3.94.89 \(\int \frac {-5196890+1036324 \log (3)}{19431075+30540 x+12 x^2+(-7772430-6108 x) \log (3)+777243 \log ^2(3)} \, dx\)

Optimal. Leaf size=21 \[ \frac {5-\frac {4 x}{3}}{5+\frac {2 x}{509}-\log (3)} \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 1981, 27, 32} \begin {gather*} \frac {509 (5105-1018 \log (3))}{3 (2 x+509 (5-\log (3)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5196890 + 1036324*Log[3])/(19431075 + 30540*x + 12*x^2 + (-7772430 - 6108*x)*Log[3] + 777243*Log[3]^2),x
]

[Out]

(509*(5105 - 1018*Log[3]))/(3*(2*x + 509*(5 - Log[3])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left ((1018 (5105-1018 \log (3))) \int \frac {1}{19431075+30540 x+12 x^2+(-7772430-6108 x) \log (3)+777243 \log ^2(3)} \, dx\right )\\ &=-\left ((1018 (5105-1018 \log (3))) \int \frac {1}{12 x^2+6108 x (5-\log (3))+777243 (5-\log (3))^2} \, dx\right )\\ &=-\left ((1018 (5105-1018 \log (3))) \int \frac {1}{3 (2545+2 x-509 \log (3))^2} \, dx\right )\\ &=-\left (\frac {1}{3} (1018 (5105-1018 \log (3))) \int \frac {1}{(2545+2 x-509 \log (3))^2} \, dx\right )\\ &=\frac {509 (5105-1018 \log (3))}{3 (2 x+509 (5-\log (3)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.05 \begin {gather*} -\frac {509 (-5105+1018 \log (3))}{3 (2 x-509 (-5+\log (3)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5196890 + 1036324*Log[3])/(19431075 + 30540*x + 12*x^2 + (-7772430 - 6108*x)*Log[3] + 777243*Log[3
]^2),x]

[Out]

(-509*(-5105 + 1018*Log[3]))/(3*(2*x - 509*(-5 + Log[3])))

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fricas [A]  time = 0.74, size = 19, normalized size = 0.90 \begin {gather*} -\frac {509 \, {\left (1018 \, \log \relax (3) - 5105\right )}}{3 \, {\left (2 \, x - 509 \, \log \relax (3) + 2545\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1036324*log(3)-5196890)/(777243*log(3)^2+(-6108*x-7772430)*log(3)+12*x^2+30540*x+19431075),x, algor
ithm="fricas")

[Out]

-509/3*(1018*log(3) - 5105)/(2*x - 509*log(3) + 2545)

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giac [A]  time = 0.12, size = 19, normalized size = 0.90 \begin {gather*} -\frac {509 \, {\left (1018 \, \log \relax (3) - 5105\right )}}{3 \, {\left (2 \, x - 509 \, \log \relax (3) + 2545\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1036324*log(3)-5196890)/(777243*log(3)^2+(-6108*x-7772430)*log(3)+12*x^2+30540*x+19431075),x, algor
ithm="giac")

[Out]

-509/3*(1018*log(3) - 5105)/(2*x - 509*log(3) + 2545)

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maple [A]  time = 0.63, size = 19, normalized size = 0.90

method	result	size
norman	\(\frac {-\frac {2598445}{3}+\frac {518162 \ln \relax (3)}{3}}{509 \ln \relax (3)-2 x -2545}\)	\(19\)
gosper	\(\frac {-\frac {2598445}{3}+\frac {518162 \ln \relax (3)}{3}}{509 \ln \relax (3)-2 x -2545}\)	\(20\)
default	\(-\frac {1036324 \ln \relax (3)-5196890}{6 \left (-509 \ln \relax (3)+2 x +2545\right )}\)	\(20\)
risch	\(\frac {1018 \ln \relax (3)}{3 \left (\ln \relax (3)-\frac {2 x}{509}-5\right )}-\frac {5105}{3 \left (\ln \relax (3)-\frac {2 x}{509}-5\right )}\)	\(26\)
meijerg	\(-\frac {5105 x}{3 \left (-\frac {509 \ln \relax (3)}{2}+\frac {2545}{2}\right ) \left (1+\frac {2 x}{509 \left (5-\ln \relax (3)\right )}\right ) \left (5-\ln \relax (3)\right )}+\frac {1018 \ln \relax (3) x}{3 \left (-\frac {509 \ln \relax (3)}{2}+\frac {2545}{2}\right ) \left (1+\frac {2 x}{509 \left (5-\ln \relax (3)\right )}\right ) \left (5-\ln \relax (3)\right )}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1036324*ln(3)-5196890)/(777243*ln(3)^2+(-6108*x-7772430)*ln(3)+12*x^2+30540*x+19431075),x,method=_RETURNV
ERBOSE)

[Out]

(-2598445/3+518162/3*ln(3))/(509*ln(3)-2*x-2545)

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maxima [A]  time = 0.36, size = 19, normalized size = 0.90 \begin {gather*} -\frac {509 \, {\left (1018 \, \log \relax (3) - 5105\right )}}{3 \, {\left (2 \, x - 509 \, \log \relax (3) + 2545\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1036324*log(3)-5196890)/(777243*log(3)^2+(-6108*x-7772430)*log(3)+12*x^2+30540*x+19431075),x, algor
ithm="maxima")

[Out]

-509/3*(1018*log(3) - 5105)/(2*x - 509*log(3) + 2545)

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mupad [B]  time = 7.58, size = 19, normalized size = 0.90 \begin {gather*} -\frac {\frac {518162\,\ln \relax (3)}{3}-\frac {2598445}{3}}{2\,x-509\,\ln \relax (3)+2545} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1036324*log(3) - 5196890)/(30540*x - log(3)*(6108*x + 7772430) + 777243*log(3)^2 + 12*x^2 + 19431075),x)

[Out]

-((518162*log(3))/3 - 2598445/3)/(2*x - 509*log(3) + 2545)

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sympy [A]  time = 0.17, size = 17, normalized size = 0.81 \begin {gather*} - \frac {-5196890 + 1036324 \log {\relax (3 )}}{12 x - 3054 \log {\relax (3 )} + 15270} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1036324*ln(3)-5196890)/(777243*ln(3)**2+(-6108*x-7772430)*ln(3)+12*x**2+30540*x+19431075),x)

[Out]

-(-5196890 + 1036324*log(3))/(12*x - 3054*log(3) + 15270)

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