3.18.45 \(\int \frac {-3+2 x \log (4 \log ^2(5))}{2 x \log (4 \log ^2(5))} \, dx\)

Optimal. Leaf size=17 \[ x-\frac {3 \log (x)}{2 \log \left (4 \log ^2(5)\right )} \]

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Rubi [B]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 2.24, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 43} \begin {gather*} \frac {x (\log (16)+4 \log (\log (5)))}{2 \log \left (4 \log ^2(5)\right )}-\frac {3 \log (x)}{2 \log \left (4 \log ^2(5)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + 2*x*Log[4*Log[5]^2])/(2*x*Log[4*Log[5]^2]),x]

[Out]

(-3*Log[x])/(2*Log[4*Log[5]^2]) + (x*(Log[16] + 4*Log[Log[5]]))/(2*Log[4*Log[5]^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{x} \, dx}{2 \log \left (4 \log ^2(5)\right )}\\ &=\frac {\int \left (-\frac {3}{x}+\log (16)+4 \log (\log (5))\right ) \, dx}{2 \log \left (4 \log ^2(5)\right )}\\ &=-\frac {3 \log (x)}{2 \log \left (4 \log ^2(5)\right )}+\frac {x (\log (16)+4 \log (\log (5)))}{2 \log \left (4 \log ^2(5)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.53 \begin {gather*} x-\frac {3 \log (x \log (16)+4 x \log (\log (5)))}{\log (16)+4 \log (\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + 2*x*Log[4*Log[5]^2])/(2*x*Log[4*Log[5]^2]),x]

[Out]

x - (3*Log[x*Log[16] + 4*x*Log[Log[5]]])/(Log[16] + 4*Log[Log[5]])

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fricas [A]  time = 0.67, size = 26, normalized size = 1.53 \begin {gather*} \frac {2 \, x \log \left (4 \, \log \relax (5)^{2}\right ) - 3 \, \log \relax (x)}{2 \, \log \left (4 \, \log \relax (5)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x*log(4*log(5)^2)-3)/x/log(4*log(5)^2),x, algorithm="fricas")

[Out]

1/2*(2*x*log(4*log(5)^2) - 3*log(x))/log(4*log(5)^2)

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giac [A]  time = 0.22, size = 27, normalized size = 1.59 \begin {gather*} \frac {2 \, x \log \left (4 \, \log \relax (5)^{2}\right ) - 3 \, \log \left ({\left | x \right |}\right )}{2 \, \log \left (4 \, \log \relax (5)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x*log(4*log(5)^2)-3)/x/log(4*log(5)^2),x, algorithm="giac")

[Out]

1/2*(2*x*log(4*log(5)^2) - 3*log(abs(x)))/log(4*log(5)^2)

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maple [A]  time = 0.05, size = 15, normalized size = 0.88

method	result	size
norman	\(x -\frac {3 \ln \relax (x )}{4 \left (\ln \relax (2)+\ln \left (\ln \relax (5)\right )\right )}\)	\(15\)
default	\(\frac {2 x \ln \left (4 \ln \relax (5)^{2}\right )-3 \ln \relax (x )}{2 \ln \left (4 \ln \relax (5)^{2}\right )}\)	\(27\)
risch	\(\frac {2 x \ln \relax (2)}{2 \ln \relax (2)+2 \ln \left (\ln \relax (5)\right )}+\frac {2 x \ln \left (\ln \relax (5)\right )}{2 \ln \relax (2)+2 \ln \left (\ln \relax (5)\right )}-\frac {3 \ln \relax (x )}{2 \left (2 \ln \relax (2)+2 \ln \left (\ln \relax (5)\right )\right )}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*x*ln(4*ln(5)^2)-3)/x/ln(4*ln(5)^2),x,method=_RETURNVERBOSE)

[Out]

x-3/4/(ln(2)+ln(ln(5)))*ln(x)

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maxima [A]  time = 0.39, size = 26, normalized size = 1.53 \begin {gather*} \frac {2 \, x \log \left (4 \, \log \relax (5)^{2}\right ) - 3 \, \log \relax (x)}{2 \, \log \left (4 \, \log \relax (5)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x*log(4*log(5)^2)-3)/x/log(4*log(5)^2),x, algorithm="maxima")

[Out]

1/2*(2*x*log(4*log(5)^2) - 3*log(x))/log(4*log(5)^2)

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mupad [B]  time = 0.05, size = 15, normalized size = 0.88 \begin {gather*} x-\frac {3\,\ln \relax (x)}{2\,\ln \left (4\,{\ln \relax (5)}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(4*log(5)^2) - 3/2)/(x*log(4*log(5)^2)),x)

[Out]

x - (3*log(x))/(2*log(4*log(5)^2))

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sympy [A]  time = 0.10, size = 29, normalized size = 1.71 \begin {gather*} \frac {x \left (4 \log {\left (\log {\relax (5 )} \right )} + 4 \log {\relax (2 )}\right ) - 3 \log {\relax (x )}}{4 \log {\left (\log {\relax (5 )} \right )} + 4 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x*ln(4*ln(5)**2)-3)/x/ln(4*ln(5)**2),x)

[Out]

(x*(4*log(log(5)) + 4*log(2)) - 3*log(x))/(4*log(log(5)) + 4*log(2))

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