3.8.42 \(\int \frac {2 x+4 x^2-x \log (6)+(4 x+4 x^2-2 x \log (6)) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx\)

Optimal. Leaf size=25 \[ \frac {x \log (x)}{5 \left (4+\frac {4 (2+3 x-\log (6))}{x}\right )} \]

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Rubi [A]  time = 0.33, antiderivative size = 32, normalized size of antiderivative = 1.28, number of steps used = 12, number of rules used = 9, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.155, Rules used = {6, 6688, 12, 6742, 43, 2357, 2295, 2314, 31} \begin {gather*} \frac {1}{80} x \log (x)-\frac {x (2-\log (6)) \log (x)}{80 (4 x+2-\log (6))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x + 4*x^2 - x*Log[6] + (4*x + 4*x^2 - 2*x*Log[6])*Log[x])/(80 + 320*x + 320*x^2 + (-80 - 160*x)*Log[6]
+ 20*Log[6]^2),x]

[Out]

(x*Log[x])/80 - (x*(2 - Log[6])*Log[x])/(80*(2 + 4*x - Log[6]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2314

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

Rule 2357

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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x^2+x (2-\log (6))+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx\\ &=\int \frac {x \left (4 x+2 \left (1-\frac {\log (6)}{2}\right )+(4+4 x-2 \log (6)) \log (x)\right )}{20 (2+4 x-\log (6))^2} \, dx\\ &=\frac {1}{20} \int \frac {x \left (4 x+2 \left (1-\frac {\log (6)}{2}\right )+(4+4 x-2 \log (6)) \log (x)\right )}{(2+4 x-\log (6))^2} \, dx\\ &=\frac {1}{20} \int \left (\frac {x}{2+4 x-\log (6)}+\frac {2 x (2+2 x-\log (6)) \log (x)}{(2+4 x-\log (6))^2}\right ) \, dx\\ &=\frac {1}{20} \int \frac {x}{2+4 x-\log (6)} \, dx+\frac {1}{10} \int \frac {x (2+2 x-\log (6)) \log (x)}{(2+4 x-\log (6))^2} \, dx\\ &=\frac {1}{20} \int \left (\frac {1}{4}+\frac {-2+\log (6)}{4 (2+4 x-\log (6))}\right ) \, dx+\frac {1}{10} \int \left (\frac {\log (x)}{8}-\frac {(-2+\log (6))^2 \log (x)}{8 (-2-4 x+\log (6))^2}\right ) \, dx\\ &=\frac {x}{80}-\frac {1}{320} (2-\log (6)) \log (2+4 x-\log (6))+\frac {1}{80} \int \log (x) \, dx-\frac {1}{80} (2-\log (6))^2 \int \frac {\log (x)}{(-2-4 x+\log (6))^2} \, dx\\ &=\frac {1}{80} x \log (x)-\frac {x (2-\log (6)) \log (x)}{80 (2+4 x-\log (6))}-\frac {1}{320} (2-\log (6)) \log (2+4 x-\log (6))+\frac {1}{80} (-2+\log (6)) \int \frac {1}{-2-4 x+\log (6)} \, dx\\ &=\frac {1}{80} x \log (x)-\frac {x (2-\log (6)) \log (x)}{80 (2+4 x-\log (6))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 20, normalized size = 0.80 \begin {gather*} \frac {x^2 \log (x)}{20 (2+4 x-\log (6))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x + 4*x^2 - x*Log[6] + (4*x + 4*x^2 - 2*x*Log[6])*Log[x])/(80 + 320*x + 320*x^2 + (-80 - 160*x)*L
og[6] + 20*Log[6]^2),x]

[Out]

(x^2*Log[x])/(20*(2 + 4*x - Log[6]))

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fricas [A]  time = 0.54, size = 18, normalized size = 0.72 \begin {gather*} \frac {x^{2} \log \relax (x)}{20 \, {\left (4 \, x - \log \relax (6) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(6)+4*x^2+4*x)*log(x)-x*log(6)+4*x^2+2*x)/(20*log(6)^2+(-160*x-80)*log(6)+320*x^2+320*x+80
),x, algorithm="fricas")

[Out]

1/20*x^2*log(x)/(4*x - log(6) + 2)

