3.56.46 \(\int \frac {(20 x-4 x^2) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{(-5 x+x^2) \log (-5+x) \log (x)} \, dx\) [5546]

Optimal. Leaf size=23 \[ 4 \left (x-2 \left (x+\log (5)+(\log (\log (-5+x))-\log (\log (x)))^2\right )\right ) \]

[Out]

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

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Rubi [F]
time = 0.58, 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 {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{\left (-5 x+x^2\right ) \log (-5+x) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((20*x - 4*x^2)*Log[-5 + x]*Log[x] + ((-80 + 16*x)*Log[-5 + x] - 16*x*Log[x])*Log[Log[-5 + x]] + ((80 - 16
*x)*Log[-5 + x] + 16*x*Log[x])*Log[Log[x]])/((-5*x + x^2)*Log[-5 + x]*Log[x]),x]

[Out]

-4*x - 8*Log[Log[-5 + x]]^2 - 8*Log[Log[x]]^2 + 16*Defer[Int][Log[Log[-5 + x]]/(x*Log[x]), x] + 16*Defer[Int][
Log[Log[x]]/((-5 + x)*Log[-5 + x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (20 x-4 x^2\right ) \log (-5+x) \log (x)+((-80+16 x) \log (-5+x)-16 x \log (x)) \log (\log (-5+x))+((80-16 x) \log (-5+x)+16 x \log (x)) \log (\log (x))}{(-5+x) x \log (-5+x) \log (x)} \, dx\\ &=\int \left (-\frac {16 (\log (\log (-5+x))-\log (\log (x)))}{(-5+x) \log (-5+x)}-\frac {4 (x \log (x)-4 \log (\log (-5+x))+4 \log (\log (x)))}{x \log (x)}\right ) \, dx\\ &=-\left (4 \int \frac {x \log (x)-4 \log (\log (-5+x))+4 \log (\log (x))}{x \log (x)} \, dx\right )-16 \int \frac {\log (\log (-5+x))-\log (\log (x))}{(-5+x) \log (-5+x)} \, dx\\ &=-\left (4 \int \left (\frac {x \log (x)-4 \log (\log (-5+x))}{x \log (x)}+\frac {4 \log (\log (x))}{x \log (x)}\right ) \, dx\right )-16 \int \left (\frac {\log (\log (-5+x))}{(-5+x) \log (-5+x)}-\frac {\log (\log (x))}{(-5+x) \log (-5+x)}\right ) \, dx\\ &=-\left (4 \int \frac {x \log (x)-4 \log (\log (-5+x))}{x \log (x)} \, dx\right )-16 \int \frac {\log (\log (-5+x))}{(-5+x) \log (-5+x)} \, dx+16 \int \frac {\log (\log (x))}{(-5+x) \log (-5+x)} \, dx-16 \int \frac {\log (\log (x))}{x \log (x)} \, dx\\ &=-8 \log ^2(\log (-5+x))-4 \int \left (1-\frac {4 \log (\log (-5+x))}{x \log (x)}\right ) \, dx+16 \int \frac {\log (\log (x))}{(-5+x) \log (-5+x)} \, dx-16 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,\log (x)\right )\\ &=-4 x-8 \log ^2(\log (-5+x))-8 \log ^2(\log (x))+16 \int \frac {\log (\log (-5+x))}{x \log (x)} \, dx+16 \int \frac {\log (\log (x))}{(-5+x) \log (-5+x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.07, size = 30, normalized size = 1.30 \begin {gather*} -4 x-8 \log ^2(\log (-5+x))+16 \log (\log (-5+x)) \log (\log (x))-8 \log ^2(\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((20*x - 4*x^2)*Log[-5 + x]*Log[x] + ((-80 + 16*x)*Log[-5 + x] - 16*x*Log[x])*Log[Log[-5 + x]] + ((8
0 - 16*x)*Log[-5 + x] + 16*x*Log[x])*Log[Log[x]])/((-5*x + x^2)*Log[-5 + x]*Log[x]),x]

[Out]

-4*x - 8*Log[Log[-5 + x]]^2 + 16*Log[Log[-5 + x]]*Log[Log[x]] - 8*Log[Log[x]]^2

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Maple [A]
time = 1.25, size = 31, normalized size = 1.35

method	result	size
risch	\(-8 \ln \left (\ln \left (x \right )\right )^{2}+16 \ln \left (\ln \left (x \right )\right ) \ln \left (\ln \left (x -5\right )\right )-8 \ln \left (\ln \left (x -5\right )\right )^{2}-4 x\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x*ln(x)+(-16*x+80)*ln(x-5))*ln(ln(x))+(-16*x*ln(x)+(16*x-80)*ln(x-5))*ln(ln(x-5))+(-4*x^2+20*x)*ln(x-
5)*ln(x))/(x^2-5*x)/ln(x-5)/ln(x),x,method=_RETURNVERBOSE)

