3.64.83 \(\int \frac {(5+(-2-x) \log (x)+(2+x) \log (\frac {x}{4+2 x})) \log (\log (5))}{(1250+525 x-48 x^2+x^3) \log (x)+(-1250-525 x+48 x^2-x^3) \log (\frac {x}{4+2 x})+((500+230 x-10 x^2) \log (x)+(-500-230 x+10 x^2) \log (\frac {x}{4+2 x})) \log (-\log (x)+\log (\frac {x}{4+2 x}))+((50+25 x) \log (x)+(-50-25 x) \log (\frac {x}{4+2 x})) \log ^2(-\log (x)+\log (\frac {x}{4+2 x}))} \, dx\)

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

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Rubi [A]  time = 0.25, antiderivative size = 31, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, integrand size = 169, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {12, 6688, 6686} \begin {gather*} -\frac {\log (\log (5))}{-x+5 \log \left (\log \left (\frac {x}{2 (x+2)}\right )-\log (x)\right )+25} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((5 + (-2 - x)*Log[x] + (2 + x)*Log[x/(4 + 2*x)])*Log[Log[5]])/((1250 + 525*x - 48*x^2 + x^3)*Log[x] + (-1
250 - 525*x + 48*x^2 - x^3)*Log[x/(4 + 2*x)] + ((500 + 230*x - 10*x^2)*Log[x] + (-500 - 230*x + 10*x^2)*Log[x/
(4 + 2*x)])*Log[-Log[x] + Log[x/(4 + 2*x)]] + ((50 + 25*x)*Log[x] + (-50 - 25*x)*Log[x/(4 + 2*x)])*Log[-Log[x]
+ Log[x/(4 + 2*x)]]^2),x]

[Out]

-(Log[Log[5]]/(25 - x + 5*Log[-Log[x] + Log[x/(2*(2 + 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 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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (\log (5)) \int \frac {5+(-2-x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )}{\left (1250+525 x-48 x^2+x^3\right ) \log (x)+\left (-1250-525 x+48 x^2-x^3\right ) \log \left (\frac {x}{4+2 x}\right )+\left (\left (500+230 x-10 x^2\right ) \log (x)+\left (-500-230 x+10 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right ) \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )+\left ((50+25 x) \log (x)+(-50-25 x) \log \left (\frac {x}{4+2 x}\right )\right ) \log ^2\left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \, dx\\ &=\log (\log (5)) \int \frac {5-(2+x) \log (x)+(2+x) \log \left (\frac {x}{4+2 x}\right )}{(2+x) \left (\log (x)-\log \left (\frac {x}{4+2 x}\right )\right ) \left (25-x+5 \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )\right )^2} \, dx\\ &=-\frac {\log (\log (5))}{25-x+5 \log \left (-\log (x)+\log \left (\frac {x}{2 (2+x)}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 0.96 \begin {gather*} \frac {\log (\log (5))}{-25+x-5 \log \left (-\log (x)+\log \left (\frac {x}{4+2 x}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((5 + (-2 - x)*Log[x] + (2 + x)*Log[x/(4 + 2*x)])*Log[Log[5]])/((1250 + 525*x - 48*x^2 + x^3)*Log[x]
+ (-1250 - 525*x + 48*x^2 - x^3)*Log[x/(4 + 2*x)] + ((500 + 230*x - 10*x^2)*Log[x] + (-500 - 230*x + 10*x^2)*
Log[x/(4 + 2*x)])*Log[-Log[x] + Log[x/(4 + 2*x)]] + ((50 + 25*x)*Log[x] + (-50 - 25*x)*Log[x/(4 + 2*x)])*Log[-
Log[x] + Log[x/(4 + 2*x)]]^2),x]

[Out]

Log[Log[5]]/(-25 + x - 5*Log[-Log[x] + Log[x/(4 + 2*x)]])

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fricas [A]  time = 0.88, size = 26, normalized size = 0.93 \begin {gather*} \frac {\log \left (\log \relax (5)\right )}{x - 5 \, \log \left (-\log \relax (x) + \log \left (\frac {x}{2 \, {\left (x + 2\right )}}\right )\right ) - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*log(x)+log(x/(2*x+4))*(2+x)+5)*log(log(5))/(((25*x+50)*log(x)+(-25*x-50)*log(x/(2*x+4)))*log
(-log(x)+log(x/(2*x+4)))^2+((-10*x^2+230*x+500)*log(x)+(10*x^2-230*x-500)*log(x/(2*x+4)))*log(-log(x)+log(x/(2
*x+4)))+(x^3-48*x^2+525*x+1250)*log(x)+(-x^3+48*x^2-525*x-1250)*log(x/(2*x+4))),x, algorithm="fricas")

[Out]

log(log(5))/(x - 5*log(-log(x) + log(1/2*x/(x + 2))) - 25)

