3.83.73 \(\int \frac {(-12 x^3+24 x^4+(-12 x^3+24 x^4) \log (3)) \log (4)}{9-48 x+64 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {x^3 (1+\log (3)) \log (4)}{2 \left (4-\frac {3}{2 x}\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 27, 1588} \begin {gather*} -\frac {x^4 (1+\log (3)) \log (4)}{3-8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-12*x^3 + 24*x^4 + (-12*x^3 + 24*x^4)*Log[3])*Log[4])/(9 - 48*x + 64*x^2),x]

[Out]

-((x^4*(1 + Log[3])*Log[4])/(3 - 8*x))

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 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (4) \int \frac {-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)}{9-48 x+64 x^2} \, dx\\ &=\log (4) \int \frac {-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)}{(-3+8 x)^2} \, dx\\ &=-\frac {x^4 (1+\log (3)) \log (4)}{3-8 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.12 \begin {gather*} \frac {\left (81-216 x+1024 x^4\right ) (1+\log (3)) \log (4)}{1024 (-3+8 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-12*x^3 + 24*x^4 + (-12*x^3 + 24*x^4)*Log[3])*Log[4])/(9 - 48*x + 64*x^2),x]

[Out]

((81 - 216*x + 1024*x^4)*(1 + Log[3])*Log[4])/(1024*(-3 + 8*x))

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fricas [A]  time = 0.55, size = 38, normalized size = 1.58 \begin {gather*} \frac {{\left (4096 \, x^{4} - 216 \, x + 81\right )} \log \relax (3) \log \relax (2) + {\left (4096 \, x^{4} - 216 \, x + 81\right )} \log \relax (2)}{2048 \, {\left (8 \, x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*((24*x^4-12*x^3)*log(3)+24*x^4-12*x^3)*log(2)/(64*x^2-48*x+9),x, algorithm="fricas")

[Out]

1/2048*((4096*x^4 - 216*x + 81)*log(3)*log(2) + (4096*x^4 - 216*x + 81)*log(2))/(8*x - 3)

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giac [B]  time = 0.12, size = 50, normalized size = 2.08 \begin {gather*} \frac {1}{2048} \, {\left (512 \, x^{3} \log \relax (3) + 512 \, x^{3} + 192 \, x^{2} \log \relax (3) + 192 \, x^{2} + 72 \, x \log \relax (3) + 72 \, x + \frac {81 \, {\left (\log \relax (3) + 1\right )}}{8 \, x - 3}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*((24*x^4-12*x^3)*log(3)+24*x^4-12*x^3)*log(2)/(64*x^2-48*x+9),x, algorithm="giac")

[Out]

1/2048*(512*x^3*log(3) + 512*x^3 + 192*x^2*log(3) + 192*x^2 + 72*x*log(3) + 72*x + 81*(log(3) + 1)/(8*x - 3))*
log(2)

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maple [A]  time = 0.39, size = 19, normalized size = 0.79

method	result	size
gosper	\(\frac {2 x^{4} \left (\ln \relax (3)+1\right ) \ln \relax (2)}{8 x -3}\)	\(19\)
norman	\(\frac {\left (2 \ln \relax (2) \ln \relax (3)+2 \ln \relax (2)\right ) x^{4}}{8 x -3}\)	\(23\)
default	\(2 \ln \relax (2) \left (12 \ln \relax (3)+12\right ) \left (\frac {x^{3}}{96}+\frac {x^{2}}{256}+\frac {3 x}{2048}+\frac {27}{16384 \left (8 x -3\right )}\right )\)	\(34\)
risch	\(\frac {x^{3} \ln \relax (3) \ln \relax (2)}{4}+\frac {3 \ln \relax (2) \ln \relax (3) x^{2}}{32}+\frac {x^{3} \ln \relax (2)}{4}+\frac {9 x \ln \relax (2) \ln \relax (3)}{256}+\frac {3 x^{2} \ln \relax (2)}{32}+\frac {9 x \ln \relax (2)}{256}+\frac {81 \ln \relax (2)}{16384 \left (x -\frac {3}{8}\right )}+\frac {81 \ln \relax (2) \ln \relax (3)}{16384 \left (x -\frac {3}{8}\right )}\)	\(66\)
meijerg	\(-\frac {27 \left (24 \ln \relax (3)+24\right ) \ln \relax (2) \left (-\frac {8 x \left (-\frac {2560}{27} x^{3}-\frac {640}{9} x^{2}-80 x +60\right )}{45 \left (1-\frac {8 x}{3}\right )}-4 \ln \left (1-\frac {8 x}{3}\right )\right )}{16384}+\frac {9 \left (-12 \ln \relax (3)-12\right ) \ln \relax (2) \left (\frac {2 x \left (-\frac {128}{9} x^{2}-16 x +12\right )}{3 \left (1-\frac {8 x}{3}\right )}+3 \ln \left (1-\frac {8 x}{3}\right )\right )}{2048}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*((24*x^4-12*x^3)*ln(3)+24*x^4-12*x^3)*ln(2)/(64*x^2-48*x+9),x,method=_RETURNVERBOSE)

