3.6.74 \(\int \frac {-8 x-4 x^2+(8+4 x) \log (3)+(-8 x-4 x \log (3)) \log (x)+(-4 x-2 x^2+(4+2 x) \log (3)) \log (\frac {x^2-2 x \log (3)+\log ^2(3)}{4+4 x+x^2})}{-2 x^2-x^3+(2 x+x^2) \log (3)} \, dx\) [574]

Optimal. Leaf size=29 \[ 2 \left (1+\log (x)-\log (x) \left (-1-\log \left (\frac {(x-\log (3))^2}{(2+x)^2}\right )\right )\right ) \]

[Out]

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

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Rubi [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 0.21, antiderivative size = 153, normalized size of antiderivative = 5.28, number of steps used = 13, number of rules used = 7, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6820, 2404, 2354, 2438, 2353, 2352, 2546} \begin {gather*} 4 \text {Li}_2\left (-\frac {x}{2}\right )-\frac {(8+\log (81)) \text {Li}_2\left (-\frac {x}{2}\right )}{2+\log (3)}-\frac {(8+\log (81)) \text {Li}_2\left (1-\frac {x}{\log (3)}\right )}{2+\log (3)}+4 \text {Li}_2\left (1-\frac {x}{\log (3)}\right )-\frac {(8+\log (81)) \log \left (\frac {x}{2}+1\right ) \log (x)}{2+\log (3)}+4 \log \left (\frac {x}{2}+1\right ) \log (x)+2 \left (\log \left (\frac {(x-\log (3))^2}{(x+2)^2}\right )+2\right ) \log (x)+\frac {(8+\log (81)) \log (\log (3)) \log (x-\log (3))}{2+\log (3)}-4 \log (\log (3)) \log (x-\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4*Log[1 + x/2]*Log[x] - ((8 + Log[81])*Log[1 + x/2]*Log[x])/(2 + Log[3]) + 2*Log[x]*(2 + Log[(x - Log[3])^2/(2
+ x)^2]) - 4*Log[x - Log[3]]*Log[Log[3]] + ((8 + Log[81])*Log[x - Log[3]]*Log[Log[3]])/(2 + Log[3]) + 4*PolyL
og[2, -1/2*x] - ((8 + Log[81])*PolyLog[2, -1/2*x])/(2 + Log[3]) + 4*PolyLog[2, 1 - x/Log[3]] - ((8 + Log[81])*
PolyLog[2, 1 - x/Log[3]])/(2 + Log[3])

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(a + b*Log[(-c)*(d/e)])*(Log[d + e*
x]/e), x] + Dist[b, Int[Log[(-e)*(x/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[(-c)*(d/e), 0]

Rule 2354

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

Rule 2404

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2546

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))/((f_.) + (g_.)*(x_)), x_S
ymbol] :> Simp[Log[f + g*x]*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])/g), x] + (-Dist[b*B*(n/g), Int[Log[f + g
*x]/(a + b*x), x], x] + Dist[B*d*(n/g), Int[Log[f + g*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, A,
B, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0]