[Out]

-8*ln(ln(x))^2+16*ln(ln(x))*ln(ln(x-5))-8*ln(ln(x-5))^2-4*x

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Maxima [A]
time = 0.39, size = 30, normalized size = 1.30 \begin {gather*} -8 \, \log \left (\log \left (x - 5\right )\right )^{2} + 16 \, \log \left (\log \left (x - 5\right )\right ) \log \left (\log \left (x\right )\right ) - 8 \, \log \left (\log \left (x\right )\right )^{2} - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*log(x)+(-16*x+80)*log(-5+x))*log(log(x))+(-16*x*log(x)+(16*x-80)*log(-5+x))*log(log(-5+x))+(-
4*x^2+20*x)*log(-5+x)*log(x))/(x^2-5*x)/log(-5+x)/log(x),x, algorithm="maxima")

[Out]

-8*log(log(x - 5))^2 + 16*log(log(x - 5))*log(log(x)) - 8*log(log(x))^2 - 4*x

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Fricas [A]
time = 0.36, size = 30, normalized size = 1.30 \begin {gather*} -8 \, \log \left (\log \left (x - 5\right )\right )^{2} + 16 \, \log \left (\log \left (x - 5\right )\right ) \log \left (\log \left (x\right )\right ) - 8 \, \log \left (\log \left (x\right )\right )^{2} - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*log(x)+(-16*x+80)*log(-5+x))*log(log(x))+(-16*x*log(x)+(16*x-80)*log(-5+x))*log(log(-5+x))+(-
4*x^2+20*x)*log(-5+x)*log(x))/(x^2-5*x)/log(-5+x)/log(x),x, algorithm="fricas")

[Out]

-8*log(log(x - 5))^2 + 16*log(log(x - 5))*log(log(x)) - 8*log(log(x))^2 - 4*x

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Sympy [A]
time = 1.03, size = 34, normalized size = 1.48 \begin {gather*} - 4 x - 8 \log {\left (\log {\left (x \right )} \right )}^{2} + 16 \log {\left (\log {\left (x \right )} \right )} \log {\left (\log {\left (x - 5 \right )} \right )} - 8 \log {\left (\log {\left (x - 5 \right )} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*ln(x)+(-16*x+80)*ln(-5+x))*ln(ln(x))+(-16*x*ln(x)+(16*x-80)*ln(-5+x))*ln(ln(-5+x))+(-4*x**2+2
0*x)*ln(-5+x)*ln(x))/(x**2-5*x)/ln(-5+x)/ln(x),x)

[Out]

-4*x - 8*log(log(x))**2 + 16*log(log(x))*log(log(x - 5)) - 8*log(log(x - 5))**2

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Giac [A]
time = 0.42, size = 30, normalized size = 1.30 \begin {gather*} -8 \, \log \left (\log \left (x - 5\right )\right )^{2} + 16 \, \log \left (\log \left (x - 5\right )\right ) \log \left (\log \left (x\right )\right ) - 8 \, \log \left (\log \left (x\right )\right )^{2} - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x*log(x)+(-16*x+80)*log(-5+x))*log(log(x))+(-16*x*log(x)+(16*x-80)*log(-5+x))*log(log(-5+x))+(-
4*x^2+20*x)*log(-5+x)*log(x))/(x^2-5*x)/log(-5+x)/log(x),x, algorithm="giac")

[Out]

-8*log(log(x - 5))^2 + 16*log(log(x - 5))*log(log(x)) - 8*log(log(x))^2 - 4*x

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Mupad [B]
time = 4.06, size = 30, normalized size = 1.30 \begin {gather*} -8\,{\ln \left (\ln \left (x\right )\right )}^2+16\,\ln \left (\ln \left (x\right )\right )\,\ln \left (\ln \left (x-5\right )\right )-8\,{\ln \left (\ln \left (x-5\right )\right )}^2-4\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(x))*(16*x*log(x) - log(x - 5)*(16*x - 80)) - log(log(x - 5))*(16*x*log(x) - log(x - 5)*(16*x - 8
0)) + log(x - 5)*log(x)*(20*x - 4*x^2))/(log(x - 5)*log(x)*(5*x - x^2)),x)

[Out]

16*log(log(x))*log(log(x - 5)) - 8*log(log(x - 5))^2 - 4*x - 8*log(log(x))^2

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