3.63.24 \(\int \frac {30+3 x^2-24 x^3+(-24-6 x+36 x^2) \log (9)-12 \log ^2(9)}{2 x^2-4 x \log (9)+2 \log ^2(9)} \, dx\)

Optimal. Leaf size=25 \[ \frac {3}{2} (-5+4 x) \left (-1+\frac {-2+x^2}{-x+\log (9)}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 43, normalized size of antiderivative = 1.72, number of steps used = 4, number of rules used = 3, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {27, 12, 1850} \begin {gather*} -6 x^2-\frac {3 (5-4 \log (9)) \left (2-\log ^2(9)\right )}{2 (x-\log (9))}+\frac {3}{2} x (1-4 \log (9)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(30 + 3*x^2 - 24*x^3 + (-24 - 6*x + 36*x^2)*Log[9] - 12*Log[9]^2)/(2*x^2 - 4*x*Log[9] + 2*Log[9]^2),x]

[Out]

-6*x^2 + (3*x*(1 - 4*Log[9]))/2 - (3*(5 - 4*Log[9])*(2 - Log[9]^2))/(2*(x - Log[9]))

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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {30+3 x^2-24 x^3+\left (-24-6 x+36 x^2\right ) \log (9)-12 \log ^2(9)}{2 (x-\log (9))^2} \, dx\\ &=\frac {1}{2} \int \frac {30+3 x^2-24 x^3+\left (-24-6 x+36 x^2\right ) \log (9)-12 \log ^2(9)}{(x-\log (9))^2} \, dx\\ &=\frac {1}{2} \int \left (-24 x-3 (-1+4 \log (9))+\frac {3 (5-4 \log (9)) \left (2-\log ^2(9)\right )}{(x-\log (9))^2}\right ) \, dx\\ &=-6 x^2+\frac {3}{2} x (1-4 \log (9))-\frac {3 (5-4 \log (9)) \left (2-\log ^2(9)\right )}{2 (x-\log (9))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 45, normalized size = 1.80 \begin {gather*} -\frac {3}{2} \left (4 x^2+x (-1+4 \log (9))+\frac {10-8 \log (9)-5 \log ^2(9)+4 \log ^3(9)}{x-\log (9)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30 + 3*x^2 - 24*x^3 + (-24 - 6*x + 36*x^2)*Log[9] - 12*Log[9]^2)/(2*x^2 - 4*x*Log[9] + 2*Log[9]^2),
x]

[Out]

(-3*(4*x^2 + x*(-1 + 4*Log[9]) + (10 - 8*Log[9] - 5*Log[9]^2 + 4*Log[9]^3)/(x - Log[9])))/2

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fricas [A]  time = 0.70, size = 46, normalized size = 1.84 \begin {gather*} -\frac {3 \, {\left (4 \, x^{3} - 4 \, {\left (4 \, x + 5\right )} \log \relax (3)^{2} + 32 \, \log \relax (3)^{3} - x^{2} + 2 \, {\left (x - 8\right )} \log \relax (3) + 10\right )}}{2 \, {\left (x - 2 \, \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*log(3)^2+2*(36*x^2-6*x-24)*log(3)-24*x^3+3*x^2+30)/(8*log(3)^2-8*x*log(3)+2*x^2),x, algorithm="
fricas")

[Out]

-3/2*(4*x^3 - 4*(4*x + 5)*log(3)^2 + 32*log(3)^3 - x^2 + 2*(x - 8)*log(3) + 10)/(x - 2*log(3))

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giac [A]  time = 0.20, size = 42, normalized size = 1.68 \begin {gather*} -6 \, x^{2} - 12 \, x \log \relax (3) + \frac {3}{2} \, x - \frac {3 \, {\left (16 \, \log \relax (3)^{3} - 10 \, \log \relax (3)^{2} - 8 \, \log \relax (3) + 5\right )}}{x - 2 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*log(3)^2+2*(36*x^2-6*x-24)*log(3)-24*x^3+3*x^2+30)/(8*log(3)^2-8*x*log(3)+2*x^2),x, algorithm="
giac")

[Out]

-6*x^2 - 12*x*log(3) + 3/2*x - 3*(16*log(3)^3 - 10*log(3)^2 - 8*log(3) + 5)/(x - 2*log(3))