Rule 6820

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {(8+\log (81)) \log (x)}{(2+x) (x-\log (3))}+\frac {2 \left (2+\log \left (\frac {(x-\log (3))^2}{(2+x)^2}\right )\right )}{x}\right ) \, dx\\ &=2 \int \frac {2+\log \left (\frac {(x-\log (3))^2}{(2+x)^2}\right )}{x} \, dx+(8+\log (81)) \int \frac {\log (x)}{(2+x) (x-\log (3))} \, dx\\ &=2 \log (x) \left (2+\log \left (\frac {(x-\log (3))^2}{(2+x)^2}\right )\right )-2 \int \frac {(2+x)^2 \left (\frac {2 (x-\log (3))}{(2+x)^2}-\frac {2 (x-\log (3))^2}{(2+x)^3}\right ) \log (x)}{(x-\log (3))^2} \, dx+(8+\log (81)) \int \left (-\frac {\log (x)}{(2+x) (2+\log (3))}+\frac {\log (x)}{(x-\log (3)) (2+\log (3))}\right ) \, dx\\ &=2 \log (x) \left (2+\log \left (\frac {(x-\log (3))^2}{(2+x)^2}\right )\right )-2 \int \left (-\frac {2 \log (x)}{2+x}+\frac {2 \log (x)}{x-\log (3)}\right ) \, dx-\frac {(8+\log (81)) \int \frac {\log (x)}{2+x} \, dx}{2+\log (3)}+\frac {(8+\log (81)) \int \frac {\log (x)}{x-\log (3)} \, dx}{2+\log (3)}\\ &=-\frac {(8+\log (81)) \log \left (1+\frac {x}{2}\right ) \log (x)}{2+\log (3)}+2 \log (x) \left (2+\log \left (\frac {(x-\log (3))^2}{(2+x)^2}\right )\right )+\frac {(8+\log (81)) \log (x-\log (3)) \log (\log (3))}{2+\log (3)}+4 \int \frac {\log (x)}{2+x} \, dx-4 \int \frac {\log (x)}{x-\log (3)} \, dx+\frac {(8+\log (81)) \int \frac {\log \left (1+\frac {x}{2}\right )}{x} \, dx}{2+\log (3)}+\frac {(8+\log (81)) \int \frac {\log \left (\frac {x}{\log (3)}\right )}{x-\log (3)} \, dx}{2+\log (3)}\\ &=4 \log \left (1+\frac {x}{2}\right ) \log (x)-\frac {(8+\log (81)) \log \left (1+\frac {x}{2}\right ) \log (x)}{2+\log (3)}+2 \log (x) \left (2+\log \left (\frac {(x-\log (3))^2}{(2+x)^2}\right )\right )-4 \log (x-\log (3)) \log (\log (3))+\frac {(8+\log (81)) \log (x-\log (3)) \log (\log (3))}{2+\log (3)}-\frac {(8+\log (81)) \text {Li}_2\left (-\frac {x}{2}\right )}{2+\log (3)}-\frac {(8+\log (81)) \text {Li}_2\left (1-\frac {x}{\log (3)}\right )}{2+\log (3)}-4 \int \frac {\log \left (1+\frac {x}{2}\right )}{x} \, dx-4 \int \frac {\log \left (\frac {x}{\log (3)}\right )}{x-\log (3)} \, dx\\ &=4 \log \left (1+\frac {x}{2}\right ) \log (x)-\frac {(8+\log (81)) \log \left (1+\frac {x}{2}\right ) \log (x)}{2+\log (3)}+2 \log (x) \left (2+\log \left (\frac {(x-\log (3))^2}{(2+x)^2}\right )\right )-4 \log (x-\log (3)) \log (\log (3))+\frac {(8+\log (81)) \log (x-\log (3)) \log (\log (3))}{2+\log (3)}+4 \text {Li}_2\left (-\frac {x}{2}\right )-\frac {(8+\log (81)) \text {Li}_2\left (-\frac {x}{2}\right )}{2+\log (3)}+4 \text {Li}_2\left (1-\frac {x}{\log (3)}\right )-\frac {(8+\log (81)) \text {Li}_2\left (1-\frac {x}{\log (3)}\right )}{2+\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.05, size = 38, normalized size = 1.31 \begin {gather*} \frac {\log (x) \left (8+\log (81)+(4+\log (9)) \log \left (\frac {x^2+\log ^2(3)-x \log (9)}{(2+x)^2}\right )\right )}{2+\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[x]*(8 + Log[81] + (4 + Log[9])*Log[(x^2 + Log[3]^2 - x*Log[9])/(2 + x)^2]))/(2 + Log[3])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.28, size = 628, normalized size = 21.66