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giac [A]  time = 0.28, size = 39, normalized size = 1.56 \begin {gather*} \frac {1}{320} \, {\left (4 \, x + \frac {\log \relax (6)^{2} - 4 \, \log \relax (6) + 4}{4 \, x - \log \relax (6) + 2}\right )} \log \relax (x) + \frac {1}{320} \, {\left (\log \relax (6) - 2\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(6)+4*x^2+4*x)*log(x)-x*log(6)+4*x^2+2*x)/(20*log(6)^2+(-160*x-80)*log(6)+320*x^2+320*x+80
),x, algorithm="giac")

[Out]

1/320*(4*x + (log(6)^2 - 4*log(6) + 4)/(4*x - log(6) + 2))*log(x) + 1/320*(log(6) - 2)*log(x)

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maple [A]  time = 0.23, size = 17, normalized size = 0.68

method	result	size
norman	\(-\frac {x^{2} \ln \relax (x )}{20 \left (\ln \relax (6)-4 x -2\right )}\)	\(17\)
risch	\(-\frac {\left (\ln \relax (2)^{2}+2 \ln \relax (2) \ln \relax (3)-4 x \ln \relax (2)+\ln \relax (3)^{2}-4 x \ln \relax (3)+16 x^{2}-4 \ln \relax (2)-4 \ln \relax (3)+8 x +4\right ) \ln \relax (x )}{320 \left (\ln \relax (2)+\ln \relax (3)-4 x -2\right )}-\frac {\ln \relax (x )}{160}+\frac {\ln \relax (2) \ln \relax (x )}{320}+\frac {\ln \relax (3) \ln \relax (x )}{320}\)	\(75\)
default	\(\frac {\ln \left (-\ln \relax (6)+4 x +2\right ) \ln \relax (6)}{320}-\frac {\ln \left (-\ln \relax (6)+4 x +2\right )}{160}+\frac {x \ln \relax (x )}{80}-\frac {\ln \relax (6)^{2} \ln \left (\ln \relax (6)-4 x -2\right )}{320 \left (\ln \relax (6)-2\right )}-\frac {\ln \relax (6)^{2} \ln \relax (x ) x}{80 \left (\ln \relax (6)-2\right ) \left (\ln \relax (6)-4 x -2\right )}+\frac {\ln \relax (6) \ln \left (\ln \relax (6)-4 x -2\right )}{80 \ln \relax (6)-160}+\frac {\ln \relax (6) \ln \relax (x ) x}{20 \left (\ln \relax (6)-2\right ) \left (\ln \relax (6)-4 x -2\right )}-\frac {\ln \left (\ln \relax (6)-4 x -2\right )}{80 \left (\ln \relax (6)-2\right )}-\frac {\ln \relax (x ) x}{20 \left (\ln \relax (6)-2\right ) \left (\ln \relax (6)-4 x -2\right )}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*ln(6)+4*x^2+4*x)*ln(x)-x*ln(6)+4*x^2+2*x)/(20*ln(6)^2+(-160*x-80)*ln(6)+320*x^2+320*x+80),x,method=
_RETURNVERBOSE)

[Out]

-1/20*x^2*ln(x)/(ln(6)-4*x-2)

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maxima [B]  time = 0.81, size = 209, normalized size = 8.36 \begin {gather*} \frac {1}{320} \, {\left (\frac {\log \relax (6) - 2}{4 \, x - \log \relax (6) + 2} - \log \left (4 \, x - \log \relax (6) + 2\right )\right )} \log \relax (6) + \frac {1}{160} \, {\left (\log \relax (6) - 2\right )} \log \left (4 \, x - \log \relax (6) + 2\right ) - \frac {1}{320} \, {\left (\log \relax (3) + \log \relax (2) - 2\right )} \log \left (4 \, x - \log \relax (3) - \log \relax (2) + 2\right ) + \frac {1}{320} \, {\left (\log \relax (3) + \log \relax (2) - 2\right )} \log \relax (x) + \frac {1}{80} \, x - \frac {\log \relax (6)^{2} - 4 \, \log \relax (6) + 4}{320 \, {\left (4 \, x - \log \relax (6) + 2\right )}} - \frac {16 \, x^{2} - 4 \, x {\left (\log \relax (3) + \log \relax (2) - 2\right )} - {\left (16 \, x^{2} - 4 \, x {\left (\log \relax (3) + \log \relax (2) - 2\right )} + \log \relax (3)^{2} + 2 \, {\left (\log \relax (3) - 2\right )} \log \relax (2) + \log \relax (2)^{2} - 4 \, \log \relax (3) + 4\right )} \log \relax (x)}{320 \, {\left (4 \, x - \log \relax (3) - \log \relax (2) + 2\right )}} - \frac {\log \relax (6) - 2}{160 \, {\left (4 \, x - \log \relax (6) + 2\right )}} + \frac {1}{160} \, \log \left (4 \, x - \log \relax (6) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(6)+4*x^2+4*x)*log(x)-x*log(6)+4*x^2+2*x)/(20*log(6)^2+(-160*x-80)*log(6)+320*x^2+320*x+80
),x, algorithm="maxima")

