+3 x) \log (\frac {10 \log (3)}{-10+3 x})}{-10 x^3+3 x^4+(-10 x+3 x^2) \log (\frac {10 \log (3)}{-10+3 x})} \, dx\)

Optimal. Leaf size=23 \[ 3+\log \left (x \left (x^2+\log \left (\frac {4 \log (3)}{-4+\frac {6 x}{5}}\right )\right )\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 2, number of rules used = 2, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6741, 6685} \begin {gather*} \log \left (x \left (x^2+\log \left (-\frac {10 \log (3)}{10-3 x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x - 30*x^2 + 9*x^3 + (-10 + 3*x)*Log[(10*Log[3])/(-10 + 3*x)])/(-10*x^3 + 3*x^4 + (-10*x + 3*x^2)*Log[
(10*Log[3])/(-10 + 3*x)]),x]

[Out]

Log[x*(x^2 + Log[(-10*Log[3])/(10 - 3*x)])]

Rule 6685

Int[(u_)/((w_)*(y_)), x_Symbol] :> With[{q = DerivativeDivides[y*w, u, x]}, Simp[q*Log[RemoveContent[y*w, x]],
x] /;  !FalseQ[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 {3 x+30 x^2-9 x^3-(-10+3 x) \log \left (\frac {10 \log (3)}{-10+3 x}\right )}{(10-3 x) x \left (x^2+\log \left (\frac {10 \log (3)}{-10+3 x}\right )\right )} \, dx\\ &=\log \left (x \left (x^2+\log \left (-\frac {10 \log (3)}{10-3 x}\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 20, normalized size = 0.87 \begin {gather*} \log (x)+\log \left (x^2+\log \left (\frac {10 \log (3)}{-10+3 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x - 30*x^2 + 9*x^3 + (-10 + 3*x)*Log[(10*Log[3])/(-10 + 3*x)])/(-10*x^3 + 3*x^4 + (-10*x + 3*x^2
)*Log[(10*Log[3])/(-10 + 3*x)]),x]

[Out]

Log[x] + Log[x^2 + Log[(10*Log[3])/(-10 + 3*x)]]

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fricas [A]  time = 0.69, size = 20, normalized size = 0.87 \begin {gather*} \log \left (x^{2} + \log \left (\frac {10 \, \log \relax (3)}{3 \, x - 10}\right )\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-10)*log(10*log(3)/(3*x-10))+9*x^3-30*x^2-3*x)/((3*x^2-10*x)*log(10*log(3)/(3*x-10))+3*x^4-10*x
^3),x, algorithm="fricas")

[Out]

log(x^2 + log(10*log(3)/(3*x - 10))) + log(x)

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giac [B]  time = 0.22, size = 102, normalized size = 4.43 \begin {gather*} \frac {\log \relax (3) \log \left (-100 \, \log \relax (3)^{2} - \frac {2000 \, \log \relax (3)^{2}}{3 \, x - 10} - \frac {900 \, \log \relax (3)^{2} \log \left (\frac {10 \, \log \relax (3)}{3 \, x - 10}\right )}{{\left (3 \, x - 10\right )}^{2}} - \frac {10000 \, \log \relax (3)^{2}}{{\left (3 \, x - 10\right )}^{2}}\right ) - 3 \, \log \relax (3) \log \left (\frac {10 \, \log \relax (3)}{3 \, x - 10}\right ) + \log \relax (3) \log \left (\frac {10 \, \log \relax (3)}{3 \, x - 10} + \log \relax (3)\right )}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-10)*log(10*log(3)/(3*x-10))+9*x^3-30*x^2-3*x)/((3*x^2-10*x)*log(10*log(3)/(3*x-10))+3*x^4-10*x
^3),x, algorithm="giac")

[Out]

(log(3)*log(-100*log(3)^2 - 2000*log(3)^2/(3*x - 10) - 900*log(3)^2*log(10*log(3)/(3*x - 10))/(3*x - 10)^2 - 1
0000*log(3)^2/(3*x - 10)^2) - 3*log(3)*log(10*log(3)/(3*x - 10)) + log(3)*log(10*log(3)/(3*x - 10) + log(3)))/
log(3)

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maple [A]  time = 0.13, size = 21, normalized size = 0.91

method	result	size
norman	\(\ln \relax (x )+\ln \left (x^{2}+\ln \left (\frac {10 \ln \relax (3)}{3 x -10}\right )\right )\)	\(21\)
risch	\(\ln \relax (x )+\ln \left (x^{2}+\ln \left (\frac {10 \ln \relax (3)}{3 x -10}\right )\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x-10)*ln(10*ln(3)/(3*x-10))+9*x^3-30*x^2-3*x)/((3*x^2-10*x)*ln(10*ln(3)/(3*x-10))+3*x^4-10*x^3),x,meth
od=_RETURNVERBOSE)

[Out]

ln(x)+ln(x^2+ln(10*ln(3)/(3*x-10)))

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maxima [A]  time = 0.48, size = 29, normalized size = 1.26 \begin {gather*} \log \left (-x^{2} - \log \relax (5) - \log \relax (2) + \log \left (3 \, x - 10\right ) - \log \left (\log \relax (3)\right )\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-10)*log(10*log(3)/(3*x-10))+9*x^3-30*x^2-3*x)/((3*x^2-10*x)*log(10*log(3)/(3*x-10))+3*x^4-10*x
^3),x, algorithm="maxima")

[Out]

log(-x^2 - log(5) - log(2) + log(3*x - 10) - log(log(3))) + log(x)

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mupad [B]  time = 5.69, size = 21, normalized size = 0.91 \begin {gather*} \ln \left (\ln \left (\frac {1}{3\,x-10}\right )+\ln \left (10\,\ln \relax (3)\right )+x^2\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x - log((10*log(3))/(3*x - 10))*(3*x - 10) + 30*x^2 - 9*x^3)/(log((10*log(3))/(3*x - 10))*(10*x - 3*x^2
) + 10*x^3 - 3*x^4),x)

[Out]

log(log(1/(3*x - 10)) + log(10*log(3)) + x^2) + log(x)

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sympy [A]  time = 0.22, size = 19, normalized size = 0.83 \begin {gather*} \log {\relax (x )} + \log {\left (x^{2} + \log {\left (\frac {10 \log {\relax (3 )}}{3 x - 10} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-10)*ln(10*ln(3)/(3*x-10))+9*x**3-30*x**2-3*x)/((3*x**2-10*x)*ln(10*ln(3)/(3*x-10))+3*x**4-10*x
**3),x)

[Out]

log(x) + log(x**2 + log(10*log(3)/(3*x - 10)))

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