\(\int \frac {5+(-3+2 x^2) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx\) [6146]

Optimal result

Integrand size = 24, antiderivative size = 23 \[ \int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx=\frac {3}{x}+2 x-\frac {3+\frac {5}{x}}{\log ^2(\log (3))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14} \[ \int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx=2 x+\frac {3-\frac {5}{\log ^2(\log (3))}}{x} \]

[In]

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

[Out]

2*x + (3 - 5/Log[Log[3]]^2)/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 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 {\int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2} \, dx}{\log ^2(\log (3))} \\ & = \frac {\int \left (2 \log ^2(\log (3))+\frac {5-3 \log ^2(\log (3))}{x^2}\right ) \, dx}{\log ^2(\log (3))} \\ & = 2 x+\frac {3-\frac {5}{\log ^2(\log (3))}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx=\frac {3}{x}+2 x-\frac {5}{x \log ^2(\log (3))} \]

[In]

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

[Out]

3/x + 2*x - 5/(x*Log[Log[3]]^2)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\(2 x +\frac {3}{x}-\frac {5}{\ln \left (\ln \left (3\right )\right )^{2} x}\)	\(20\)
gosper	\(\frac {2 x^{2} \ln \left (\ln \left (3\right )\right )^{2}-5+3 \ln \left (\ln \left (3\right )\right )^{2}}{\ln \left (\ln \left (3\right )\right )^{2} x}\)	\(29\)
parallelrisch	\(\frac {2 x^{2} \ln \left (\ln \left (3\right )\right )^{2}-5+3 \ln \left (\ln \left (3\right )\right )^{2}}{\ln \left (\ln \left (3\right )\right )^{2} x}\)	\(29\)
default	\(\frac {2 x \ln \left (\ln \left (3\right )\right )^{2}-\frac {-3 \ln \left (\ln \left (3\right )\right )^{2}+5}{x}}{\ln \left (\ln \left (3\right )\right )^{2}}\)	\(30\)
norman	\(\frac {\frac {3 \ln \left (\ln \left (3\right )\right )^{2}-5}{\ln \left (\ln \left (3\right )\right )}+2 x^{2} \ln \left (\ln \left (3\right )\right )}{\ln \left (\ln \left (3\right )\right ) x}\)	\(34\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx=\frac {{\left (2 \, x^{2} + 3\right )} \log \left (\log \left (3\right )\right )^{2} - 5}{x \log \left (\log \left (3\right )\right )^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx=\frac {2 x \log {\left (\log {\left (3 \right )} \right )}^{2} + \frac {-5 + 3 \log {\left (\log {\left (3 \right )} \right )}^{2}}{x}}{\log {\left (\log {\left (3 \right )} \right )}^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx=\frac {2 \, x \log \left (\log \left (3\right )\right )^{2} + \frac {3 \, \log \left (\log \left (3\right )\right )^{2} - 5}{x}}{\log \left (\log \left (3\right )\right )^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx=\frac {2 \, x \log \left (\log \left (3\right )\right )^{2} + \frac {3 \, \log \left (\log \left (3\right )\right )^{2} - 5}{x}}{\log \left (\log \left (3\right )\right )^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {5+\left (-3+2 x^2\right ) \log ^2(\log (3))}{x^2 \log ^2(\log (3))} \, dx=2\,x+\frac {3\,{\ln \left (\ln \left (3\right )\right )}^2-5}{x\,{\ln \left (\ln \left (3\right )\right )}^2} \]

[In]

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

[Out]

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