\(\int \frac {(-12 x^3+24 x^4+(-12 x^3+24 x^4) \log (3)) \log (4)}{9-48 x+64 x^2} \, dx\) [8273]

Optimal result

Integrand size = 40, antiderivative size = 24 \[ \int \frac {\left (-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)\right ) \log (4)}{9-48 x+64 x^2} \, dx=\frac {x^3 (1+\log (3)) \log (4)}{2 \left (4-\frac {3}{2 x}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 27, 1602} \[ \int \frac {\left (-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)\right ) \log (4)}{9-48 x+64 x^2} \, dx=-\frac {x^4 (1+\log (3)) \log (4)}{3-8 x} \]

[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 1602

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {\left (-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)\right ) \log (4)}{9-48 x+64 x^2} \, dx=\frac {\left (81-216 x+1024 x^4\right ) (1+\log (3)) \log (4)}{1024 (-3+8 x)} \]

[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))

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
gosper	\(\frac {2 x^{4} \left (\ln \left (3\right )+1\right ) \ln \left (2\right )}{8 x -3}\)	\(19\)
norman	\(\frac {\left (2 \ln \left (2\right ) \ln \left (3\right )+2 \ln \left (2\right )\right ) x^{4}}{8 x -3}\)	\(23\)
parallelrisch	\(\frac {\ln \left (2\right ) \left (8 x^{4} \ln \left (3\right )+8 x^{4}\right )}{32 x -12}\)	\(25\)
default	\(2 \ln \left (2\right ) \left (12 \ln \left (3\right )+12\right ) \left (\frac {x^{3}}{96}+\frac {x^{2}}{256}+\frac {3 x}{2048}+\frac {27}{16384 \left (8 x -3\right )}\right )\)	\(34\)
risch	\(\frac {x^{3} \ln \left (3\right ) \ln \left (2\right )}{4}+\frac {3 \ln \left (2\right ) \ln \left (3\right ) x^{2}}{32}+\frac {9 x \ln \left (2\right ) \ln \left (3\right )}{256}+\frac {x^{3} \ln \left (2\right )}{4}+\frac {3 x^{2} \ln \left (2\right )}{32}+\frac {9 x \ln \left (2\right )}{256}+\frac {81 \ln \left (2\right )}{16384 \left (x -\frac {3}{8}\right )}+\frac {81 \ln \left (2\right ) \ln \left (3\right )}{16384 \left (x -\frac {3}{8}\right )}\)	\(66\)
meijerg	\(-\frac {243 \left (\frac {16 \ln \left (3\right )}{3}+\frac {16}{3}\right ) \ln \left (2\right ) \left (-\frac {8 x \left (-\frac {2560}{27} x^{3}-\frac {640}{9} x^{2}-80 x +60\right )}{45 \left (1-\frac {8 x}{3}\right )}-4 \ln \left (1-\frac {8 x}{3}\right )\right )}{32768}+\frac {81 \left (-\frac {8 \ln \left (3\right )}{3}-\frac {8}{3}\right ) \ln \left (2\right ) \left (\frac {2 x \left (-\frac {128}{9} x^{2}-16 x +12\right )}{3 \left (1-\frac {8 x}{3}\right )}+3 \ln \left (1-\frac {8 x}{3}\right )\right )}{4096}\)	\(85\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {\left (-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)\right ) \log (4)}{9-48 x+64 x^2} \, dx=\frac {{\left (4096 \, x^{4} - 216 \, x + 81\right )} \log \left (3\right ) \log \left (2\right ) + {\left (4096 \, x^{4} - 216 \, x + 81\right )} \log \left (2\right )}{2048 \, {\left (8 \, x - 3\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (17) = 34\).

Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.04 \[ \int \frac {\left (-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)\right ) \log (4)}{9-48 x+64 x^2} \, dx=x^{3} \left (\frac {\log {\left (2 \right )}}{4} + \frac {\log {\left (2 \right )} \log {\left (3 \right )}}{4}\right ) + x^{2} \cdot \left (\frac {3 \log {\left (2 \right )}}{32} + \frac {3 \log {\left (2 \right )} \log {\left (3 \right )}}{32}\right ) + x \left (\frac {9 \log {\left (2 \right )}}{256} + \frac {9 \log {\left (2 \right )} \log {\left (3 \right )}}{256}\right ) + \frac {81 \log {\left (2 \right )} + 81 \log {\left (2 \right )} \log {\left (3 \right )}}{16384 x - 6144} \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (20) = 40\).

Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {\left (-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)\right ) \log (4)}{9-48 x+64 x^2} \, dx=\frac {1}{2048} \, {\left (512 \, x^{3} {\left (\log \left (3\right ) + 1\right )} + 192 \, x^{2} {\left (\log \left (3\right ) + 1\right )} + 72 \, x {\left (\log \left (3\right ) + 1\right )} + \frac {81 \, {\left (\log \left (3\right ) + 1\right )}}{8 \, x - 3}\right )} \log \left (2\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (20) = 40\).

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {\left (-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)\right ) \log (4)}{9-48 x+64 x^2} \, dx=\frac {1}{2048} \, {\left (512 \, x^{3} \log \left (3\right ) + 512 \, x^{3} + 192 \, x^{2} \log \left (3\right ) + 192 \, x^{2} + 72 \, x \log \left (3\right ) + 72 \, x + \frac {81 \, {\left (\log \left (3\right ) + 1\right )}}{8 \, x - 3}\right )} \log \left (2\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 13.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {\left (-12 x^3+24 x^4+\left (-12 x^3+24 x^4\right ) \log (3)\right ) \log (4)}{9-48 x+64 x^2} \, dx=\frac {\frac {81\,\ln \left (2\right )}{8}+\frac {81\,\ln \left (2\right )\,\ln \left (3\right )}{8}}{2048\,x-768}+\frac {9\,x\,\ln \left (2\right )\,\left (\ln \left (3\right )+1\right )}{256}+\frac {3\,x^2\,\ln \left (2\right )\,\left (\ln \left (3\right )+1\right )}{32}+\frac {x^3\,\ln \left (2\right )\,\left (\ln \left (3\right )+1\right )}{4} \]

[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