3.10.98 \(\int \frac {-\log (\frac {x}{2}) \log (x)+\log (\frac {x}{2}) (1+\log (\frac {x}{e^4}))+\log (x) (1+\log (\frac {x}{e^4})) \log (\frac {-4-4 \log (\frac {x}{e^4})}{\log (x)})}{\log (x) (x+x \log (\frac {x}{e^4})) \log ^2(\frac {-4-4 \log (\frac {x}{e^4})}{\log (x)})} \, dx\)

Optimal. Leaf size=26 \[ \frac {\log \left (\frac {x}{2}\right )}{\log \left (\frac {4 \left (-1-\log \left (\frac {x}{e^4}\right )\right )}{\log (x)}\right )} \]

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Rubi [F]  time = 1.65, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-\log \left (\frac {x}{2}\right ) \log (x)+\log \left (\frac {x}{2}\right ) \left (1+\log \left (\frac {x}{e^4}\right )\right )+\log (x) \left (1+\log \left (\frac {x}{e^4}\right )\right ) \log \left (\frac {-4-4 \log \left (\frac {x}{e^4}\right )}{\log (x)}\right )}{\log (x) \left (x+x \log \left (\frac {x}{e^4}\right )\right ) \log ^2\left (\frac {-4-4 \log \left (\frac {x}{e^4}\right )}{\log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(Log[x/2]*Log[x]) + Log[x/2]*(1 + Log[x/E^4]) + Log[x]*(1 + Log[x/E^4])*Log[(-4 - 4*Log[x/E^4])/Log[x]])
/(Log[x]*(x + x*Log[x/E^4])*Log[(-4 - 4*Log[x/E^4])/Log[x]]^2),x]

[Out]

Defer[Subst][Defer[Int][Log[-4 + 12/x]^(-1), x], x, Log[x]] + Defer[Subst][Defer[Int][(-3*x + Log[8])/((-3 + x
)*x*Log[(12 - 4*x)/x]^2), x], x, Log[x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\log \left (\frac {x}{2}\right ) \log (x)+\log \left (\frac {x}{2}\right ) \left (1+\log \left (\frac {x}{e^4}\right )\right )+\log (x) \left (1+\log \left (\frac {x}{e^4}\right )\right ) \log \left (\frac {-4-4 \log \left (\frac {x}{e^4}\right )}{\log (x)}\right )}{x \log (x) \left (1+\log \left (\frac {x}{e^4}\right )\right ) \log ^2\left (\frac {-4-4 \log \left (\frac {x}{e^4}\right )}{\log (x)}\right )} \, dx\\ &=\int \frac {-\log (8)-\log ^2(x) \log \left (-4+\frac {12}{\log (x)}\right )+3 \log (x) \left (1+\log \left (-4+\frac {12}{\log (x)}\right )\right )}{x (3-\log (x)) \log (x) \log ^2\left (-\frac {4 (-3+\log (x))}{\log (x)}\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {-3 x+\log (8)-3 x \log \left (-4+\frac {12}{x}\right )+x^2 \log \left (-4+\frac {12}{x}\right )}{(-3+x) x \log ^2\left (-\frac {4 (-3+x)}{x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {3 x-\log (8)+3 x \log \left (-4+\frac {12}{x}\right )-x^2 \log \left (-4+\frac {12}{x}\right )}{(3-x) x \log ^2\left (-4+\frac {12}{x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {3 x-\log (8)-(-3+x) x \log \left (-4+\frac {12}{x}\right )}{(3-x) x \log ^2\left (-4+\frac {12}{x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {-3 x+\log (8)}{(-3+x) x \log ^2\left (-4+\frac {12}{x}\right )}+\frac {1}{\log \left (-4+\frac {12}{x}\right )}\right ) \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {-3 x+\log (8)}{(-3+x) x \log ^2\left (-4+\frac {12}{x}\right )} \, dx,x,\log (x)\right )+\operatorname {Subst}\left (\int \frac {1}{\log \left (-4+\frac {12}{x}\right )} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{\log \left (-4+\frac {12}{x}\right )} \, dx,x,\log (x)\right )+\operatorname {Subst}\left (\int \frac {-3 x+\log (8)}{(-3+x) x \log ^2\left (\frac {12-4 x}{x}\right )} \, dx,x,\log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 24, normalized size = 0.92 \begin {gather*} \frac {-\log (8)+3 \log (x)}{3 \log \left (-4+\frac {12}{\log (x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(Log[x/2]*Log[x]) + Log[x/2]*(1 + Log[x/E^4]) + Log[x]*(1 + Log[x/E^4])*Log[(-4 - 4*Log[x/E^4])/Lo
g[x]])/(Log[x]*(x + x*Log[x/E^4])*Log[(-4 - 4*Log[x/E^4])/Log[x]]^2),x]

[Out]

(-Log[8] + 3*Log[x])/(3*Log[-4 + 12/Log[x]])