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maple [A]  time = 0.55, size = 33, normalized size = 1.32

method	result	size
gosper	\(-\frac {3 \left (-4 x^{3}+x^{2}-10+16 \ln \relax (3)^{2}+16 \ln \relax (3)\right )}{2 \left (2 \ln \relax (3)-x \right )}\)	\(33\)
norman	\(\frac {-\frac {3 x^{2}}{2}+6 x^{3}+15-24 \ln \relax (3)^{2}-24 \ln \relax (3)}{2 \ln \relax (3)-x}\)	\(34\)
default	\(-6 x^{2}-12 x \ln \relax (3)+\frac {3 x}{2}-\frac {3 \left (32 \ln \relax (3)^{3}-20 \ln \relax (3)^{2}-16 \ln \relax (3)+10\right )}{2 \left (-2 \ln \relax (3)+x \right )}\)	\(43\)
risch	\(-12 x \ln \relax (3)-6 x^{2}+\frac {3 x}{2}+\frac {24 \ln \relax (3)^{3}}{\ln \relax (3)-\frac {x}{2}}-\frac {15 \ln \relax (3)^{2}}{\ln \relax (3)-\frac {x}{2}}-\frac {12 \ln \relax (3)}{\ln \relax (3)-\frac {x}{2}}+\frac {15}{2 \left (\ln \relax (3)-\frac {x}{2}\right )}\)	\(65\)
meijerg	\(-\frac {6 x}{1-\frac {x}{2 \ln \relax (3)}}+\frac {15 x}{4 \ln \relax (3)^{2} \left (1-\frac {x}{2 \ln \relax (3)}\right )}-48 \ln \relax (3)^{2} \left (\frac {x \left (-\frac {x^{2}}{2 \ln \relax (3)^{2}}-\frac {3 x}{\ln \relax (3)}+12\right )}{8 \ln \relax (3) \left (1-\frac {x}{2 \ln \relax (3)}\right )}+3 \ln \left (1-\frac {x}{2 \ln \relax (3)}\right )\right )-2 \left (36 \ln \relax (3)+\frac {3}{2}\right ) \ln \relax (3) \left (-\frac {x \left (-\frac {3 x}{2 \ln \relax (3)}+6\right )}{6 \ln \relax (3) \left (1-\frac {x}{2 \ln \relax (3)}\right )}-2 \ln \left (1-\frac {x}{2 \ln \relax (3)}\right )\right )-6 \ln \relax (3) \left (\frac {x}{2 \ln \relax (3) \left (1-\frac {x}{2 \ln \relax (3)}\right )}+\ln \left (1-\frac {x}{2 \ln \relax (3)}\right )\right )-\frac {6 x}{\ln \relax (3) \left (1-\frac {x}{2 \ln \relax (3)}\right )}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-48*ln(3)^2+2*(36*x^2-6*x-24)*ln(3)-24*x^3+3*x^2+30)/(8*ln(3)^2-8*x*ln(3)+2*x^2),x,method=_RETURNVERBOSE)

[Out]

-3/2*(-4*x^3+x^2-10+16*ln(3)^2+16*ln(3))/(2*ln(3)-x)

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maxima [A]  time = 0.34, size = 43, normalized size = 1.72 \begin {gather*} -6 \, x^{2} - \frac {3}{2} \, x {\left (8 \, \log \relax (3) - 1\right )} - \frac {3 \, {\left (16 \, \log \relax (3)^{3} - 10 \, \log \relax (3)^{2} - 8 \, \log \relax (3) + 5\right )}}{x - 2 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*log(3)^2+2*(36*x^2-6*x-24)*log(3)-24*x^3+3*x^2+30)/(8*log(3)^2-8*x*log(3)+2*x^2),x, algorithm="
maxima")

[Out]

-6*x^2 - 3/2*x*(8*log(3) - 1) - 3*(16*log(3)^3 - 10*log(3)^2 - 8*log(3) + 5)/(x - 2*log(3))

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mupad [B]  time = 0.15, size = 42, normalized size = 1.68 \begin {gather*} \frac {24\,\ln \relax (3)+30\,{\ln \relax (3)}^2-48\,{\ln \relax (3)}^3-15}{x-2\,\ln \relax (3)}-6\,x^2-x\,\left (12\,\ln \relax (3)-\frac {3}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(3)*(6*x - 36*x^2 + 24) + 48*log(3)^2 - 3*x^2 + 24*x^3 - 30)/(8*log(3)^2 - 8*x*log(3) + 2*x^2),x)

[Out]

(24*log(3) + 30*log(3)^2 - 48*log(3)^3 - 15)/(x - 2*log(3)) - 6*x^2 - x*(12*log(3) - 3/2)

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sympy [B]  time = 0.24, size = 42, normalized size = 1.68 \begin {gather*} - 6 x^{2} - x \left (- \frac {3}{2} + 12 \log {\relax (3 )}\right ) - \frac {- 30 \log {\relax (3 )}^{2} - 24 \log {\relax (3 )} + 15 + 48 \log {\relax (3 )}^{3}}{x - 2 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*ln(3)**2+2*(36*x**2-6*x-24)*ln(3)-24*x**3+3*x**2+30)/(8*ln(3)**2-8*x*ln(3)+2*x**2),x)

[Out]

-6*x**2 - x*(-3/2 + 12*log(3)) - (-30*log(3)**2 - 24*log(3) + 15 + 48*log(3)**3)/(x - 2*log(3))

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