3.61.12 \(\int \frac {18-9 x+(-2+x) \log (4-2 x)+2 \log ^2(2) \log (\frac {1}{5} (9-\log (4-2 x)))}{18-9 x+(-2+x) \log (4-2 x)} \, dx\)

Optimal. Leaf size=33 \[ x-\log ^2(\log (4))+\log ^2(2) \log ^2\left (\frac {1}{5} (9-\log (2 (2-x)))\right ) \]

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Rubi [A]  time = 0.32, antiderivative size = 24, normalized size of antiderivative = 0.73, number of steps used = 4, number of rules used = 7, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6741, 6742, 2390, 12, 2302, 29, 6686} \begin {gather*} x+\log ^2(2) \log ^2\left (\frac {1}{5} (9-\log (4-2 x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18 - 9*x + (-2 + x)*Log[4 - 2*x] + 2*Log[2]^2*Log[(9 - Log[4 - 2*x])/5])/(18 - 9*x + (-2 + x)*Log[4 - 2*x
]),x]

[Out]

x + Log[2]^2*Log[(9 - Log[4 - 2*x])/5]^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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6741

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

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 {18-9 x+(-2+x) \log (4-2 x)+2 \log ^2(2) \log \left (\frac {1}{5} (9-\log (4-2 x))\right )}{(2-x) (9-\log (4-2 x))} \, dx\\ &=\int \left (1+\frac {2 \log ^2(2) \log \left (\frac {1}{5} (9-\log (4-2 x))\right )}{(-2+x) (-9+\log (4-2 x))}\right ) \, dx\\ &=x+\left (2 \log ^2(2)\right ) \int \frac {\log \left (\frac {1}{5} (9-\log (4-2 x))\right )}{(-2+x) (-9+\log (4-2 x))} \, dx\\ &=x+\log ^2(2) \log ^2\left (\frac {1}{5} (9-\log (4-2 x))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 24, normalized size = 0.73 \begin {gather*} x+\log ^2(2) \log ^2\left (\frac {1}{5} (9-\log (4-2 x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 - 9*x + (-2 + x)*Log[4 - 2*x] + 2*Log[2]^2*Log[(9 - Log[4 - 2*x])/5])/(18 - 9*x + (-2 + x)*Log[4
- 2*x]),x]

[Out]

x + Log[2]^2*Log[(9 - Log[4 - 2*x])/5]^2

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fricas [A]  time = 0.61, size = 20, normalized size = 0.61 \begin {gather*} \log \relax (2)^{2} \log \left (-\frac {1}{5} \, \log \left (-2 \, x + 4\right ) + \frac {9}{5}\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(2)^2*log(-1/5*log(4-2*x)+9/5)+(x-2)*log(4-2*x)-9*x+18)/((x-2)*log(4-2*x)-9*x+18),x, algorithm
="fricas")

[Out]

log(2)^2*log(-1/5*log(-2*x + 4) + 9/5)^2 + x

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giac [B]  time = 0.14, size = 68, normalized size = 2.06 \begin {gather*} -2 \, \log \relax (5) \log \relax (2)^{2} \log \left (\log \relax (2) + \log \left (-x + 2\right ) - 9\right ) - \log \relax (2)^{2} \log \left (\log \relax (2) + \log \left (-x + 2\right ) - 9\right )^{2} + 2 \, \log \relax (2)^{2} \log \left (\log \relax (2) + \log \left (-x + 2\right ) - 9\right ) \log \left (-\log \left (-2 \, x + 4\right ) + 9\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(2)^2*log(-1/5*log(4-2*x)+9/5)+(x-2)*log(4-2*x)-9*x+18)/((x-2)*log(4-2*x)-9*x+18),x, algorithm
="giac")

[Out]

-2*log(5)*log(2)^2*log(log(2) + log(-x + 2) - 9) - log(2)^2*log(log(2) + log(-x + 2) - 9)^2 + 2*log(2)^2*log(l
og(2) + log(-x + 2) - 9)*log(-log(-2*x + 4) + 9) + x

