\(\int \frac {-3+2 x \log (4 \log ^2(5))}{2 x \log (4 \log ^2(5))} \, dx\) [1745]

Optimal result

Integrand size = 28, antiderivative size = 17 \[ \int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{2 x \log \left (4 \log ^2(5)\right )} \, dx=x-\frac {3 \log (x)}{2 \log \left (4 \log ^2(5)\right )} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(38\) vs. \(2(17)=34\).

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.24, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 45} \[ \int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{2 x \log \left (4 \log ^2(5)\right )} \, dx=\frac {x (\log (16)+4 \log (\log (5)))}{2 \log \left (4 \log ^2(5)\right )}-\frac {3 \log (x)}{2 \log \left (4 \log ^2(5)\right )} \]

[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 45

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{2 x \log \left (4 \log ^2(5)\right )} \, dx=x-\frac {3 \log (x \log (16)+4 x \log (\log (5)))}{\log (16)+4 \log (\log (5))} \]

[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]])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{2 x \log \left (4 \log ^2(5)\right )} \, dx=\frac {2 \, x \log \left (4 \, \log \left (5\right )^{2}\right ) - 3 \, \log \left (x\right )}{2 \, \log \left (4 \, \log \left (5\right )^{2}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71 \[ \int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{2 x \log \left (4 \log ^2(5)\right )} \, dx=\frac {x \left (4 \log {\left (\log {\left (5 \right )} \right )} + 4 \log {\left (2 \right )}\right ) - 3 \log {\left (x \right )}}{4 \log {\left (\log {\left (5 \right )} \right )} + 4 \log {\left (2 \right )}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{2 x \log \left (4 \log ^2(5)\right )} \, dx=\frac {2 \, x \log \left (4 \, \log \left (5\right )^{2}\right ) - 3 \, \log \left (x\right )}{2 \, \log \left (4 \, \log \left (5\right )^{2}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{2 x \log \left (4 \log ^2(5)\right )} \, dx=\frac {2 \, x \log \left (4 \, \log \left (5\right )^{2}\right ) - 3 \, \log \left ({\left | x \right |}\right )}{2 \, \log \left (4 \, \log \left (5\right )^{2}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-3+2 x \log \left (4 \log ^2(5)\right )}{2 x \log \left (4 \log ^2(5)\right )} \, dx=x-\frac {3\,\ln \left (x\right )}{2\,\ln \left (4\,{\ln \left (5\right )}^2\right )} \]

[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))