method	result	size
risch	\(4 \ln \left (x \right ) \ln \left (\ln \left (3\right )-x \right )-i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i}{\left (2+x \right )^{2}}\right ) \mathrm {csgn}\left (i \left (\ln \left (3\right )-x \right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (3\right )-x \right )^{2}}{\left (2+x \right )^{2}}\right )+i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i}{\left (2+x \right )^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (3\right )-x \right )^{2}}{\left (2+x \right )^{2}}\right )^{2}+i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )-2 i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )^{2}+i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (2+x \right )^{2}\right )^{3}-i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (\ln \left (3\right )-x \right )\right )^{2} \mathrm {csgn}\left (i \left (\ln \left (3\right )-x \right )^{2}\right )+2 i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (\ln \left (3\right )-x \right )\right ) \mathrm {csgn}\left (i \left (\ln \left (3\right )-x \right )^{2}\right )^{2}-i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (\ln \left (3\right )-x \right )^{2}\right )^{3}+i \pi \ln \left (x \right ) \mathrm {csgn}\left (i \left (\ln \left (3\right )-x \right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (3\right )-x \right )^{2}}{\left (2+x \right )^{2}}\right )^{2}-i \pi \ln \left (x \right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (3\right )-x \right )^{2}}{\left (2+x \right )^{2}}\right )^{3}-4 \ln \left (x \right ) \ln \left (2+x \right )+4 \ln \left (x \right )\)	\(310\)
default	\(4 \ln \left (x \right )-2 \ln \left (\frac {1}{2+x}\right ) \ln \left (\frac {\ln \left (3\right )^{2}}{\left (2+x \right )^{2}}+\frac {4 \ln \left (3\right )}{\left (2+x \right )^{2}}-\frac {2 \ln \left (3\right )}{2+x}+\frac {4}{\left (2+x \right )^{2}}-\frac {4}{2+x}+1\right )+\frac {8 \ln \left (-\frac {2+\ln \left (3\right )}{2+x}+1\right ) \ln \left (\frac {1}{2+x}\right )}{2+\ln \left (3\right )}-\frac {8 \ln \left (-\frac {2+\ln \left (3\right )}{2+x}+1\right ) \ln \left (\frac {2+\ln \left (3\right )}{2+x}\right )}{2+\ln \left (3\right )}-\frac {8 \dilog \left (\frac {2+\ln \left (3\right )}{2+x}\right )}{2+\ln \left (3\right )}+\frac {4 \ln \left (3\right ) \ln \left (-\frac {2+\ln \left (3\right )}{2+x}+1\right ) \ln \left (\frac {1}{2+x}\right )}{2+\ln \left (3\right )}-\frac {4 \ln \left (3\right ) \ln \left (-\frac {2+\ln \left (3\right )}{2+x}+1\right ) \ln \left (\frac {2+\ln \left (3\right )}{2+x}\right )}{2+\ln \left (3\right )}-\frac {4 \ln \left (3\right ) \dilog \left (\frac {2+\ln \left (3\right )}{2+x}\right )}{2+\ln \left (3\right )}+2 \ln \left (\frac {2}{2+x}-1\right ) \ln \left (\frac {\ln \left (3\right )^{2}}{\left (2+x \right )^{2}}+\frac {4 \ln \left (3\right )}{\left (2+x \right )^{2}}-\frac {2 \ln \left (3\right )}{2+x}+\frac {4}{\left (2+x \right )^{2}}-\frac {4}{2+x}+1\right )-\frac {4 \ln \left (3\right ) \dilog \left (\frac {\left (2+\ln \left (3\right )\right ) \left (\frac {2}{2+x}-1\right )+\ln \left (3\right )}{\ln \left (3\right )}\right )}{2+\ln \left (3\right )}-\frac {4 \ln \left (3\right ) \ln \left (\frac {2}{2+x}-1\right ) \ln \left (\frac {\left (2+\ln \left (3\right )\right ) \left (\frac {2}{2+x}-1\right )+\ln \left (3\right )}{\ln \left (3\right )}\right )}{2+\ln \left (3\right )}-\frac {8 \dilog \left (\frac {\left (2+\ln \left (3\right )\right ) \left (\frac {2}{2+x}-1\right )+\ln \left (3\right )}{\ln \left (3\right )}\right )}{2+\ln \left (3\right )}-\frac {8 \ln \left (\frac {2}{2+x}-1\right ) \ln \left (\frac {\left (2+\ln \left (3\right )\right ) \left (\frac {2}{2+x}-1\right )+\ln \left (3\right )}{\ln \left (3\right )}\right )}{2+\ln \left (3\right )}-\frac {4 \ln \left (3\right ) \ln \left (\frac {\ln \left (3\right )-x}{\ln \left (3\right )}\right ) \ln \left (\frac {x}{\ln \left (3\right )}\right )}{2+\ln \left (3\right )}+\frac {4 \ln \left (3\right ) \ln \left (\frac {\ln \left (3\right )-x}{\ln \left (3\right )}\right ) \ln \left (x \right )}{2+\ln \left (3\right )}-\frac {4 \ln \left (3\right ) \dilog \left (\frac {x}{\ln \left (3\right )}\right )}{2+\ln \left (3\right )}-\frac {4 \ln \left (3\right ) \ln \left (x \right ) \ln \left (1+\frac {x}{2}\right )}{2+\ln \left (3\right )}-\frac {4 \ln \left (3\right ) \dilog \left (1+\frac {x}{2}\right )}{2+\ln \left (3\right )}-\frac {8 \ln \left (\frac {\ln \left (3\right )-x}{\ln \left (3\right )}\right ) \ln \left (\frac {x}{\ln \left (3\right )}\right )}{2+\ln \left (3\right )}+\frac {8 \ln \left (\frac {\ln \left (3\right )-x}{\ln \left (3\right )}\right ) \ln \left (x \right )}{2+\ln \left (3\right )}-\frac {8 \dilog \left (\frac {x}{\ln \left (3\right )}\right )}{2+\ln \left (3\right )}-\frac {8 \ln \left (x \right ) \ln \left (1+\frac {x}{2}\right )}{2+\ln \left (3\right )}-\frac {8 \dilog \left (1+\frac {x}{2}\right )}{2+\ln \left (3\right )}\)	\(628\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*ln(x)-2*ln(1/(2+x))*ln(ln(3)^2/(2+x)^2+4*ln(3)/(2+x)^2-2/(2+x)*ln(3)+4/(2+x)^2-4/(2+x)+1)+8*ln(-(2+ln(3))/(2
+x)+1)/(2+ln(3))*ln(1/(2+x))-8*ln(-(2+ln(3))/(2+x)+1)/(2+ln(3))*ln((2+ln(3))/(2+x))-8*dilog((2+ln(3))/(2+x))/(
2+ln(3))+4*ln(3)*ln(-(2+ln(3))/(2+x)+1)/(2+ln(3))*ln(1/(2+x))-4*ln(3)*ln(-(2+ln(3))/(2+x)+1)/(2+ln(3))*ln((2+l
n(3))/(2+x))-4*ln(3)*dilog((2+ln(3))/(2+x))/(2+ln(3))+2*ln(2/(2+x)-1)*ln(ln(3)^2/(2+x)^2+4*ln(3)/(2+x)^2-2/(2+
x)*ln(3)+4/(2+x)^2-4/(2+x)+1)-4*ln(3)*dilog(((2+ln(3))*(2/(2+x)-1)+ln(3))/ln(3))/(2+ln(3))-4*ln(3)*ln(2/(2+x)-
1)*ln(((2+ln(3))*(2/(2+x)-1)+ln(3))/ln(3))/(2+ln(3))-8*dilog(((2+ln(3))*(2/(2+x)-1)+ln(3))/ln(3))/(2+ln(3))-8*
ln(2/(2+x)-1)*ln(((2+ln(3))*(2/(2+x)-1)+ln(3))/ln(3))/(2+ln(3))-4*ln(3)/(2+ln(3))*ln((ln(3)-x)/ln(3))*ln(x/ln(
3))+4*ln(3)/(2+ln(3))*ln((ln(3)-x)/ln(3))*ln(x)-4*ln(3)/(2+ln(3))*dilog(x/ln(3))-4*ln(3)/(2+ln(3))*ln(x)*ln(1+
1/2*x)-4*ln(3)/(2+ln(3))*dilog(1+1/2*x)-8/(2+ln(3))*ln((ln(3)-x)/ln(3))*ln(x/ln(3))+8/(2+ln(3))*ln((ln(3)-x)/l
n(3))*ln(x)-8/(2+ln(3))*dilog(x/ln(3))-8/(2+ln(3))*ln(x)*ln(1+1/2*x)-8/(2+ln(3))*dilog(1+1/2*x)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (27) = 54\).
time = 0.82, size = 127, normalized size = 4.38 \begin {gather*} -4 \, {\left (\frac {2 \, \log \left (x - \log \left (3\right )\right )}{\log \left (3\right )^{2} + 2 \, \log \left (3\right )} + \frac {\log \left (x + 2\right )}{\log \left (3\right ) + 2} - \frac {\log \left (x\right )}{\log \left (3\right )}\right )} \log \left (3\right ) - 4 \, {\left (\frac {\log \left (x - \log \left (3\right )\right )}{\log \left (3\right ) + 2} - \frac {\log \left (x + 2\right )}{\log \left (3\right ) + 2}\right )} \log \left (3\right ) + 4 \, \log \left (x - \log \left (3\right )\right ) \log \left (x\right ) - 4 \, \log \left (x + 2\right ) \log \left (x\right ) + \frac {4 \, \log \left (3\right ) \log \left (x - \log \left (3\right )\right )}{\log \left (3\right ) + 2} + \frac {8 \, \log \left (x - \log \left (3\right )\right )}{\log \left (3\right ) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+4)*log(3)-2*x^2-4*x)*log((log(3)^2-2*x*log(3)+x^2)/(x^2+4*x+4))+(-4*x*log(3)-8*x)*log(x)+(4*x
+8)*log(3)-4*x^2-8*x)/((x^2+2*x)*log(3)-x^3-2*x^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 34, normalized size = 1.17 \begin {gather*} 2 \, \log \left (x\right ) \log \left (\frac {x^{2} - 2 \, x \log \left (3\right ) + \log \left (3\right )^{2}}{x^{2} + 4 \, x + 4}\right ) + 4 \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+4)*log(3)-2*x^2-4*x)*log((log(3)^2-2*x*log(3)+x^2)/(x^2+4*x+4))+(-4*x*log(3)-8*x)*log(x)+(4*x
+8)*log(3)-4*x^2-8*x)/((x^2+2*x)*log(3)-x^3-2*x^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.17, size = 34, normalized size = 1.17 \begin {gather*} 2 \log {\left (x \right )} \log {\left (\frac {x^{2} - 2 x \log {\left (3 \right )} + \log {\left (3 \right )}^{2}}{x^{2} + 4 x + 4} \right )} + 4 \log {\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x)*log((x**2 - 2*x*log(3) + log(3)**2)/(x**2 + 4*x + 4)) + 4*log(x)

