3.57.68 \(\int \frac {4-4 \log (x)+(2-2 \log (x)) \log (\frac {4 x \log (4)-\log (4) \log (x)}{3 x})}{-4 x^2 \log ^8(5)+x \log ^8(5) \log (x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {\left (2+\log \left (\frac {\log (4) \left (x+\frac {1}{3} (x-\log (x))\right )}{x}\right )\right )^2}{\log ^8(5)} \]

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Rubi [A]  time = 0.28, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2561, 6741, 12, 6686} \begin {gather*} \frac {\left (\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )+2\right )^2}{\log ^8(5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 4*Log[x] + (2 - 2*Log[x])*Log[(4*x*Log[4] - Log[4]*Log[x])/(3*x)])/(-4*x^2*Log[5]^8 + x*Log[5]^8*Log[
x]),x]

[Out]

(2 + Log[(Log[4]*(4*x - Log[x]))/(3*x)])^2/Log[5]^8

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )}{x \left (-4 x \log ^8(5)+\log ^8(5) \log (x)\right )} \, dx\\ &=\int \frac {2 (1-\log (x)) \left (-2-\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )\right )}{x \log ^8(5) (4 x-\log (x))} \, dx\\ &=\frac {2 \int \frac {(1-\log (x)) \left (-2-\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )\right )}{x (4 x-\log (x))} \, dx}{\log ^8(5)}\\ &=\frac {\left (2+\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )\right )^2}{\log ^8(5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 0.96 \begin {gather*} \frac {\left (2+\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )\right )^2}{\log ^8(5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 4*Log[x] + (2 - 2*Log[x])*Log[(4*x*Log[4] - Log[4]*Log[x])/(3*x)])/(-4*x^2*Log[5]^8 + x*Log[5]^
8*Log[x]),x]

[Out]

(2 + Log[(Log[4]*(4*x - Log[x]))/(3*x)])^2/Log[5]^8

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fricas [A]  time = 0.67, size = 46, normalized size = 1.64 \begin {gather*} \frac {\log \left (\frac {2 \, {\left (4 \, x \log \relax (2) - \log \relax (2) \log \relax (x)\right )}}{3 \, x}\right )^{2} + 4 \, \log \left (\frac {2 \, {\left (4 \, x \log \relax (2) - \log \relax (2) \log \relax (x)\right )}}{3 \, x}\right )}{\log \relax (5)^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)+2)*log(1/3*(-2*log(2)*log(x)+8*x*log(2))/x)+4-4*log(x))/(x*log(5)^8*log(x)-4*x^2*log(5)^
8),x, algorithm="fricas")

[Out]

(log(2/3*(4*x*log(2) - log(2)*log(x))/x)^2 + 4*log(2/3*(4*x*log(2) - log(2)*log(x))/x))/log(5)^8

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giac [B]  time = 0.14, size = 88, normalized size = 3.14 \begin {gather*} \frac {\log \left (4 \, x \log \relax (2) - \log \relax (2) \log \relax (x)\right )^{2}}{\log \relax (5)^{8}} + \frac {2 \, {\left (\log \relax (3) - \log \relax (2) - 2\right )} \log \relax (x)}{\log \relax (5)^{8}} - \frac {2 \, \log \left (4 \, x \log \relax (2) - \log \relax (2) \log \relax (x)\right ) \log \relax (x)}{\log \relax (5)^{8}} + \frac {\log \relax (x)^{2}}{\log \relax (5)^{8}} - \frac {2 \, {\left (\log \relax (3) - \log \relax (2) - 2\right )} \log \left (-4 \, x + \log \relax (x)\right )}{\log \relax (5)^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)+2)*log(1/3*(-2*log(2)*log(x)+8*x*log(2))/x)+4-4*log(x))/(x*log(5)^8*log(x)-4*x^2*log(5)^
8),x, algorithm="giac")