[Out]

2*x^4*(ln(3)+1)*ln(2)/(8*x-3)

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maxima [B]  time = 0.36, size = 43, normalized size = 1.79 \begin {gather*} \frac {1}{2048} \, {\left (512 \, x^{3} {\left (\log \relax (3) + 1\right )} + 192 \, x^{2} {\left (\log \relax (3) + 1\right )} + 72 \, x {\left (\log \relax (3) + 1\right )} + \frac {81 \, {\left (\log \relax (3) + 1\right )}}{8 \, x - 3}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*((24*x^4-12*x^3)*log(3)+24*x^4-12*x^3)*log(2)/(64*x^2-48*x+9),x, algorithm="maxima")

[Out]

1/2048*(512*x^3*(log(3) + 1) + 192*x^2*(log(3) + 1) + 72*x*(log(3) + 1) + 81*(log(3) + 1)/(8*x - 3))*log(2)

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mupad [B]  time = 5.23, size = 51, normalized size = 2.12 \begin {gather*} \frac {\frac {81\,\ln \relax (2)}{8}+\frac {81\,\ln \relax (2)\,\ln \relax (3)}{8}}{2048\,x-768}+\frac {9\,x\,\ln \relax (2)\,\left (\ln \relax (3)+1\right )}{256}+\frac {3\,x^2\,\ln \relax (2)\,\left (\ln \relax (3)+1\right )}{32}+\frac {x^3\,\ln \relax (2)\,\left (\ln \relax (3)+1\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2)*(log(3)*(12*x^3 - 24*x^4) + 12*x^3 - 24*x^4))/(64*x^2 - 48*x + 9),x)

[Out]

((81*log(2))/8 + (81*log(2)*log(3))/8)/(2048*x - 768) + (9*x*log(2)*(log(3) + 1))/256 + (3*x^2*log(2)*(log(3)
+ 1))/32 + (x^3*log(2)*(log(3) + 1))/4

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sympy [B]  time = 0.23, size = 73, normalized size = 3.04 \begin {gather*} x^{3} \left (\frac {\log {\relax (2 )}}{4} + \frac {\log {\relax (2 )} \log {\relax (3 )}}{4}\right ) + x^{2} \left (\frac {3 \log {\relax (2 )}}{32} + \frac {3 \log {\relax (2 )} \log {\relax (3 )}}{32}\right ) + x \left (\frac {9 \log {\relax (2 )}}{256} + \frac {9 \log {\relax (2 )} \log {\relax (3 )}}{256}\right ) + \frac {81 \log {\relax (2 )} + 81 \log {\relax (2 )} \log {\relax (3 )}}{16384 x - 6144} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*((24*x**4-12*x**3)*ln(3)+24*x**4-12*x**3)*ln(2)/(64*x**2-48*x+9),x)

[Out]

x**3*(log(2)/4 + log(2)*log(3)/4) + x**2*(3*log(2)/32 + 3*log(2)*log(3)/32) + x*(9*log(2)/256 + 9*log(2)*log(3
)/256) + (81*log(2) + 81*log(2)*log(3))/(16384*x - 6144)

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