\(\int \frac {2-4 \log (x)+(4-2 x+4 x^2) \log (x) \log (\frac {\log (x)}{x^2})}{x \log (x) \log (\frac {\log (x)}{x^2})} \, dx\) [1408]

Optimal result

Integrand size = 43, antiderivative size = 30 \[ \int \frac {2-4 \log (x)+\left (4-2 x+4 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \log \left (\frac {\log (x)}{x^2}\right )} \, dx=\log \left (9 e^{-2 x+2 x^2} x^4 \log ^2(5) \log ^2\left (\frac {\log (x)}{x^2}\right )\right ) \]

[Out]

ln(9*exp(x^2-x)^2*ln(ln(x)/x^2)^2*x^4*ln(5)^2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6874, 14, 6816} \[ \int \frac {2-4 \log (x)+\left (4-2 x+4 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \log \left (\frac {\log (x)}{x^2}\right )} \, dx=2 x^2+2 \log \left (\log \left (\frac {\log (x)}{x^2}\right )\right )-2 x+4 \log (x) \]

[In]

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

[Out]

-2*x + 2*x^2 + 4*Log[x] + 2*Log[Log[Log[x]/x^2]]

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]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {2-4 \log (x)+\left (4-2 x+4 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \log \left (\frac {\log (x)}{x^2}\right )} \, dx=-2 x+2 x^2+4 \log (x)+2 \log \left (\log \left (\frac {\log (x)}{x^2}\right )\right ) \]

[In]

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

[Out]

-2*x + 2*x^2 + 4*Log[x] + 2*Log[Log[Log[x]/x^2]]

Maple [A] (verified)

Time = 2.77 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80

method	result	size
norman	\(4 \ln \left (x \right )-2 x +2 x^{2}+2 \ln \left (\ln \left (\frac {\ln \left (x \right )}{x^{2}}\right )\right )\)	\(24\)
parallelrisch	\(4 \ln \left (x \right )-2 x +2 x^{2}+2 \ln \left (\ln \left (\frac {\ln \left (x \right )}{x^{2}}\right )\right )\)	\(24\)
risch	\(2 x^{2}-2 x +4 \ln \left (x \right )+2 \ln \left (\ln \left (\ln \left (x \right )\right )+\frac {i \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x^{2}}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x^{2}}\right )^{2}+\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x^{2}}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x^{2}}\right )^{3}+4 i \ln \left (x \right )\right )}{2}\right )\)	\(152\)

[In]

int(((4*x^2-2*x+4)*ln(x)*ln(ln(x)/x^2)-4*ln(x)+2)/x/ln(x)/ln(ln(x)/x^2),x,method=_RETURNVERBOSE)

[Out]

4*ln(x)-2*x+2*x^2+2*ln(ln(ln(x)/x^2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {2-4 \log (x)+\left (4-2 x+4 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \log \left (\frac {\log (x)}{x^2}\right )} \, dx=2 \, x^{2} - 2 \, x + 4 \, \log \left (x\right ) + 2 \, \log \left (\log \left (\frac {\log \left (x\right )}{x^{2}}\right )\right ) \]

[In]

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

[Out]

2*x^2 - 2*x + 4*log(x) + 2*log(log(log(x)/x^2))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {2-4 \log (x)+\left (4-2 x+4 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \log \left (\frac {\log (x)}{x^2}\right )} \, dx=2 x^{2} - 2 x + 4 \log {\left (x \right )} + 2 \log {\left (\log {\left (\frac {\log {\left (x \right )}}{x^{2}} \right )} \right )} \]

[In]

integrate(((4*x**2-2*x+4)*ln(x)*ln(ln(x)/x**2)-4*ln(x)+2)/x/ln(x)/ln(ln(x)/x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {2-4 \log (x)+\left (4-2 x+4 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \log \left (\frac {\log (x)}{x^2}\right )} \, dx=2 \, x^{2} - 2 \, x + 4 \, \log \left (x\right ) + 2 \, \log \left (-2 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right )\right ) \]

[In]

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

[Out]

2*x^2 - 2*x + 4*log(x) + 2*log(-2*log(x) + log(log(x)))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {2-4 \log (x)+\left (4-2 x+4 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \log \left (\frac {\log (x)}{x^2}\right )} \, dx=2 \, x^{2} - 2 \, x + 4 \, \log \left (x\right ) + 2 \, \log \left (2 \, \log \left (x\right ) - \log \left (\log \left (x\right )\right )\right ) \]

[In]

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

[Out]

2*x^2 - 2*x + 4*log(x) + 2*log(2*log(x) - log(log(x)))

Mupad [B] (verification not implemented)

Time = 7.90 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {2-4 \log (x)+\left (4-2 x+4 x^2\right ) \log (x) \log \left (\frac {\log (x)}{x^2}\right )}{x \log (x) \log \left (\frac {\log (x)}{x^2}\right )} \, dx=4\,\ln \left (x\right )-2\,x+2\,\ln \left (\ln \left (\frac {\ln \left (x\right )}{x^2}\right )\right )+2\,x^2 \]

[In]

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

[Out]

4*log(x) - 2*x + 2*log(log(log(x)/x^2)) + 2*x^2