3.59.40 \(\int \frac {64-24 x+4 x^2+(16-3 x) \log (2)+\log ^2(2)+(16-3 x+2 \log (2)) \log (3)+\log ^2(3)}{64 x-32 x^2+4 x^3+(16 x-4 x^2) \log (2)+x \log ^2(2)+(16 x-4 x^2+2 x \log (2)) \log (3)+x \log ^2(3)} \, dx\)

Optimal. Leaf size=16 \[ \frac {x}{8-2 x+\log (2)+\log (3)}+\log (x) \]

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Rubi [A]  time = 0.12, antiderivative size = 20, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 2, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6, 2074} \begin {gather*} \log (x)+\frac {8+\log (6)}{2 (-2 x+8+\log (6))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(64 - 24*x + 4*x^2 + (16 - 3*x)*Log[2] + Log[2]^2 + (16 - 3*x + 2*Log[2])*Log[3] + Log[3]^2)/(64*x - 32*x^
2 + 4*x^3 + (16*x - 4*x^2)*Log[2] + x*Log[2]^2 + (16*x - 4*x^2 + 2*x*Log[2])*Log[3] + x*Log[3]^2),x]

[Out]

(8 + Log[6])/(2*(8 - 2*x + Log[6])) + Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 \frac {64-24 x+4 x^2+(16-3 x) \log (2)+\log ^2(2)+(16-3 x+2 \log (2)) \log (3)+\log ^2(3)}{-32 x^2+4 x^3+\left (16 x-4 x^2\right ) \log (2)+x \left (64+\log ^2(2)\right )+\left (16 x-4 x^2+2 x \log (2)\right ) \log (3)+x \log ^2(3)} \, dx\\ &=\int \frac {64-24 x+4 x^2+(16-3 x) \log (2)+\log ^2(2)+(16-3 x+2 \log (2)) \log (3)+\log ^2(3)}{-32 x^2+4 x^3+\left (16 x-4 x^2\right ) \log (2)+\left (16 x-4 x^2+2 x \log (2)\right ) \log (3)+x \left (64+\log ^2(2)+\log ^2(3)\right )} \, dx\\ &=\int \left (\frac {1}{x}+\frac {8+\log (6)}{(8-2 x+\log (6))^2}\right ) \, dx\\ &=\frac {8+\log (6)}{2 (8-2 x+\log (6))}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 86, normalized size = 5.38 \begin {gather*} \frac {-\frac {(8+\log (6)) \left (64+2 \log ^2(2)+2 \log ^2(3)+2 \log ^2(6)+\log (4) \log (9)-\log (6) \log (216)+\log (2821109907456)\right )}{-8+2 x-\log (6)}+2 \left (64+16 \log (2)+\log ^2(2)+\log ^2(3)+\log (3) (16+\log (4))\right ) \log (x)}{2 (8+\log (6))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64 - 24*x + 4*x^2 + (16 - 3*x)*Log[2] + Log[2]^2 + (16 - 3*x + 2*Log[2])*Log[3] + Log[3]^2)/(64*x -
32*x^2 + 4*x^3 + (16*x - 4*x^2)*Log[2] + x*Log[2]^2 + (16*x - 4*x^2 + 2*x*Log[2])*Log[3] + x*Log[3]^2),x]

[Out]

(-(((8 + Log[6])*(64 + 2*Log[2]^2 + 2*Log[3]^2 + 2*Log[6]^2 + Log[4]*Log[9] - Log[6]*Log[216] + Log[2821109907
456]))/(-8 + 2*x - Log[6])) + 2*(64 + 16*Log[2] + Log[2]^2 + Log[3]^2 + Log[3]*(16 + Log[4]))*Log[x])/(2*(8 +
Log[6])^2)

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fricas [B]  time = 0.75, size = 44, normalized size = 2.75 \begin {gather*} \frac {2 \, {\left (2 \, x - \log \relax (3) - \log \relax (2) - 8\right )} \log \relax (x) - \log \relax (3) - \log \relax (2) - 8}{2 \, {\left (2 \, x - \log \relax (3) - \log \relax (2) - 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)^2+(2*log(2)-3*x+16)*log(3)+log(2)^2+(-3*x+16)*log(2)+4*x^2-24*x+64)/(x*log(3)^2+(2*x*log(2)-
4*x^2+16*x)*log(3)+x*log(2)^2+(-4*x^2+16*x)*log(2)+4*x^3-32*x^2+64*x),x, algorithm="fricas")

[Out]

1/2*(2*(2*x - log(3) - log(2) - 8)*log(x) - log(3) - log(2) - 8)/(2*x - log(3) - log(2) - 8)

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giac [A]  time = 0.14, size = 27, normalized size = 1.69 \begin {gather*} -\frac {\log \relax (3) + \log \relax (2) + 8}{2 \, {\left (2 \, x - \log \relax (3) - \log \relax (2) - 8\right )}} + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)^2+(2*log(2)-3*x+16)*log(3)+log(2)^2+(-3*x+16)*log(2)+4*x^2-24*x+64)/(x*log(3)^2+(2*x*log(2)-
4*x^2+16*x)*log(3)+x*log(2)^2+(-4*x^2+16*x)*log(2)+4*x^3-32*x^2+64*x),x, algorithm="giac")