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fricas [A]  time = 0.72, size = 34, normalized size = 1.31 \begin {gather*} -\frac {\log \relax (2) - \log \left (x e^{\left (-4\right )}\right ) - 4}{\log \left (-\frac {4 \, {\left (\log \left (x e^{\left (-4\right )}\right ) + 1\right )}}{\log \left (x e^{\left (-4\right )}\right ) + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.38, size = 55, normalized size = 2.12 \begin {gather*} -\frac {\frac {{\left (\log \relax (x) - 3\right )} \log \relax (2)}{\log \relax (x)} - \log \relax (2) + 3}{\frac {{\left (\log \relax (x) - 3\right )} \log \left (-\frac {4 \, {\left (\log \relax (x) - 3\right )}}{\log \relax (x)}\right )}{\log \relax (x)} - \log \left (-\frac {4 \, {\left (\log \relax (x) - 3\right )}}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.22, size = 275, normalized size = 10.58

method	result	size
risch	\(-\frac {2 \ln \relax (2)-2 \ln \relax (x )}{2 \ln \relax (2)+i \pi -2 \ln \left (\ln \relax (x )\right )+2 \ln \left (-6 i+2 i \ln \relax (x )\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i \left (-6 i+2 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (\frac {i \left (-6 i+2 i \ln \relax (x )\right )}{\ln \relax (x )}\right )+i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i \left (-6 i+2 i \ln \relax (x )\right )}{\ln \relax (x )}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (-6 i+2 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (\frac {i \left (-6 i+2 i \ln \relax (x )\right )}{\ln \relax (x )}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i \left (-6 i+2 i \ln \relax (x )\right )}{\ln \relax (x )}\right )^{3}-i \pi \,\mathrm {csgn}\left (\frac {i \left (-6 i+2 i \ln \relax (x )\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {-6 i+2 i \ln \relax (x )}{\ln \relax (x )}\right )-i \pi \mathrm {csgn}\left (\frac {-6 i+2 i \ln \relax (x )}{\ln \relax (x )}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i \left (-6 i+2 i \ln \relax (x )\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {-6 i+2 i \ln \relax (x )}{\ln \relax (x )}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {-6 i+2 i \ln \relax (x )}{\ln \relax (x )}\right )^{3}}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*ln(2)-2*ln(x))/(2*ln(2)+I*Pi-2*ln(ln(x))+2*ln(-6*I+2*I*ln(x))-I*Pi*csgn(I/ln(x))*csgn(I*(-6*I+2*I*ln(x)))*
csgn(I/ln(x)*(-6*I+2*I*ln(x)))+I*Pi*csgn(I/ln(x))*csgn(I/ln(x)*(-6*I+2*I*ln(x)))^2+I*Pi*csgn(I*(-6*I+2*I*ln(x)
))*csgn(I/ln(x)*(-6*I+2*I*ln(x)))^2-I*Pi*csgn(I/ln(x)*(-6*I+2*I*ln(x)))^3-I*Pi*csgn(I/ln(x)*(-6*I+2*I*ln(x)))*
csgn(1/ln(x)*(-6*I+2*I*ln(x)))-I*Pi*csgn(1/ln(x)*(-6*I+2*I*ln(x)))^2+I*Pi*csgn(I/ln(x)*(-6*I+2*I*ln(x)))*csgn(
1/ln(x)*(-6*I+2*I*ln(x)))^2+I*Pi*csgn(1/ln(x)*(-6*I+2*I*ln(x)))^3)

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maxima [C]  time = 1.02, size = 29, normalized size = 1.12 \begin {gather*} -\frac {\log \relax (2) - \log \relax (x)}{i \, \pi + 2 \, \log \relax (2) + \log \left (\log \relax (x) - 3\right ) - \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(2) - log(x))/(I*pi + 2*log(2) + log(log(x) - 3) - log(log(x)))

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mupad [B]  time = 1.28, size = 20, normalized size = 0.77 \begin {gather*} \frac {\ln \left (\frac {x}{2}\right )}{\ln \left (-\frac {4\,\ln \relax (x)-12}{\ln \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x/2)*(log(x*exp(-4)) + 1) - log(x/2)*log(x) + log(-(4*log(x*exp(-4)) + 4)/log(x))*log(x)*(log(x*exp(-
4)) + 1))/(log(-(4*log(x*exp(-4)) + 4)/log(x))^2*log(x)*(x + x*log(x*exp(-4)))),x)

[Out]

log(x/2)/log(-(4*log(x) - 12)/log(x))

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sympy [A]  time = 0.29, size = 17, normalized size = 0.65 \begin {gather*} \frac {\log {\relax (x )} - \log {\relax (2 )}}{\log {\left (\frac {12 - 4 \log {\relax (x )}}{\log {\relax (x )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x/exp(4))+1)*ln(x)*ln((-4*ln(x/exp(4))-4)/ln(x))-ln(1/2*x)*ln(x)+(ln(x/exp(4))+1)*ln(1/2*x))/(x
*ln(x/exp(4))+x)/ln(x)/ln((-4*ln(x/exp(4))-4)/ln(x))**2,x)

[Out]

(log(x) - log(2))/log((12 - 4*log(x))/log(x))

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