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Giac [A]
time = 0.43, size = 36, normalized size = 1.24 \begin {gather*} 2 \, \log \left (x^{2} - 2 \, x \log \left (3\right ) + \log \left (3\right )^{2}\right ) \log \left (x\right ) - 2 \, \log \left (x^{2} + 4 \, x + 4\right ) \log \left (x\right ) + 4 \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+4)*log(3)-2*x^2-4*x)*log((log(3)^2-2*x*log(3)+x^2)/(x^2+4*x+4))+(-4*x*log(3)-8*x)*log(x)+(4*x
+8)*log(3)-4*x^2-8*x)/((x^2+2*x)*log(3)-x^3-2*x^2),x, algorithm="giac")

[Out]

2*log(x^2 - 2*x*log(3) + log(3)^2)*log(x) - 2*log(x^2 + 4*x + 4)*log(x) + 4*log(x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {8\,x-\ln \left (3\right )\,\left (4\,x+8\right )+\ln \left (\frac {x^2-2\,\ln \left (3\right )\,x+{\ln \left (3\right )}^2}{x^2+4\,x+4}\right )\,\left (4\,x-\ln \left (3\right )\,\left (2\,x+4\right )+2\,x^2\right )+\ln \left (x\right )\,\left (8\,x+4\,x\,\ln \left (3\right )\right )+4\,x^2}{2\,x^2+x^3-\ln \left (3\right )\,\left (x^2+2\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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