[Out]

-1/2*(log(3) + log(2) + 8)/(2*x - log(3) - log(2) - 8) + log(abs(x))

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maple [A]  time = 0.11, size = 26, normalized size = 1.62

method	result	size
norman	\(\frac {\frac {\ln \relax (3)}{2}+\frac {\ln \relax (2)}{2}+4}{\ln \relax (3)+8-2 x +\ln \relax (2)}+\ln \relax (x )\)	\(26\)
default	\(-\frac {\frac {\ln \relax (3)}{2}+\frac {\ln \relax (2)}{2}+4}{-\ln \relax (3)-8+2 x -\ln \relax (2)}+\ln \relax (x )\)	\(31\)
risch	\(\frac {\ln \relax (3)}{2 \ln \relax (3)+16-4 x +2 \ln \relax (2)}+\frac {\ln \relax (2)}{2 \ln \relax (3)+16-4 x +2 \ln \relax (2)}+\frac {4}{\ln \relax (3)+8-2 x +\ln \relax (2)}+\ln \relax (x )\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(3)^2+(2*ln(2)-3*x+16)*ln(3)+ln(2)^2+(-3*x+16)*ln(2)+4*x^2-24*x+64)/(x*ln(3)^2+(2*x*ln(2)-4*x^2+16*x)*l
n(3)+x*ln(2)^2+(-4*x^2+16*x)*ln(2)+4*x^3-32*x^2+64*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 26, normalized size = 1.62 \begin {gather*} -\frac {\log \relax (3) + \log \relax (2) + 8}{2 \, {\left (2 \, x - \log \relax (3) - \log \relax (2) - 8\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)^2+(2*log(2)-3*x+16)*log(3)+log(2)^2+(-3*x+16)*log(2)+4*x^2-24*x+64)/(x*log(3)^2+(2*x*log(2)-
4*x^2+16*x)*log(3)+x*log(2)^2+(-4*x^2+16*x)*log(2)+4*x^3-32*x^2+64*x),x, algorithm="maxima")

[Out]

-1/2*(log(3) + log(2) + 8)/(2*x - log(3) - log(2) - 8) + log(x)

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mupad [B]  time = 4.51, size = 114, normalized size = 7.12 \begin {gather*} \frac {\ln \relax (x)\,\left (16\,\ln \relax (6)+2\,\ln \relax (2)\,\ln \relax (3)+{\ln \relax (2)}^2+{\ln \relax (3)}^2+64\right )}{{\left (\ln \relax (6)+8\right )}^2}+\frac {16\,\ln \relax (6)+4\,\ln \relax (2)\,\ln \relax (3)+2\,{\ln \relax (2)}^2+2\,{\ln \relax (3)}^2-{\ln \relax (6)}^2+64}{2\,\left (\ln \relax (6)+8\right )\,\left (\ln \relax (6)-2\,x+8\right )}-\frac {\ln \left (x-\frac {\ln \relax (6)}{2}-4\right )\,\left (2\,\ln \relax (2)\,\ln \relax (3)+{\ln \relax (2)}^2+{\ln \relax (3)}^2-{\ln \relax (6)}^2\right )}{{\left (\ln \relax (6)+8\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)^2 - log(2)*(3*x - 16) - 24*x + log(3)^2 + log(3)*(2*log(2) - 3*x + 16) + 4*x^2 + 64)/(64*x + log(2
)*(16*x - 4*x^2) + log(3)*(16*x + 2*x*log(2) - 4*x^2) + x*log(2)^2 + x*log(3)^2 - 32*x^2 + 4*x^3),x)

[Out]

(log(x)*(16*log(6) + 2*log(2)*log(3) + log(2)^2 + log(3)^2 + 64))/(log(6) + 8)^2 + (16*log(6) + 4*log(2)*log(3
) + 2*log(2)^2 + 2*log(3)^2 - log(6)^2 + 64)/(2*(log(6) + 8)*(log(6) - 2*x + 8)) - (log(x - log(6)/2 - 4)*(2*l
og(2)*log(3) + log(2)^2 + log(3)^2 - log(6)^2))/(log(6) + 8)^2

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sympy [A]  time = 0.98, size = 27, normalized size = 1.69 \begin {gather*} \log {\relax (x )} + \frac {-8 - \log {\relax (3 )} - \log {\relax (2 )}}{4 x - 16 - 2 \log {\relax (3 )} - 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(3)**2+(2*ln(2)-3*x+16)*ln(3)+ln(2)**2+(-3*x+16)*ln(2)+4*x**2-24*x+64)/(x*ln(3)**2+(2*x*ln(2)-4*x
**2+16*x)*ln(3)+x*ln(2)**2+(-4*x**2+16*x)*ln(2)+4*x**3-32*x**2+64*x),x)

[Out]

log(x) + (-8 - log(3) - log(2))/(4*x - 16 - 2*log(3) - 2*log(2))

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