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maple [A]  time = 0.09, size = 21, normalized size = 0.64

method	result	size
norman	\(x +\ln \relax (2)^{2} \ln \left (-\frac {\ln \left (4-2 x \right )}{5}+\frac {9}{5}\right )^{2}\)	\(21\)
risch	\(x +\ln \relax (2)^{2} \ln \left (-\frac {\ln \left (4-2 x \right )}{5}+\frac {9}{5}\right )^{2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(2)^2*ln(-1/5*ln(4-2*x)+9/5)+(x-2)*ln(4-2*x)-9*x+18)/((x-2)*ln(4-2*x)-9*x+18),x,method=_RETURNVERBOSE
)

[Out]

x+ln(2)^2*ln(-1/5*ln(4-2*x)+9/5)^2

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maxima [C]  time = 0.47, size = 162, normalized size = 4.91 \begin {gather*} 2 \, \log \relax (2)^{2} \log \left (i \, \pi + \log \relax (2) + \log \left (x - 2\right ) - 9\right ) \log \left (-\frac {1}{5} \, \log \left (-2 \, x + 4\right ) + \frac {9}{5}\right ) - {\left (2 \, \log \relax (5) \log \left (-i \, \pi - \log \relax (2) - \log \left (x - 2\right ) + 9\right ) - \log \left (-i \, \pi - \log \relax (2) - \log \left (x - 2\right ) + 9\right )^{2} + 2 \, \log \left (i \, \pi + \log \relax (2) + \log \left (x - 2\right ) - 9\right ) \log \left (-\frac {1}{5} \, \log \left (-2 \, x + 4\right ) + \frac {9}{5}\right )\right )} \log \relax (2)^{2} + 2 \, {\left (i \, \pi + \log \relax (2) + \log \left (x - 2\right ) - 9\right )} \log \left (i \, \pi + \log \relax (2) + \log \left (x - 2\right ) - 9\right ) - 2 \, \log \left (i \, \pi + \log \relax (2) + \log \left (x - 2\right ) - 9\right ) \log \left (-2 \, x + 4\right ) + x + 18 \, \log \left (i \, \pi + \log \relax (2) + \log \left (x - 2\right ) - 9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(2)^2*log(-1/5*log(4-2*x)+9/5)+(x-2)*log(4-2*x)-9*x+18)/((x-2)*log(4-2*x)-9*x+18),x, algorithm
="maxima")

[Out]

2*log(2)^2*log(I*pi + log(2) + log(x - 2) - 9)*log(-1/5*log(-2*x + 4) + 9/5) - (2*log(5)*log(-I*pi - log(2) -
log(x - 2) + 9) - log(-I*pi - log(2) - log(x - 2) + 9)^2 + 2*log(I*pi + log(2) + log(x - 2) - 9)*log(-1/5*log(
-2*x + 4) + 9/5))*log(2)^2 + 2*(I*pi + log(2) + log(x - 2) - 9)*log(I*pi + log(2) + log(x - 2) - 9) - 2*log(I*
pi + log(2) + log(x - 2) - 9)*log(-2*x + 4) + x + 18*log(I*pi + log(2) + log(x - 2) - 9)

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mupad [B]  time = 4.62, size = 20, normalized size = 0.61 \begin {gather*} {\ln \relax (2)}^2\,{\ln \left (\frac {9}{5}-\frac {\ln \left (4-2\,x\right )}{5}\right )}^2+x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4 - 2*x)*(x - 2) - 9*x + 2*log(2)^2*log(9/5 - log(4 - 2*x)/5) + 18)/(log(4 - 2*x)*(x - 2) - 9*x + 18)
,x)

[Out]

x + log(2)^2*log(9/5 - log(4 - 2*x)/5)^2

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sympy [A]  time = 0.33, size = 20, normalized size = 0.61 \begin {gather*} x + \log {\relax (2 )}^{2} \log {\left (\frac {9}{5} - \frac {\log {\left (4 - 2 x \right )}}{5} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(2)**2*ln(-1/5*ln(4-2*x)+9/5)+(x-2)*ln(4-2*x)-9*x+18)/((x-2)*ln(4-2*x)-9*x+18),x)

[Out]

x + log(2)**2*log(9/5 - log(4 - 2*x)/5)**2

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