3.12.35 \(\int \frac {64 \log ^2(1+e-\log (16))}{x \log (x)} \, dx\)

Optimal. Leaf size=15 \[ 64 \log ^2(1+e-\log (16)) \log (\log (x)) \]

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2302, 29} \begin {gather*} 64 \log ^2(1+e-\log (16)) \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(64*Log[1 + E - Log[16]]^2)/(x*Log[x]),x]

[Out]

64*Log[1 + E - Log[16]]^2*Log[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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (64 \log ^2(1+e-\log (16))\right ) \int \frac {1}{x \log (x)} \, dx\\ &=\left (64 \log ^2(1+e-\log (16))\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=64 \log ^2(1+e-\log (16)) \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} 64 \log ^2(1+e-\log (16)) \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64*Log[1 + E - Log[16]]^2)/(x*Log[x]),x]

[Out]

64*Log[1 + E - Log[16]]^2*Log[Log[x]]

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fricas [A]  time = 0.70, size = 16, normalized size = 1.07 \begin {gather*} 64 \, \log \left (e - 4 \, \log \relax (2) + 1\right )^{2} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(64*log(-4*log(2)+1+exp(1))^2/x/log(x),x, algorithm="fricas")

[Out]

64*log(e - 4*log(2) + 1)^2*log(log(x))

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giac [A]  time = 0.32, size = 17, normalized size = 1.13 \begin {gather*} 64 \, \log \left (e - 4 \, \log \relax (2) + 1\right )^{2} \log \left ({\left | \log \relax (x) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(64*log(-4*log(2)+1+exp(1))^2/x/log(x),x, algorithm="giac")

[Out]

64*log(e - 4*log(2) + 1)^2*log(abs(log(x)))

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maple [A]  time = 0.03, size = 17, normalized size = 1.13

method	result	size
derivativedivides	\(64 \ln \left (\ln \relax (x )\right ) \ln \left (-4 \ln \relax (2)+1+{\mathrm e}\right )^{2}\)	\(17\)
default	\(64 \ln \left (\ln \relax (x )\right ) \ln \left (-4 \ln \relax (2)+1+{\mathrm e}\right )^{2}\)	\(17\)
norman	\(64 \ln \left (\ln \relax (x )\right ) \ln \left (-4 \ln \relax (2)+1+{\mathrm e}\right )^{2}\)	\(17\)
risch	\(64 \ln \left (\ln \relax (x )\right ) \ln \left (-4 \ln \relax (2)+1+{\mathrm e}\right )^{2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(64*ln(-4*ln(2)+1+exp(1))^2/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

64*ln(ln(x))*ln(-4*ln(2)+1+exp(1))^2

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maxima [A]  time = 0.36, size = 16, normalized size = 1.07 \begin {gather*} 64 \, \log \left (e - 4 \, \log \relax (2) + 1\right )^{2} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(64*log(-4*log(2)+1+exp(1))^2/x/log(x),x, algorithm="maxima")

[Out]

64*log(e - 4*log(2) + 1)^2*log(log(x))

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mupad [B]  time = 0.73, size = 16, normalized size = 1.07 \begin {gather*} 64\,\ln \left (\ln \relax (x)\right )\,{\ln \left (\mathrm {e}-\ln \left (16\right )+1\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((64*log(exp(1) - 4*log(2) + 1)^2)/(x*log(x)),x)

[Out]

64*log(log(x))*log(exp(1) - log(16) + 1)^2

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sympy [A]  time = 0.11, size = 19, normalized size = 1.27 \begin {gather*} 64 \log {\left (- 4 \log {\relax (2 )} + 1 + e \right )}^{2} \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(64*ln(-4*ln(2)+1+exp(1))**2/x/ln(x),x)

[Out]

64*log(-4*log(2) + 1 + E)**2*log(log(x))

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