\(\int \frac {2 \log (\frac {2 x}{\log (3)})-\log ^2(\frac {2 x}{\log (3)})}{x^2+(8 x-2 e x) \log ^2(\frac {2 x}{\log (3)})+(16-8 e+e^2) \log ^4(\frac {2 x}{\log (3)})} \, dx\) [5052]

Optimal result

Integrand size = 68, antiderivative size = 19 \[ \int \frac {2 \log \left (\frac {2 x}{\log (3)}\right )-\log ^2\left (\frac {2 x}{\log (3)}\right )}{x^2+(8 x-2 e x) \log ^2\left (\frac {2 x}{\log (3)}\right )+\left (16-8 e+e^2\right ) \log ^4\left (\frac {2 x}{\log (3)}\right )} \, dx=\frac {1}{4-e+\frac {x}{\log ^2\left (\frac {2 x}{\log (3)}\right )}} \]

[Out]

1/(4-exp(1)+x/ln(2*x/ln(3))^2)

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6820, 2627, 6843, 32} \[ \int \frac {2 \log \left (\frac {2 x}{\log (3)}\right )-\log ^2\left (\frac {2 x}{\log (3)}\right )}{x^2+(8 x-2 e x) \log ^2\left (\frac {2 x}{\log (3)}\right )+\left (16-8 e+e^2\right ) \log ^4\left (\frac {2 x}{\log (3)}\right )} \, dx=\frac {1}{\frac {x}{\log ^2\left (\frac {2 x}{\log (3)}\right )}-e+4} \]

[In]

Int[(2*Log[(2*x)/Log[3]] - Log[(2*x)/Log[3]]^2)/(x^2 + (8*x - 2*E*x)*Log[(2*x)/Log[3]]^2 + (16 - 8*E + E^2)*Lo
g[(2*x)/Log[3]]^4),x]

[Out]

(4 - E + x/Log[(2*x)/Log[3]]^2)^(-1)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2627

Int[(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))/(Log[(c_.)*(x_)^(n_.)]^(q_)*(b_.) + (a_.)*(x_))^2, x_Symbol] :> Simp[
(-e)*(Log[c*x^n]/(a*(a*x + b*Log[c*x^n]^q))), x] + Dist[(d + e*n)/a, Int[1/(x*(a*x + b*Log[c*x^n]^q)), x], x]
/; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[d + e*n*q, 0]

Rule 6820

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

Rule 6843

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

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

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {2 \log \left (\frac {2 x}{\log (3)}\right )-\log ^2\left (\frac {2 x}{\log (3)}\right )}{x^2+(8 x-2 e x) \log ^2\left (\frac {2 x}{\log (3)}\right )+\left (16-8 e+e^2\right ) \log ^4\left (\frac {2 x}{\log (3)}\right )} \, dx=-\frac {x}{(-4+e) \left (-x+(-4+e) \log ^2\left (\frac {2 x}{\log (3)}\right )\right )} \]

[In]

Integrate[(2*Log[(2*x)/Log[3]] - Log[(2*x)/Log[3]]^2)/(x^2 + (8*x - 2*E*x)*Log[(2*x)/Log[3]]^2 + (16 - 8*E + E
^2)*Log[(2*x)/Log[3]]^4),x]

[Out]

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

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16

method	result	size
risch	\(-\frac {x}{\left ({\mathrm e}-4\right ) \left ({\mathrm e} \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}-4 \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}-x \right )}\)	\(41\)
norman	\(-\frac {\ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}}{{\mathrm e} \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}-4 \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}-x}\)	\(44\)
parallelrisch	\(-\frac {\ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}}{{\mathrm e} \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}-4 \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}-x}\)	\(44\)
derivativedivides	\(\frac {2 \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}}{-2 \,{\mathrm e} \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}+2 x +8 \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}}\)	\(45\)
default	\(\frac {2 \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}}{-2 \,{\mathrm e} \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}+2 x +8 \ln \left (\frac {2 x}{\ln \left (3\right )}\right )^{2}}\)	\(45\)

