3.98.25 \(\int \frac {-x+(-5+2 x) \log (-5+2 x)}{(-20 x+8 x^2) \log (5) \log (-5+2 x)+(-10 x+4 x^2) \log (-5+2 x) \log (-\frac {8 x^2}{5 \log (-5+2 x)})} \, dx\)

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

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Rubi [A]  time = 0.19, antiderivative size = 19, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 3, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6741, 12, 6684} \begin {gather*} \frac {1}{4} \log \left (\log \left (-\frac {40 x^2}{\log (2 x-5)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + (-5 + 2*x)*Log[-5 + 2*x])/((-20*x + 8*x^2)*Log[5]*Log[-5 + 2*x] + (-10*x + 4*x^2)*Log[-5 + 2*x]*Log[
(-8*x^2)/(5*Log[-5 + 2*x])]),x]

[Out]

Log[Log[(-40*x^2)/Log[-5 + 2*x]]]/4

Rule 12

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {x-(-5+2 x) \log (-5+2 x)}{2 (5-2 x) x \log (-5+2 x) \log \left (-\frac {40 x^2}{\log (-5+2 x)}\right )} \, dx\\ &=\frac {1}{2} \int \frac {x-(-5+2 x) \log (-5+2 x)}{(5-2 x) x \log (-5+2 x) \log \left (-\frac {40 x^2}{\log (-5+2 x)}\right )} \, dx\\ &=\frac {1}{4} \log \left (\log \left (-\frac {40 x^2}{\log (-5+2 x)}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 19, normalized size = 0.68 \begin {gather*} \frac {1}{4} \log \left (\log \left (-\frac {40 x^2}{\log (-5+2 x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + (-5 + 2*x)*Log[-5 + 2*x])/((-20*x + 8*x^2)*Log[5]*Log[-5 + 2*x] + (-10*x + 4*x^2)*Log[-5 + 2*x
]*Log[(-8*x^2)/(5*Log[-5 + 2*x])]),x]

[Out]

Log[Log[(-40*x^2)/Log[-5 + 2*x]]]/4

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fricas [A]  time = 0.71, size = 22, normalized size = 0.79 \begin {gather*} \frac {1}{4} \, \log \left (2 \, \log \relax (5) + \log \left (-\frac {8 \, x^{2}}{5 \, \log \left (2 \, x - 5\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-5)*log(2*x-5)-x)/((4*x^2-10*x)*log(2*x-5)*log(-8/5*x^2/log(2*x-5))+(8*x^2-20*x)*log(5)*log(2*x
-5)),x, algorithm="fricas")

[Out]

1/4*log(2*log(5) + log(-8/5*x^2/log(2*x - 5)))

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giac [A]  time = 0.66, size = 23, normalized size = 0.82 \begin {gather*} \frac {1}{4} \, \log \left (-\log \relax (5) - \log \left (-8 \, x^{2}\right ) + \log \left (\log \left (2 \, x - 5\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-5)*log(2*x-5)-x)/((4*x^2-10*x)*log(2*x-5)*log(-8/5*x^2/log(2*x-5))+(8*x^2-20*x)*log(5)*log(2*x
-5)),x, algorithm="giac")

[Out]

1/4*log(-log(5) - log(-8*x^2) + log(log(2*x - 5)))

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maple [A]  time = 0.34, size = 23, normalized size = 0.82

method	result	size
norman	\(\frac {\ln \left (2 \ln \relax (5)+\ln \left (-\frac {8 x^{2}}{5 \ln \left (2 x -5\right )}\right )\right )}{4}\)	\(23\)
risch	\(\frac {\ln \left (\ln \left (\ln \left (2 x -5\right )\right )+\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (2 x -5\right )}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{\ln \left (2 x -5\right )}\right )-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{\ln \left (2 x -5\right )}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i x^{2}}{\ln \left (2 x -5\right )}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\ln \left (2 x -5\right )}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{\ln \left (2 x -5\right )}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i x^{2}}{\ln \left (2 x -5\right )}\right )^{3}+6 i \ln \relax (2)+2 i \ln \relax (5)+4 i \ln \relax (x )-2 \pi \right )}{2}\right )}{4}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x-5)*ln(2*x-5)-x)/((4*x^2-10*x)*ln(2*x-5)*ln(-8/5*x^2/ln(2*x-5))+(8*x^2-20*x)*ln(5)*ln(2*x-5)),x,metho
d=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 25, normalized size = 0.89 \begin {gather*} \frac {1}{4} \, \log \left (-\log \relax (5) - 3 \, \log \relax (2) - 2 \, \log \relax (x) + \log \left (-\log \left (2 \, x - 5\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-5)*log(2*x-5)-x)/((4*x^2-10*x)*log(2*x-5)*log(-8/5*x^2/log(2*x-5))+(8*x^2-20*x)*log(5)*log(2*x
-5)),x, algorithm="maxima")

[Out]

1/4*log(-log(5) - 3*log(2) - 2*log(x) + log(-log(2*x - 5)))

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mupad [B]  time = 6.74, size = 17, normalized size = 0.61 \begin {gather*} \frac {\ln \left (\ln \left (-\frac {40\,x^2}{\ln \left (2\,x-5\right )}\right )\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log(2*x - 5)*(2*x - 5))/(log(2*x - 5)*log(-(8*x^2)/(5*log(2*x - 5)))*(10*x - 4*x^2) + log(5)*log(2*x
- 5)*(20*x - 8*x^2)),x)

[Out]

log(log(-(40*x^2)/log(2*x - 5)))/4

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sympy [A]  time = 0.63, size = 24, normalized size = 0.86 \begin {gather*} \frac {\log {\left (\log {\left (- \frac {8 x^{2}}{5 \log {\left (2 x - 5 \right )}} \right )} + 2 \log {\relax (5 )} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-5)*ln(2*x-5)-x)/((4*x**2-10*x)*ln(2*x-5)*ln(-8/5*x**2/ln(2*x-5))+(8*x**2-20*x)*ln(5)*ln(2*x-5)
),x)

[Out]

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

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