3.102.28 \(\int \frac {3+(-25+24 x+4 x^2) \log (3)}{-9-3 x+(12-17 x+5 x^2+4 x^3) \log (3)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (\frac {4 (-1+x)^2+x-\frac {3}{\log (3)}}{3+x}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {2074, 628} \begin {gather*} \log \left (x^2 (-\log (81))+7 x \log (3)+3-\log (81)\right )-\log (x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + (-25 + 24*x + 4*x^2)*Log[3])/(-9 - 3*x + (12 - 17*x + 5*x^2 + 4*x^3)*Log[3]),x]

[Out]

-Log[3 + x] + Log[3 + 7*x*Log[3] - Log[81] - x^2*Log[81]]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{-3-x}+\frac {(-7+8 x) \log (3)}{-3-7 x \log (3)+\log (81)+x^2 \log (81)}\right ) \, dx\\ &=-\log (3+x)+\log (3) \int \frac {-7+8 x}{-3-7 x \log (3)+\log (81)+x^2 \log (81)} \, dx\\ &=-\log (3+x)+\log \left (3+7 x \log (3)-\log (81)-x^2 \log (81)\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.16, size = 154, normalized size = 7.00 \begin {gather*} \frac {-\frac {6 \tanh ^{-1}\left (\frac {-7 \log (3)+2 x \log (81)}{\sqrt {49 \log ^2(3)-4 (-3+\log (81)) \log (81)}}\right ) (-1+7 \log (3)) \left (28 \log ^2(3)+17 \log (3) \log (81)-6 \log ^2(81)\right )}{\log (81) \sqrt {49 \log ^2(3)-4 (-3+\log (81)) \log (81)}}+2 (3-61 \log (3)) \log (3+x)+\frac {\left (84 \log ^2(3)-3 \log (81)+\log (3) (-12+101 \log (81))\right ) \log \left (3+7 x \log (3)-\log (81)-x^2 \log (81)\right )}{\log (81)}}{2 (-3+21 \log (3)+10 \log (81))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + (-25 + 24*x + 4*x^2)*Log[3])/(-9 - 3*x + (12 - 17*x + 5*x^2 + 4*x^3)*Log[3]),x]

[Out]

((-6*ArcTanh[(-7*Log[3] + 2*x*Log[81])/Sqrt[49*Log[3]^2 - 4*(-3 + Log[81])*Log[81]]]*(-1 + 7*Log[3])*(28*Log[3
]^2 + 17*Log[3]*Log[81] - 6*Log[81]^2))/(Log[81]*Sqrt[49*Log[3]^2 - 4*(-3 + Log[81])*Log[81]]) + 2*(3 - 61*Log
[3])*Log[3 + x] + ((84*Log[3]^2 - 3*Log[81] + Log[3]*(-12 + 101*Log[81]))*Log[3 + 7*x*Log[3] - Log[81] - x^2*L
og[81]])/Log[81])/(2*(-3 + 21*Log[3] + 10*Log[81]))

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fricas [A]  time = 0.54, size = 23, normalized size = 1.05 \begin {gather*} \log \left ({\left (4 \, x^{2} - 7 \, x + 4\right )} \log \relax (3) - 3\right ) - \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+24*x-25)*log(3)+3)/((4*x^3+5*x^2-17*x+12)*log(3)-3*x-9),x, algorithm="fricas")

[Out]

log((4*x^2 - 7*x + 4)*log(3) - 3) - log(x + 3)

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giac [A]  time = 0.24, size = 28, normalized size = 1.27 \begin {gather*} \log \left ({\left | 4 \, x^{2} \log \relax (3) - 7 \, x \log \relax (3) + 4 \, \log \relax (3) - 3 \right |}\right ) - \log \left ({\left | x + 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+24*x-25)*log(3)+3)/((4*x^3+5*x^2-17*x+12)*log(3)-3*x-9),x, algorithm="giac")

[Out]

log(abs(4*x^2*log(3) - 7*x*log(3) + 4*log(3) - 3)) - log(abs(x + 3))

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maple [A]  time = 0.08, size = 27, normalized size = 1.23

method	result	size
default	\(\ln \left (4 x^{2} \ln \relax (3)-7 x \ln \relax (3)+4 \ln \relax (3)-3\right )-\ln \left (3+x \right )\)	\(27\)
norman	\(\ln \left (4 x^{2} \ln \relax (3)-7 x \ln \relax (3)+4 \ln \relax (3)-3\right )-\ln \left (3+x \right )\)	\(27\)
risch	\(-\ln \left (-3-x \right )+\ln \left (-4 x^{2} \ln \relax (3)+7 x \ln \relax (3)-4 \ln \relax (3)+3\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2+24*x-25)*ln(3)+3)/((4*x^3+5*x^2-17*x+12)*ln(3)-3*x-9),x,method=_RETURNVERBOSE)

[Out]

ln(4*x^2*ln(3)-7*x*ln(3)+4*ln(3)-3)-ln(3+x)

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maxima [A]  time = 0.35, size = 26, normalized size = 1.18 \begin {gather*} \log \left (4 \, x^{2} \log \relax (3) - 7 \, x \log \relax (3) + 4 \, \log \relax (3) - 3\right ) - \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+24*x-25)*log(3)+3)/((4*x^3+5*x^2-17*x+12)*log(3)-3*x-9),x, algorithm="maxima")

[Out]

log(4*x^2*log(3) - 7*x*log(3) + 4*log(3) - 3) - log(x + 3)

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mupad [B]  time = 0.33, size = 26, normalized size = 1.18 \begin {gather*} \ln \left (-8\,\ln \relax (3)\,x^2+14\,\ln \relax (3)\,x-8\,\ln \relax (3)+6\right )-\ln \left (x+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(24*x + 4*x^2 - 25) + 3)/(3*x - log(3)*(5*x^2 - 17*x + 4*x^3 + 12) + 9),x)

[Out]

log(14*x*log(3) - 8*log(3) - 8*x^2*log(3) + 6) - log(x + 3)

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sympy [A]  time = 0.56, size = 26, normalized size = 1.18 \begin {gather*} - \log {\left (x + 3 \right )} + \log {\left (x^{2} - \frac {7 x}{4} + \frac {-3 + 4 \log {\relax (3 )}}{4 \log {\relax (3 )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2+24*x-25)*ln(3)+3)/((4*x**3+5*x**2-17*x+12)*ln(3)-3*x-9),x)

[Out]

-log(x + 3) + log(x**2 - 7*x/4 + (-3 + 4*log(3))/(4*log(3)))

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