[In]

int((-ln(2*x/ln(3))^2+2*ln(2*x/ln(3)))/((exp(1)^2-8*exp(1)+16)*ln(2*x/ln(3))^4+(-2*x*exp(1)+8*x)*ln(2*x/ln(3))
^2+x^2),x,method=_RETURNVERBOSE)

[Out]

-x/(exp(1)-4)/(exp(1)*ln(2*x/ln(3))^2-4*ln(2*x/ln(3))^2-x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {2 \log \left (\frac {2 x}{\log (3)}\right )-\log ^2\left (\frac {2 x}{\log (3)}\right )}{x^2+(8 x-2 e x) \log ^2\left (\frac {2 x}{\log (3)}\right )+\left (16-8 e+e^2\right ) \log ^4\left (\frac {2 x}{\log (3)}\right )} \, dx=-\frac {x}{{\left (e^{2} - 8 \, e + 16\right )} \log \left (\frac {2 \, x}{\log \left (3\right )}\right )^{2} - x e + 4 \, x} \]

[In]

integrate((-log(2*x/log(3))^2+2*log(2*x/log(3)))/((exp(1)^2-8*exp(1)+16)*log(2*x/log(3))^4+(-2*x*exp(1)+8*x)*l
og(2*x/log(3))^2+x^2),x, algorithm="fricas")

[Out]

-x/((e^2 - 8*e + 16)*log(2*x/log(3))^2 - x*e + 4*x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {2 \log \left (\frac {2 x}{\log (3)}\right )-\log ^2\left (\frac {2 x}{\log (3)}\right )}{x^2+(8 x-2 e x) \log ^2\left (\frac {2 x}{\log (3)}\right )+\left (16-8 e+e^2\right ) \log ^4\left (\frac {2 x}{\log (3)}\right )} \, dx=- \frac {x}{- e x + 4 x + \left (- 8 e + e^{2} + 16\right ) \log {\left (\frac {2 x}{\log {\left (3 \right )}} \right )}^{2}} \]

[In]

integrate((-ln(2*x/ln(3))**2+2*ln(2*x/ln(3)))/((exp(1)**2-8*exp(1)+16)*ln(2*x/ln(3))**4+(-2*x*exp(1)+8*x)*ln(2
*x/ln(3))**2+x**2),x)

[Out]

-x/(-E*x + 4*x + (-8*E + exp(2) + 16)*log(2*x/log(3))**2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (20) = 40\).

Time = 0.33 (sec) , antiderivative size = 124, normalized size of antiderivative = 6.53 \[ \int \frac {2 \log \left (\frac {2 x}{\log (3)}\right )-\log ^2\left (\frac {2 x}{\log (3)}\right )}{x^2+(8 x-2 e x) \log ^2\left (\frac {2 x}{\log (3)}\right )+\left (16-8 e+e^2\right ) \log ^4\left (\frac {2 x}{\log (3)}\right )} \, dx=-\frac {x}{{\left (e^{2} - 8 \, e + 16\right )} \log \left (x\right )^{2} - x {\left (e - 4\right )} + {\left (\log \left (2\right )^{2} - 2 \, \log \left (2\right ) \log \left (\log \left (3\right )\right ) + \log \left (\log \left (3\right )\right )^{2}\right )} e^{2} - 8 \, {\left (\log \left (2\right )^{2} - 2 \, \log \left (2\right ) \log \left (\log \left (3\right )\right ) + \log \left (\log \left (3\right )\right )^{2}\right )} e + 16 \, \log \left (2\right )^{2} + 2 \, {\left ({\left (\log \left (2\right ) - \log \left (\log \left (3\right )\right )\right )} e^{2} - 8 \, {\left (\log \left (2\right ) - \log \left (\log \left (3\right )\right )\right )} e + 16 \, \log \left (2\right ) - 16 \, \log \left (\log \left (3\right )\right )\right )} \log \left (x\right ) - 32 \, \log \left (2\right ) \log \left (\log \left (3\right )\right ) + 16 \, \log \left (\log \left (3\right )\right )^{2}} \]