[Out]

1/320*((log(6) - 2)/(4*x - log(6) + 2) - log(4*x - log(6) + 2))*log(6) + 1/160*(log(6) - 2)*log(4*x - log(6) +
2) - 1/320*(log(3) + log(2) - 2)*log(4*x - log(3) - log(2) + 2) + 1/320*(log(3) + log(2) - 2)*log(x) + 1/80*x
- 1/320*(log(6)^2 - 4*log(6) + 4)/(4*x - log(6) + 2) - 1/320*(16*x^2 - 4*x*(log(3) + log(2) - 2) - (16*x^2 -
4*x*(log(3) + log(2) - 2) + log(3)^2 + 2*(log(3) - 2)*log(2) + log(2)^2 - 4*log(3) + 4)*log(x))/(4*x - log(3)
- log(2) + 2) - 1/160*(log(6) - 2)/(4*x - log(6) + 2) + 1/160*log(4*x - log(6) + 2)

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mupad [B]  time = 1.26, size = 135, normalized size = 5.40 \begin {gather*} \frac {\ln \left (\ln \relax (6)-4\,x-2\right )}{160}-\frac {\frac {{\ln \relax (6)}^2}{4}-\ln \relax (6)+1}{320\,x-80\,\ln \relax (6)+160}+\ln \left (\ln \relax (6)-4\,x-2\right )\,\left (\frac {\ln \relax (6)}{160}-\frac {1}{80}\right )-\frac {\ln \left (\ln \relax (6)-4\,x-2\right )\,\ln \relax (6)}{320}-\frac {\frac {\ln \relax (6)}{16}-\frac {1}{8}}{40\,x-10\,\ln \relax (6)+20}-\ln \left (4\,x-\ln \relax (6)+2\right )\,\left (\frac {\ln \relax (6)}{320}-\frac {1}{160}\right )+\frac {x^2\,\ln \relax (x)}{20\,\left (4\,x-\ln \relax (6)+2\right )}+\frac {\ln \relax (6)\,\left (\ln \relax (6)-2\right )}{16\,\left (80\,x-20\,\ln \relax (6)+40\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - x*log(6) + 4*x^2 + log(x)*(4*x - 2*x*log(6) + 4*x^2))/(320*x - log(6)*(160*x + 80) + 20*log(6)^2 +
320*x^2 + 80),x)

[Out]

log(log(6) - 4*x - 2)/160 - (log(6)^2/4 - log(6) + 1)/(320*x - 80*log(6) + 160) + log(log(6) - 4*x - 2)*(log(6
)/160 - 1/80) - (log(log(6) - 4*x - 2)*log(6))/320 - (log(6)/16 - 1/8)/(40*x - 10*log(6) + 20) - log(4*x - log
(6) + 2)*(log(6)/320 - 1/160) + (x^2*log(x))/(20*(4*x - log(6) + 2)) + (log(6)*(log(6) - 2))/(16*(80*x - 20*lo
g(6) + 40))

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sympy [B]  time = 0.28, size = 51, normalized size = 2.04 \begin {gather*} \left (- \frac {1}{160} + \frac {\log {\relax (6 )}}{320}\right ) \log {\relax (x )} + \frac {\left (16 x^{2} - 4 x \log {\relax (6 )} + 8 x - 4 \log {\relax (6 )} + \log {\relax (6 )}^{2} + 4\right ) \log {\relax (x )}}{1280 x - 320 \log {\relax (6 )} + 640} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*ln(6)+4*x**2+4*x)*ln(x)-x*ln(6)+4*x**2+2*x)/(20*ln(6)**2+(-160*x-80)*ln(6)+320*x**2+320*x+80)
,x)

[Out]

(-1/160 + log(6)/320)*log(x) + (16*x**2 - 4*x*log(6) + 8*x - 4*log(6) + log(6)**2 + 4)*log(x)/(1280*x - 320*lo
g(6) + 640)

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