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giac [A]  time = 0.39, size = 23, normalized size = 0.82 \begin {gather*} \frac {\log \left (\log \relax (5)\right )}{x - 5 \, \log \left (-\log \relax (2) - \log \left (x + 2\right )\right ) - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*log(x)+log(x/(2*x+4))*(2+x)+5)*log(log(5))/(((25*x+50)*log(x)+(-25*x-50)*log(x/(2*x+4)))*log
(-log(x)+log(x/(2*x+4)))^2+((-10*x^2+230*x+500)*log(x)+(10*x^2-230*x-500)*log(x/(2*x+4)))*log(-log(x)+log(x/(2
*x+4)))+(x^3-48*x^2+525*x+1250)*log(x)+(-x^3+48*x^2-525*x-1250)*log(x/(2*x+4))),x, algorithm="giac")

[Out]

log(log(5))/(x - 5*log(-log(2) - log(x + 2)) - 25)

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maple [C]  time = 0.19, size = 78, normalized size = 2.79

method	result	size
risch	\(\frac {\ln \left (\ln \relax (5)\right )}{x -5 \ln \left (-\ln \relax (2)-\ln \left (2+x \right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x}{2+x}\right ) \left (-\mathrm {csgn}\left (\frac {i x}{2+x}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x}{2+x}\right )+\mathrm {csgn}\left (\frac {i}{2+x}\right )\right )}{2}\right )-25}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x-2)*ln(x)+ln(x/(2*x+4))*(2+x)+5)*ln(ln(5))/(((25*x+50)*ln(x)+(-25*x-50)*ln(x/(2*x+4)))*ln(-ln(x)+ln(x/
(2*x+4)))^2+((-10*x^2+230*x+500)*ln(x)+(10*x^2-230*x-500)*ln(x/(2*x+4)))*ln(-ln(x)+ln(x/(2*x+4)))+(x^3-48*x^2+
525*x+1250)*ln(x)+(-x^3+48*x^2-525*x-1250)*ln(x/(2*x+4))),x,method=_RETURNVERBOSE)

[Out]

ln(ln(5))/(x-5*ln(-ln(2)-ln(2+x)-1/2*I*Pi*csgn(I*x/(2+x))*(-csgn(I*x/(2+x))+csgn(I*x))*(-csgn(I*x/(2+x))+csgn(
I/(2+x))))-25)

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maxima [A]  time = 0.60, size = 23, normalized size = 0.82 \begin {gather*} \frac {\log \left (\log \relax (5)\right )}{x - 5 \, \log \left (-\log \relax (2) - \log \left (x + 2\right )\right ) - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*log(x)+log(x/(2*x+4))*(2+x)+5)*log(log(5))/(((25*x+50)*log(x)+(-25*x-50)*log(x/(2*x+4)))*log
(-log(x)+log(x/(2*x+4)))^2+((-10*x^2+230*x+500)*log(x)+(10*x^2-230*x-500)*log(x/(2*x+4)))*log(-log(x)+log(x/(2
*x+4)))+(x^3-48*x^2+525*x+1250)*log(x)+(-x^3+48*x^2-525*x-1250)*log(x/(2*x+4))),x, algorithm="maxima")

[Out]

log(log(5))/(x - 5*log(-log(2) - log(x + 2)) - 25)

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mupad [B]  time = 4.80, size = 30, normalized size = 1.07 \begin {gather*} -\frac {\ln \left (\ln \relax (5)\right )}{5\,\ln \left (\ln \left (\frac {x}{2\,x+4}\right )-\ln \relax (x)\right )-x+25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(5))*(log(x/(2*x + 4))*(x + 2) - log(x)*(x + 2) + 5))/(log(log(x/(2*x + 4)) - log(x))*(log(x/(2*x
+ 4))*(230*x - 10*x^2 + 500) - log(x)*(230*x - 10*x^2 + 500)) + log(x/(2*x + 4))*(525*x - 48*x^2 + x^3 + 1250
) - log(x)*(525*x - 48*x^2 + x^3 + 1250) + log(log(x/(2*x + 4)) - log(x))^2*(log(x/(2*x + 4))*(25*x + 50) - lo
g(x)*(25*x + 50))),x)

[Out]

-log(log(5))/(5*log(log(x/(2*x + 4)) - log(x)) - x + 25)

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sympy [A]  time = 0.78, size = 24, normalized size = 0.86 \begin {gather*} - \frac {\log {\left (\log {\relax (5 )} \right )}}{- x + 5 \log {\left (- \log {\relax (x )} + \log {\left (\frac {x}{2 x + 4} \right )} \right )} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*ln(x)+ln(x/(2*x+4))*(2+x)+5)*ln(ln(5))/(((25*x+50)*ln(x)+(-25*x-50)*ln(x/(2*x+4)))*ln(-ln(x)
+ln(x/(2*x+4)))**2+((-10*x**2+230*x+500)*ln(x)+(10*x**2-230*x-500)*ln(x/(2*x+4)))*ln(-ln(x)+ln(x/(2*x+4)))+(x*
*3-48*x**2+525*x+1250)*ln(x)+(-x**3+48*x**2-525*x-1250)*ln(x/(2*x+4))),x)

[Out]

-log(log(5))/(-x + 5*log(-log(x) + log(x/(2*x + 4))) + 25)

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