[Out]

log(4*x*log(2) - log(2)*log(x))^2/log(5)^8 + 2*(log(3) - log(2) - 2)*log(x)/log(5)^8 - 2*log(4*x*log(2) - log(
2)*log(x))*log(x)/log(5)^8 + log(x)^2/log(5)^8 - 2*(log(3) - log(2) - 2)*log(-4*x + log(x))/log(5)^8

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maple [B]  time = 0.47, size = 56, normalized size = 2.00

method	result	size
norman	\(\frac {\frac {4 \ln \left (\frac {-2 \ln \relax (2) \ln \relax (x )+8 x \ln \relax (2)}{3 x}\right )}{\ln \relax (5)}+\frac {\ln \left (\frac {-2 \ln \relax (2) \ln \relax (x )+8 x \ln \relax (2)}{3 x}\right )^{2}}{\ln \relax (5)}}{\ln \relax (5)^{7}}\)	\(56\)
default	\(-\frac {4 \ln \relax (x )}{\ln \relax (5)^{8}}+\frac {4 \ln \left (4 x -\ln \relax (x )\right )}{\ln \relax (5)^{8}}+\frac {\ln \left (\frac {4 x -\ln \relax (x )}{x}\right )^{2}}{\ln \relax (5)^{8}}-\frac {2 \ln \left (\ln \relax (2)\right ) \ln \relax (x )}{\ln \relax (5)^{8}}+\frac {2 \ln \left (\ln \relax (2)\right ) \ln \left (4 x -\ln \relax (x )\right )}{\ln \relax (5)^{8}}-\frac {2 \ln \relax (2) \ln \relax (x )}{\ln \relax (5)^{8}}+\frac {2 \ln \relax (2) \ln \left (4 x -\ln \relax (x )\right )}{\ln \relax (5)^{8}}+\frac {2 \ln \relax (3) \ln \relax (x )}{\ln \relax (5)^{8}}-\frac {2 \ln \relax (3) \ln \left (4 x -\ln \relax (x )\right )}{\ln \relax (5)^{8}}\)	\(128\)
risch	\(\frac {\ln \left (x -\frac {\ln \relax (x )}{4}\right )^{2}}{\ln \relax (5)^{8}}-\frac {2 \ln \relax (x ) \ln \left (x -\frac {\ln \relax (x )}{4}\right )}{\ln \relax (5)^{8}}+\frac {i \pi \ln \left (-4 x +\ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (-x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{2}}{\ln \relax (5)^{8}}+\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (-x +\frac {\ln \relax (x )}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-x +\frac {\ln \relax (x )}{4}\right )}{x}\right )}{\ln \relax (5)^{8}}-\frac {i \pi \ln \left (-4 x +\ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (-x +\frac {\ln \relax (x )}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-x +\frac {\ln \relax (x )}{4}\right )}{x}\right )}{\ln \relax (5)^{8}}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i \left (-x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{3}}{\ln \relax (5)^{8}}-\frac {i \pi \ln \left (-4 x +\ln \relax (x )\right ) \mathrm {csgn}\left (i \left (-x +\frac {\ln \relax (x )}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{2}}{\ln \relax (5)^{8}}+\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (i \left (-x +\frac {\ln \relax (x )}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{2}}{\ln \relax (5)^{8}}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (-x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{2}}{\ln \relax (5)^{8}}+\frac {i \pi \ln \left (-4 x +\ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i \left (-x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{3}}{\ln \relax (5)^{8}}+\frac {\ln \relax (x )^{2}}{\ln \relax (5)^{8}}+\frac {2 \ln \relax (3) \ln \relax (x )}{\ln \relax (5)^{8}}-\frac {6 \ln \relax (2) \ln \relax (x )}{\ln \relax (5)^{8}}-\frac {2 \ln \left (\ln \relax (2)\right ) \ln \relax (x )}{\ln \relax (5)^{8}}-\frac {2 \ln \relax (3) \ln \left (-4 x +\ln \relax (x )\right )}{\ln \relax (5)^{8}}+\frac {6 \ln \relax (2) \ln \left (-4 x +\ln \relax (x )\right )}{\ln \relax (5)^{8}}+\frac {2 \ln \left (-4 x +\ln \relax (x )\right ) \ln \left (\ln \relax (2)\right )}{\ln \relax (5)^{8}}-\frac {4 \ln \relax (x )}{\ln \relax (5)^{8}}+\frac {4 \ln \left (-4 x +\ln \relax (x )\right )}{\ln \relax (5)^{8}}\)	\(446\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*ln(x)+2)*ln(1/3*(-2*ln(2)*ln(x)+8*x*ln(2))/x)+4-4*ln(x))/(x*ln(5)^8*ln(x)-4*x^2*ln(5)^8),x,method=_RE
TURNVERBOSE)

