\(\int \frac {5+3 x^2 \log (\frac {\log (4)}{2})}{3 x^2} \, dx\) [5719]

Optimal result

Integrand size = 21, antiderivative size = 21 \[ \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{3 x^2} \, dx=\frac {1}{3} \left (25-\frac {5}{x}\right )+x \log \left (\frac {\log (4)}{2}\right ) \]

[Out]

x*ln(ln(2))+25/3-5/3/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 14} \[ \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{3 x^2} \, dx=x \log (\log (2))-\frac {5}{3 x} \]

[In]

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

[Out]

-5/(3*x) + x*Log[Log[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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{x^2} \, dx \\ & = \frac {1}{3} \int \left (\frac {5}{x^2}+3 \log (\log (2))\right ) \, dx \\ & = -\frac {5}{3 x}+x \log (\log (2)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{3 x^2} \, dx=-\frac {5}{3 x}+x \log (\log (2)) \]

[In]

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

[Out]

-5/(3*x) + x*Log[Log[2]]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57

method	result	size
default	\(x \ln \left (\ln \left (2\right )\right )-\frac {5}{3 x}\)	\(12\)
risch	\(x \ln \left (\ln \left (2\right )\right )-\frac {5}{3 x}\)	\(12\)
norman	\(\frac {-\frac {5}{3}+x^{2} \ln \left (\ln \left (2\right )\right )}{x}\)	\(14\)
gosper	\(\frac {3 x^{2} \ln \left (\ln \left (2\right )\right )-5}{3 x}\)	\(16\)
parallelrisch	\(\frac {3 x^{2} \ln \left (\ln \left (2\right )\right )-5}{3 x}\)	\(16\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{3 x^2} \, dx=\frac {3 \, x^{2} \log \left (\log \left (2\right )\right ) - 5}{3 \, x} \]

[In]

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

[Out]

1/3*(3*x^2*log(log(2)) - 5)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.48 \[ \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{3 x^2} \, dx=x \log {\left (\log {\left (2 \right )} \right )} - \frac {5}{3 x} \]

[In]

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

[Out]

x*log(log(2)) - 5/(3*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52 \[ \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{3 x^2} \, dx=x \log \left (\log \left (2\right )\right ) - \frac {5}{3 \, x} \]

[In]

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

[Out]

x*log(log(2)) - 5/3/x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52 \[ \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{3 x^2} \, dx=x \log \left (\log \left (2\right )\right ) - \frac {5}{3 \, x} \]

[In]

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

[Out]

x*log(log(2)) - 5/3/x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52 \[ \int \frac {5+3 x^2 \log \left (\frac {\log (4)}{2}\right )}{3 x^2} \, dx=x\,\ln \left (\ln \left (2\right )\right )-\frac {5}{3\,x} \]

[In]

int((x^2*log(log(2)) + 5/3)/x^2,x)

[Out]

x*log(log(2)) - 5/(3*x)