3.44.45 \(\int \frac {-3 \log (144)-3 e^9 x \log (4 x)}{x \log (144) \log (4 x)} \, dx\)

Optimal. Leaf size=22 \[ 5-3 \left (e^4+\frac {e^9 x}{\log (144)}+\log (\log (4 x))\right ) \]

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Rubi [A]  time = 0.28, antiderivative size = 18, normalized size of antiderivative = 0.82, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {12, 6741, 6742, 2302, 29} \begin {gather*} -\frac {3 e^9 x}{\log (144)}-3 \log (\log (4 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*Log[144] - 3*E^9*x*Log[4*x])/(x*Log[144]*Log[4*x]),x]

[Out]

(-3*E^9*x)/Log[144] - 3*Log[Log[4*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 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} &=\frac {\int \frac {-3 \log (144)-3 e^9 x \log (4 x)}{x \log (4 x)} \, dx}{\log (144)}\\ &=\frac {\int \frac {3 \left (-\log (144)-e^9 x \log (4 x)\right )}{x \log (4 x)} \, dx}{\log (144)}\\ &=\frac {3 \int \frac {-\log (144)-e^9 x \log (4 x)}{x \log (4 x)} \, dx}{\log (144)}\\ &=\frac {3 \int \left (-e^9-\frac {\log (144)}{x \log (4 x)}\right ) \, dx}{\log (144)}\\ &=-\frac {3 e^9 x}{\log (144)}-3 \int \frac {1}{x \log (4 x)} \, dx\\ &=-\frac {3 e^9 x}{\log (144)}-3 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (4 x)\right )\\ &=-\frac {3 e^9 x}{\log (144)}-3 \log (\log (4 x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.82 \begin {gather*} -\frac {3 e^9 x}{\log (144)}-3 \log (\log (4 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*Log[144] - 3*E^9*x*Log[4*x])/(x*Log[144]*Log[4*x]),x]

[Out]

(-3*E^9*x)/Log[144] - 3*Log[Log[4*x]]

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fricas [A]  time = 0.95, size = 20, normalized size = 0.91 \begin {gather*} -\frac {3 \, {\left (x e^{9} + 2 \, \log \left (12\right ) \log \left (\log \left (4 \, x\right )\right )\right )}}{2 \, \log \left (12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x*exp(9)*log(4*x)-6*log(12))/x/log(12)/log(4*x),x, algorithm="fricas")

[Out]

-3/2*(x*e^9 + 2*log(12)*log(log(4*x)))/log(12)

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giac [A]  time = 0.15, size = 35, normalized size = 1.59 \begin {gather*} -\frac {3 \, {\left (x e^{9} + 2 \, \log \relax (3) \log \left (2 \, \log \relax (2) + \log \relax (x)\right ) + 4 \, \log \relax (2) \log \left (2 \, \log \relax (2) + \log \relax (x)\right )\right )}}{2 \, \log \left (12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x*exp(9)*log(4*x)-6*log(12))/x/log(12)/log(4*x),x, algorithm="giac")

[Out]

-3/2*(x*e^9 + 2*log(3)*log(2*log(2) + log(x)) + 4*log(2)*log(2*log(2) + log(x)))/log(12)

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maple [A]  time = 0.04, size = 18, normalized size = 0.82

method	result	size
norman	\(-\frac {3 \,{\mathrm e}^{9} x}{2 \ln \left (12\right )}-3 \ln \left (\ln \left (4 x \right )\right )\)	\(18\)
derivativedivides	\(\frac {-\frac {3 x \,{\mathrm e}^{9}}{2}-3 \ln \left (12\right ) \ln \left (\ln \left (4 x \right )\right )}{\ln \left (12\right )}\)	\(22\)
default	\(\frac {-\frac {3 x \,{\mathrm e}^{9}}{2}-3 \ln \left (12\right ) \ln \left (\ln \left (4 x \right )\right )}{\ln \left (12\right )}\)	\(22\)
risch	\(-\frac {3 x \,{\mathrm e}^{9}}{2 \left (\ln \relax (3)+2 \ln \relax (2)\right )}-\frac {6 \ln \left (\ln \left (4 x \right )\right ) \ln \relax (2)}{\ln \relax (3)+2 \ln \relax (2)}-\frac {3 \ln \left (\ln \left (4 x \right )\right ) \ln \relax (3)}{\ln \relax (3)+2 \ln \relax (2)}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-3*x*exp(9)*ln(4*x)-6*ln(12))/x/ln(12)/ln(4*x),x,method=_RETURNVERBOSE)

[Out]

-3/2*exp(9)/ln(12)*x-3*ln(ln(4*x))

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maxima [A]  time = 0.36, size = 20, normalized size = 0.91 \begin {gather*} -\frac {3 \, {\left (x e^{9} + 2 \, \log \left (12\right ) \log \left (\log \left (4 \, x\right )\right )\right )}}{2 \, \log \left (12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x*exp(9)*log(4*x)-6*log(12))/x/log(12)/log(4*x),x, algorithm="maxima")

[Out]

-3/2*(x*e^9 + 2*log(12)*log(log(4*x)))/log(12)

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mupad [B]  time = 3.07, size = 17, normalized size = 0.77 \begin {gather*} -3\,\ln \left (\ln \left (4\,x\right )\right )-\frac {3\,x\,{\mathrm {e}}^9}{2\,\ln \left (12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*log(12) + (3*x*log(4*x)*exp(9))/2)/(x*log(4*x)*log(12)),x)

[Out]

- 3*log(log(4*x)) - (3*x*exp(9))/(2*log(12))

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sympy [A]  time = 0.11, size = 20, normalized size = 0.91 \begin {gather*} - \frac {3 x e^{9}}{2 \log {\left (12 \right )}} - 3 \log {\left (\log {\left (4 x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x*exp(9)*ln(4*x)-6*ln(12))/x/ln(12)/ln(4*x),x)

[Out]

-3*x*exp(9)/(2*log(12)) - 3*log(log(4*x))

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