3.83.13 \(\int \frac {8 x-4 \log (9)-8 x \log (x)}{(-8 x^2+4 x \log (9)) \log (x)+e^{4096} x \log ^2(x)} \, dx\)

Optimal. Leaf size=19 \[ \log \left (e^{4096}-\frac {4 (2 x-\log (9))}{\log (x)}\right ) \]

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Rubi [A]  time = 0.45, antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6741, 12, 6742, 2302, 29, 6684} \begin {gather*} \log \left (8 x-e^{4096} \log (x)-4 \log (9)\right )-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*x - 4*Log[9] - 8*x*Log[x])/((-8*x^2 + 4*x*Log[9])*Log[x] + E^4096*x*Log[x]^2),x]

[Out]

-Log[Log[x]] + Log[8*x - 4*Log[9] - E^4096*Log[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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 (-2 x+\log (9)+2 x \log (x))}{x \log (x) \left (8 x-4 \log (9)-e^{4096} \log (x)\right )} \, dx\\ &=4 \int \frac {-2 x+\log (9)+2 x \log (x)}{x \log (x) \left (8 x-4 \log (9)-e^{4096} \log (x)\right )} \, dx\\ &=4 \int \left (-\frac {1}{4 x \log (x)}+\frac {e^{4096}-8 x}{4 x \left (-8 x+4 \log (9)+e^{4096} \log (x)\right )}\right ) \, dx\\ &=-\int \frac {1}{x \log (x)} \, dx+\int \frac {e^{4096}-8 x}{x \left (-8 x+4 \log (9)+e^{4096} \log (x)\right )} \, dx\\ &=\log \left (8 x-4 \log (9)-e^{4096} \log (x)\right )-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-\log (\log (x))+\log \left (8 x-4 \log (9)-e^{4096} \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 30, normalized size = 1.58 \begin {gather*} 4 \left (-\frac {1}{4} \log (\log (x))+\frac {1}{4} \log \left (8 x-4 \log (9)-e^{4096} \log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x - 4*Log[9] - 8*x*Log[x])/((-8*x^2 + 4*x*Log[9])*Log[x] + E^4096*x*Log[x]^2),x]

[Out]

4*(-1/4*Log[Log[x]] + Log[8*x - 4*Log[9] - E^4096*Log[x]]/4)

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fricas [A]  time = 0.65, size = 20, normalized size = 1.05 \begin {gather*} \log \left (e^{4096} \log \relax (x) - 8 \, x + 8 \, \log \relax (3)\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(x)-8*log(3)+8*x)/(x*exp(4096)*log(x)^2+(8*x*log(3)-8*x^2)*log(x)),x, algorithm="fricas")

[Out]

log(e^4096*log(x) - 8*x + 8*log(3)) - log(log(x))

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giac [A]  time = 0.19, size = 21, normalized size = 1.11 \begin {gather*} \log \left (-e^{4096} \log \relax (x) + 8 \, x - 8 \, \log \relax (3)\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(x)-8*log(3)+8*x)/(x*exp(4096)*log(x)^2+(8*x*log(3)-8*x^2)*log(x)),x, algorithm="giac")

[Out]

log(-e^4096*log(x) + 8*x - 8*log(3)) - log(log(x))

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maple [A]  time = 0.05, size = 21, normalized size = 1.11

method	result	size
norman	\(-\ln \left (\ln \relax (x )\right )+\ln \left ({\mathrm e}^{4096} \ln \relax (x )+8 \ln \relax (3)-8 x \right )\)	\(21\)
risch	\(-\ln \left (\ln \relax (x )\right )+\ln \left (\ln \relax (x )+8 \left (\ln \relax (3)-x \right ) {\mathrm e}^{-4096}\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x*ln(x)-8*ln(3)+8*x)/(x*exp(4096)*ln(x)^2+(8*x*ln(3)-8*x^2)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(x))+ln(exp(4096)*ln(x)+8*ln(3)-8*x)

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maxima [A]  time = 0.51, size = 23, normalized size = 1.21 \begin {gather*} \log \left ({\left (e^{4096} \log \relax (x) - 8 \, x + 8 \, \log \relax (3)\right )} e^{\left (-4096\right )}\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*log(x)-8*log(3)+8*x)/(x*exp(4096)*log(x)^2+(8*x*log(3)-8*x^2)*log(x)),x, algorithm="maxima")

[Out]

log((e^4096*log(x) - 8*x + 8*log(3))*e^(-4096)) - log(log(x))

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mupad [B]  time = 6.24, size = 19, normalized size = 1.00 \begin {gather*} \ln \left (x-\ln \relax (3)-\frac {{\mathrm {e}}^{4096}\,\ln \relax (x)}{8}\right )-\ln \left (\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*log(3) - 8*x + 8*x*log(x))/(log(x)*(8*x*log(3) - 8*x^2) + x*exp(4096)*log(x)^2),x)

[Out]

log(x - log(3) - (exp(4096)*log(x))/8) - log(log(x))

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sympy [A]  time = 0.32, size = 20, normalized size = 1.05 \begin {gather*} \log {\left (\frac {- 8 x + 8 \log {\relax (3 )}}{e^{4096}} + \log {\relax (x )} \right )} - \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*ln(x)-8*ln(3)+8*x)/(x*exp(4096)*ln(x)**2+(8*x*ln(3)-8*x**2)*ln(x)),x)

[Out]

log((-8*x + 8*log(3))*exp(-4096) + log(x)) - log(log(x))

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