3.64.78 \(\int \frac {2 x^2+8 x \log (\log (50))+(-4+x^2+8 x \log (\log (50))+16 \log ^2(\log (50))) \log (-4+x^2+8 x \log (\log (50))+16 \log ^2(\log (50)))}{-4+x^2+8 x \log (\log (50))+16 \log ^2(\log (50))} \, dx\)

Optimal. Leaf size=14 \[ x \log \left (-4+(x+4 \log (\log (50)))^2\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 24, normalized size of antiderivative = 1.71, number of steps used = 10, number of rules used = 7, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.099, Rules used = {6728, 142, 2523, 12, 773, 632, 31} \begin {gather*} x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^2 + 8*x*Log[Log[50]] + (-4 + x^2 + 8*x*Log[Log[50]] + 16*Log[Log[50]]^2)*Log[-4 + x^2 + 8*x*Log[Log[5
0]] + 16*Log[Log[50]]^2])/(-4 + x^2 + 8*x*Log[Log[50]] + 16*Log[Log[50]]^2),x]

[Out]

x*Log[x^2 + 8*x*Log[Log[50]] - 4*(1 - 4*Log[Log[50]]^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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 142

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h},
x] && (IGtQ[m, 0] || IntegersQ[m, n])

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 x (x+4 \log (\log (50)))}{(-2+x+4 \log (\log (50))) (2+x+4 \log (\log (50)))}+\log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )\right ) \, dx\\ &=2 \int \frac {x (x+4 \log (\log (50)))}{(-2+x+4 \log (\log (50))) (2+x+4 \log (\log (50)))} \, dx+\int \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right ) \, dx\\ &=x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )+2 \int \left (1+\frac {1-2 \log (\log (50))}{-2+x+4 \log (\log (50))}+\frac {-1-2 \log (\log (50))}{2+x+4 \log (\log (50))}\right ) \, dx-\int \frac {2 x (x+4 \log (\log (50)))}{x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )} \, dx\\ &=2 x+2 (1-2 \log (\log (50))) \log (-x+2 (1-2 \log (\log (50))))-2 (1+2 \log (\log (50))) \log (x+2 (1+2 \log (\log (50))))+x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )-2 \int \frac {x (x+4 \log (\log (50)))}{x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )} \, dx\\ &=2 (1-2 \log (\log (50))) \log (-x+2 (1-2 \log (\log (50))))-2 (1+2 \log (\log (50))) \log (x+2 (1+2 \log (\log (50))))+x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )-2 \int \frac {-4 x \log (\log (50))+4 \left (1-4 \log ^2(\log (50))\right )}{x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )} \, dx\\ &=2 (1-2 \log (\log (50))) \log (-x+2 (1-2 \log (\log (50))))-2 (1+2 \log (\log (50))) \log (x+2 (1+2 \log (\log (50))))+x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )-(2 (1-2 \log (\log (50)))) \int \frac {1}{-2+x+4 \log (\log (50))} \, dx+(2 (1+2 \log (\log (50)))) \int \frac {1}{2+x+4 \log (\log (50))} \, dx\\ &=x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 21, normalized size = 1.50 \begin {gather*} x \log \left (-4+x^2+8 x \log (\log (50))+16 \log ^2(\log (50))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + 8*x*Log[Log[50]] + (-4 + x^2 + 8*x*Log[Log[50]] + 16*Log[Log[50]]^2)*Log[-4 + x^2 + 8*x*Log
[Log[50]] + 16*Log[Log[50]]^2])/(-4 + x^2 + 8*x*Log[Log[50]] + 16*Log[Log[50]]^2),x]

[Out]

x*Log[-4 + x^2 + 8*x*Log[Log[50]] + 16*Log[Log[50]]^2]

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fricas [A]  time = 0.60, size = 21, normalized size = 1.50 \begin {gather*} x \log \left (x^{2} + 8 \, x \log \left (\log \left (50\right )\right ) + 16 \, \log \left (\log \left (50\right )\right )^{2} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*log(log(50))^2+8*x*log(log(50))+x^2-4)*log(16*log(log(50))^2+8*x*log(log(50))+x^2-4)+8*x*log(lo
g(50))+2*x^2)/(16*log(log(50))^2+8*x*log(log(50))+x^2-4),x, algorithm="fricas")

[Out]

x*log(x^2 + 8*x*log(log(50)) + 16*log(log(50))^2 - 4)

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giac [A]  time = 0.21, size = 21, normalized size = 1.50 \begin {gather*} x \log \left (x^{2} + 8 \, x \log \left (\log \left (50\right )\right ) + 16 \, \log \left (\log \left (50\right )\right )^{2} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*log(log(50))^2+8*x*log(log(50))+x^2-4)*log(16*log(log(50))^2+8*x*log(log(50))+x^2-4)+8*x*log(lo
g(50))+2*x^2)/(16*log(log(50))^2+8*x*log(log(50))+x^2-4),x, algorithm="giac")

[Out]

x*log(x^2 + 8*x*log(log(50)) + 16*log(log(50))^2 - 4)