[Out]

(4/ln(5)*ln(1/3*(-2*ln(2)*ln(x)+8*x*ln(2))/x)+1/ln(5)*ln(1/3*(-2*ln(2)*ln(x)+8*x*ln(2))/x)^2)/ln(5)^7

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maxima [C]  time = 0.49, size = 69, normalized size = 2.46 \begin {gather*} -\frac {2 \, {\left (i \, \pi - \log \relax (3) + \log \relax (2) + \log \left (\log \relax (2)\right ) + 2\right )} \log \relax (x) - \log \relax (x)^{2} + 2 \, {\left (-i \, \pi + \log \relax (3) - \log \relax (2) + \log \relax (x) - \log \left (\log \relax (2)\right ) - 2\right )} \log \left (-4 \, x + \log \relax (x)\right ) - \log \left (-4 \, x + \log \relax (x)\right )^{2}}{\log \relax (5)^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)+2)*log(1/3*(-2*log(2)*log(x)+8*x*log(2))/x)+4-4*log(x))/(x*log(5)^8*log(x)-4*x^2*log(5)^
8),x, algorithm="maxima")

[Out]

-(2*(I*pi - log(3) + log(2) + log(log(2)) + 2)*log(x) - log(x)^2 + 2*(-I*pi + log(3) - log(2) + log(x) - log(l
og(2)) - 2)*log(-4*x + log(x)) - log(-4*x + log(x))^2)/log(5)^8

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mupad [B]  time = 3.86, size = 38, normalized size = 1.36 \begin {gather*} \frac {{\ln \left (\frac {\frac {8\,x\,\ln \relax (2)}{3}-\frac {2\,\ln \relax (2)\,\ln \relax (x)}{3}}{x}\right )}^2+4\,\ln \left (\ln \relax (x)-4\,x\right )-4\,\ln \relax (x)}{{\ln \relax (5)}^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(x) + log(((8*x*log(2))/3 - (2*log(2)*log(x))/3)/x)*(2*log(x) - 2) - 4)/(4*x^2*log(5)^8 - x*log(5)^8
*log(x)),x)

[Out]

(4*log(log(x) - 4*x) - 4*log(x) + log(((8*x*log(2))/3 - (2*log(2)*log(x))/3)/x)^2)/log(5)^8

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sympy [B]  time = 0.42, size = 53, normalized size = 1.89 \begin {gather*} - \frac {4 \log {\relax (x )}}{\log {\relax (5 )}^{8}} + \frac {\log {\left (\frac {\frac {8 x \log {\relax (2 )}}{3} - \frac {2 \log {\relax (2 )} \log {\relax (x )}}{3}}{x} \right )}^{2}}{\log {\relax (5 )}^{8}} + \frac {4 \log {\left (- 4 x + \log {\relax (x )} \right )}}{\log {\relax (5 )}^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*ln(x)+2)*ln(1/3*(-2*ln(2)*ln(x)+8*x*ln(2))/x)+4-4*ln(x))/(x*ln(5)**8*ln(x)-4*x**2*ln(5)**8),x)

[Out]

-4*log(x)/log(5)**8 + log((8*x*log(2)/3 - 2*log(2)*log(x)/3)/x)**2/log(5)**8 + 4*log(-4*x + log(x))/log(5)**8

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