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

Optimal. Leaf size=19 \[ \frac {1}{4-e+\frac {x}{\log ^2\left (\frac {2 x}{\log (3)}\right )}} \]

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Rubi [A]  time = 0.57, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6688, 2547, 6711, 32} \begin {gather*} \frac {1}{\frac {x}{\log ^2\left (\frac {2 x}{\log (3)}\right )}-e+4} \end {gather*}

Antiderivative was successfully verified.

[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 2547

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 6688

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

Rule 6711

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 {gather*} \begin {aligned} \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\\ &=-\operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 28, normalized size = 1.47 \begin {gather*} -\frac {x}{(-4+e) \left (-x+(-4+e) \log ^2\left (\frac {2 x}{\log (3)}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[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)))

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fricas [A]  time = 0.74, size = 33, normalized size = 1.74 \begin {gather*} -\frac {x}{{\left (e^{2} - 8 \, e + 16\right )} \log \left (\frac {2 \, x}{\log \relax (3)}\right )^{2} - x e + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.31, size = 96, normalized size = 5.05 \begin {gather*} -\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 \relax (3)\right ) + 16 \, e \log \left (2 \, x\right ) \log \left (\log \relax (3)\right ) + e^{2} \log \left (\log \relax (3)\right )^{2} - 8 \, e \log \left (\log \relax (3)\right )^{2} - x e + 16 \, \log \left (2 \, x\right )^{2} - 32 \, \log \left (2 \, x\right ) \log \left (\log \relax (3)\right ) + 16 \, \log \left (\log \relax (3)\right )^{2} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.16, size = 41, normalized size = 2.16

method	result	size
risch	\(-\frac {x}{\left ({\mathrm e}-4\right ) \left ({\mathrm e} \ln \left (\frac {2 x}{\ln \relax (3)}\right )^{2}-4 \ln \left (\frac {2 x}{\ln \relax (3)}\right )^{2}-x \right )}\)	\(41\)
norman	\(-\frac {\ln \left (\frac {2 x}{\ln \relax (3)}\right )^{2}}{{\mathrm e} \ln \left (\frac {2 x}{\ln \relax (3)}\right )^{2}-4 \ln \left (\frac {2 x}{\ln \relax (3)}\right )^{2}-x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [B]  time = 0.49, size = 124, normalized size = 6.53 \begin {gather*} -\frac {x}{{\left (e^{2} - 8 \, e + 16\right )} \log \relax (x)^{2} - x {\left (e - 4\right )} + {\left (\log \relax (2)^{2} - 2 \, \log \relax (2) \log \left (\log \relax (3)\right ) + \log \left (\log \relax (3)\right )^{2}\right )} e^{2} - 8 \, {\left (\log \relax (2)^{2} - 2 \, \log \relax (2) \log \left (\log \relax (3)\right ) + \log \left (\log \relax (3)\right )^{2}\right )} e + 16 \, \log \relax (2)^{2} + 2 \, {\left ({\left (\log \relax (2) - \log \left (\log \relax (3)\right )\right )} e^{2} - 8 \, {\left (\log \relax (2) - \log \left (\log \relax (3)\right )\right )} e + 16 \, \log \relax (2) - 16 \, \log \left (\log \relax (3)\right )\right )} \log \relax (x) - 32 \, \log \relax (2) \log \left (\log \relax (3)\right ) + 16 \, \log \left (\log \relax (3)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.85, size = 38, normalized size = 2.00 \begin {gather*} \frac {x}{\left (\mathrm {e}-4\right )\,\left (x-{\ln \left (\frac {2\,x}{\ln \relax (3)}\right )}^2\,\mathrm {e}+4\,{\ln \left (\frac {2\,x}{\ln \relax (3)}\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.23, size = 31, normalized size = 1.63 \begin {gather*} - \frac {x}{- e x + 4 x + \left (- 8 e + e^{2} + 16\right ) \log {\left (\frac {2 x}{\log {\relax (3 )}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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