[In]

integrate((-log(2*x/log(3))^2+2*log(2*x/log(3)))/((exp(1)^2-8*exp(1)+16)*log(2*x/log(3))^4+(-2*x*exp(1)+8*x)*l
og(2*x/log(3))^2+x^2),x, algorithm="maxima")

[Out]

-x/((e^2 - 8*e + 16)*log(x)^2 - x*(e - 4) + (log(2)^2 - 2*log(2)*log(log(3)) + log(log(3))^2)*e^2 - 8*(log(2)^
2 - 2*log(2)*log(log(3)) + log(log(3))^2)*e + 16*log(2)^2 + 2*((log(2) - log(log(3)))*e^2 - 8*(log(2) - log(lo
g(3)))*e + 16*log(2) - 16*log(log(3)))*log(x) - 32*log(2)*log(log(3)) + 16*log(log(3))^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (20) = 40\).

Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 5.05 \[ \int \frac {2 \log \left (\frac {2 x}{\log (3)}\right )-\log ^2\left (\frac {2 x}{\log (3)}\right )}{x^2+(8 x-2 e x) \log ^2\left (\frac {2 x}{\log (3)}\right )+\left (16-8 e+e^2\right ) \log ^4\left (\frac {2 x}{\log (3)}\right )} \, dx=-\frac {x}{e^{2} \log \left (2 \, x\right )^{2} - 8 \, e \log \left (2 \, x\right )^{2} - 2 \, e^{2} \log \left (2 \, x\right ) \log \left (\log \left (3\right )\right ) + 16 \, e \log \left (2 \, x\right ) \log \left (\log \left (3\right )\right ) + e^{2} \log \left (\log \left (3\right )\right )^{2} - 8 \, e \log \left (\log \left (3\right )\right )^{2} - x e + 16 \, \log \left (2 \, x\right )^{2} - 32 \, \log \left (2 \, x\right ) \log \left (\log \left (3\right )\right ) + 16 \, \log \left (\log \left (3\right )\right )^{2} + 4 \, x} \]

[In]

integrate((-log(2*x/log(3))^2+2*log(2*x/log(3)))/((exp(1)^2-8*exp(1)+16)*log(2*x/log(3))^4+(-2*x*exp(1)+8*x)*l
og(2*x/log(3))^2+x^2),x, algorithm="giac")

[Out]

-x/(e^2*log(2*x)^2 - 8*e*log(2*x)^2 - 2*e^2*log(2*x)*log(log(3)) + 16*e*log(2*x)*log(log(3)) + e^2*log(log(3))
^2 - 8*e*log(log(3))^2 - x*e + 16*log(2*x)^2 - 32*log(2*x)*log(log(3)) + 16*log(log(3))^2 + 4*x)

Mupad [B] (verification not implemented)

Time = 12.39 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.00 \[ \int \frac {2 \log \left (\frac {2 x}{\log (3)}\right )-\log ^2\left (\frac {2 x}{\log (3)}\right )}{x^2+(8 x-2 e x) \log ^2\left (\frac {2 x}{\log (3)}\right )+\left (16-8 e+e^2\right ) \log ^4\left (\frac {2 x}{\log (3)}\right )} \, dx=\frac {x}{\left (\mathrm {e}-4\right )\,\left (x-{\ln \left (\frac {2\,x}{\ln \left (3\right )}\right )}^2\,\mathrm {e}+4\,{\ln \left (\frac {2\,x}{\ln \left (3\right )}\right )}^2\right )} \]

[In]

int((2*log((2*x)/log(3)) - log((2*x)/log(3))^2)/(log((2*x)/log(3))^2*(8*x - 2*x*exp(1)) + log((2*x)/log(3))^4*
(exp(2) - 8*exp(1) + 16) + x^2),x)

[Out]

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