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maple [A]  time = 0.53, size = 22, normalized size = 1.57

method	result	size
default	\(x \ln \left (16 \ln \left (\ln \left (50\right )\right )^{2}+8 x \ln \left (\ln \left (50\right )\right )+x^{2}-4\right )\)	\(22\)
norman	\(x \ln \left (16 \ln \left (\ln \left (50\right )\right )^{2}+8 x \ln \left (\ln \left (50\right )\right )+x^{2}-4\right )\)	\(22\)
risch	\(x \ln \left (16 \ln \left (\ln \relax (2)+2 \ln \relax (5)\right )^{2}+8 x \ln \left (\ln \relax (2)+2 \ln \relax (5)\right )+x^{2}-4\right )\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*ln(ln(50))^2+8*x*ln(ln(50))+x^2-4)*ln(16*ln(ln(50))^2+8*x*ln(ln(50))+x^2-4)+8*x*ln(ln(50))+2*x^2)/(16
*ln(ln(50))^2+8*x*ln(ln(50))+x^2-4),x,method=_RETURNVERBOSE)

[Out]

x*ln(16*ln(ln(50))^2+8*x*ln(ln(50))+x^2-4)

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maxima [B]  time = 0.58, size = 148, normalized size = 10.57 \begin {gather*} {\left (x + 4 \, \log \left (2 \, \log \relax (5) + \log \relax (2)\right ) + 2\right )} \log \left (x + 4 \, \log \left (2 \, \log \relax (5) + \log \relax (2)\right ) + 2\right ) + {\left (x + 4 \, \log \left (2 \, \log \relax (5) + \log \relax (2)\right ) - 2\right )} \log \left (x + 4 \, \log \left (2 \, \log \relax (5) + \log \relax (2)\right ) - 2\right ) - 2 \, {\left (4 \, \log \left (\log \left (50\right )\right )^{2} + 4 \, \log \left (\log \left (50\right )\right ) + 1\right )} \log \left (x + 4 \, \log \left (\log \left (50\right )\right ) + 2\right ) + 2 \, {\left (4 \, \log \left (\log \left (50\right )\right )^{2} - 4 \, \log \left (\log \left (50\right )\right ) + 1\right )} \log \left (x + 4 \, \log \left (\log \left (50\right )\right ) - 2\right ) + 4 \, {\left ({\left (2 \, \log \left (\log \left (50\right )\right ) + 1\right )} \log \left (x + 4 \, \log \left (\log \left (50\right )\right ) + 2\right ) - {\left (2 \, \log \left (\log \left (50\right )\right ) - 1\right )} \log \left (x + 4 \, \log \left (\log \left (50\right )\right ) - 2\right )\right )} \log \left (\log \left (50\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*log(log(50))^2+8*x*log(log(50))+x^2-4)*log(16*log(log(50))^2+8*x*log(log(50))+x^2-4)+8*x*log(lo
g(50))+2*x^2)/(16*log(log(50))^2+8*x*log(log(50))+x^2-4),x, algorithm="maxima")

[Out]

(x + 4*log(2*log(5) + log(2)) + 2)*log(x + 4*log(2*log(5) + log(2)) + 2) + (x + 4*log(2*log(5) + log(2)) - 2)*
log(x + 4*log(2*log(5) + log(2)) - 2) - 2*(4*log(log(50))^2 + 4*log(log(50)) + 1)*log(x + 4*log(log(50)) + 2)
+ 2*(4*log(log(50))^2 - 4*log(log(50)) + 1)*log(x + 4*log(log(50)) - 2) + 4*((2*log(log(50)) + 1)*log(x + 4*lo
g(log(50)) + 2) - (2*log(log(50)) - 1)*log(x + 4*log(log(50)) - 2))*log(log(50))

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mupad [B]  time = 0.62, size = 21, normalized size = 1.50 \begin {gather*} x\,\ln \left (x^2+8\,\ln \left (\ln \left (50\right )\right )\,x+16\,{\ln \left (\ln \left (50\right )\right )}^2-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x*log(log(50)) + log(16*log(log(50))^2 + 8*x*log(log(50)) + x^2 - 4)*(16*log(log(50))^2 + 8*x*log(log(5
0)) + x^2 - 4) + 2*x^2)/(16*log(log(50))^2 + 8*x*log(log(50)) + x^2 - 4),x)

[Out]

x*log(16*log(log(50))^2 + 8*x*log(log(50)) + x^2 - 4)

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sympy [A]  time = 0.16, size = 24, normalized size = 1.71 \begin {gather*} x \log {\left (x^{2} + 8 x \log {\left (\log {\left (50 \right )} \right )} - 4 + 16 \log {\left (\log {\left (50 \right )} \right )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*ln(ln(50))**2+8*x*ln(ln(50))+x**2-4)*ln(16*ln(ln(50))**2+8*x*ln(ln(50))+x**2-4)+8*x*ln(ln(50))+
2*x**2)/(16*ln(ln(50))**2+8*x*ln(ln(50))+x**2-4),x)

[Out]

x*log(x**2 + 8*x*log(log(50)) - 4 + 16*log(log(50))**2)

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