3.90.55 \(\int \frac {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=21 \[ 4-\frac {2 (-4+x)}{x^2 \log (5 x) \log (\log (x))} \]

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Rubi [F]  time = 1.68, 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 {(-8+2 x) \log (5 x)+((-8+2 x) \log (x)+(-16+2 x) \log (x) \log (5 x)) \log (\log (x))}{x^3 \log (x) \log ^2(5 x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-8 + 2*x)*Log[5*x] + ((-8 + 2*x)*Log[x] + (-16 + 2*x)*Log[x]*Log[5*x])*Log[Log[x]])/(x^3*Log[x]*Log[5*x]
^2*Log[Log[x]]^2),x]

[Out]

-8*Defer[Int][1/(x^3*Log[x]*Log[5*x]*Log[Log[x]]^2), x] + 2*Defer[Int][1/(x^2*Log[x]*Log[5*x]*Log[Log[x]]^2),
x] - 8*Defer[Int][1/(x^3*Log[5*x]^2*Log[Log[x]]), x] + 2*Defer[Int][1/(x^2*Log[5*x]^2*Log[Log[x]]), x] - 16*De
fer[Int][1/(x^3*Log[5*x]*Log[Log[x]]), x] + 2*Defer[Int][1/(x^2*Log[5*x]*Log[Log[x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 (-4+x)}{x^3 \log (x) \log (5 x) \log ^2(\log (x))}+\frac {2 (-4+x-8 \log (5 x)+x \log (5 x))}{x^3 \log ^2(5 x) \log (\log (x))}\right ) \, dx\\ &=2 \int \frac {-4+x}{x^3 \log (x) \log (5 x) \log ^2(\log (x))} \, dx+2 \int \frac {-4+x-8 \log (5 x)+x \log (5 x)}{x^3 \log ^2(5 x) \log (\log (x))} \, dx\\ &=2 \int \left (-\frac {4}{x^3 \log (x) \log (5 x) \log ^2(\log (x))}+\frac {1}{x^2 \log (x) \log (5 x) \log ^2(\log (x))}\right ) \, dx+2 \int \frac {-4+x+(-8+x) \log (5 x)}{x^3 \log ^2(5 x) \log (\log (x))} \, dx\\ &=2 \int \left (-\frac {4}{x^3 \log ^2(5 x) \log (\log (x))}+\frac {1}{x^2 \log ^2(5 x) \log (\log (x))}-\frac {8}{x^3 \log (5 x) \log (\log (x))}+\frac {1}{x^2 \log (5 x) \log (\log (x))}\right ) \, dx+2 \int \frac {1}{x^2 \log (x) \log (5 x) \log ^2(\log (x))} \, dx-8 \int \frac {1}{x^3 \log (x) \log (5 x) \log ^2(\log (x))} \, dx\\ &=2 \int \frac {1}{x^2 \log (x) \log (5 x) \log ^2(\log (x))} \, dx+2 \int \frac {1}{x^2 \log ^2(5 x) \log (\log (x))} \, dx+2 \int \frac {1}{x^2 \log (5 x) \log (\log (x))} \, dx-8 \int \frac {1}{x^3 \log (x) \log (5 x) \log ^2(\log (x))} \, dx-8 \int \frac {1}{x^3 \log ^2(5 x) \log (\log (x))} \, dx-16 \int \frac {1}{x^3 \log (5 x) \log (\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 21, normalized size = 1.00 \begin {gather*} \frac {2 (4-x)}{x^2 \log (5 x) \log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-8 + 2*x)*Log[5*x] + ((-8 + 2*x)*Log[x] + (-16 + 2*x)*Log[x]*Log[5*x])*Log[Log[x]])/(x^3*Log[x]*Lo
g[5*x]^2*Log[Log[x]]^2),x]

[Out]

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

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fricas [A]  time = 0.63, size = 25, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (x - 4\right )}}{{\left (x^{2} \log \relax (5) + x^{2} \log \relax (x)\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-16)*log(x)*log(5*x)+(2*x-8)*log(x))*log(log(x))+(2*x-8)*log(5*x))/x^3/log(x)/log(5*x)^2/log(l
og(x))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 26, normalized size = 1.24 \begin {gather*} -\frac {2 \, {\left (x - 4\right )}}{x^{2} \log \relax (5) \log \left (\log \relax (x)\right ) + x^{2} \log \relax (x) \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-16)*log(x)*log(5*x)+(2*x-8)*log(x))*log(log(x))+(2*x-8)*log(5*x))/x^3/log(x)/log(5*x)^2/log(l
og(x))^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.15, size = 28, normalized size = 1.33

method	result	size
risch	\(-\frac {4 i \left (x -4\right )}{x^{2} \left (2 i \ln \relax (5)+2 i \ln \relax (x )\right ) \ln \left (\ln \relax (x )\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x-16)*ln(x)*ln(5*x)+(2*x-8)*ln(x))*ln(ln(x))+(2*x-8)*ln(5*x))/x^3/ln(x)/ln(5*x)^2/ln(ln(x))^2,x,metho
d=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 25, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (x - 4\right )}}{{\left (x^{2} \log \relax (5) + x^{2} \log \relax (x)\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-16)*log(x)*log(5*x)+(2*x-8)*log(x))*log(log(x))+(2*x-8)*log(5*x))/x^3/log(x)/log(5*x)^2/log(l
og(x))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.75, size = 513, normalized size = 24.43 \begin {gather*} \frac {\frac {x-8\,\ln \left (5\,x\right )+8\,\ln \relax (x)-x\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2+x\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^3+16\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2-32\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^3+x\,\left (\ln \left (5\,x\right )-\ln \relax (x)\right )-4}{x^2}+\frac {{\ln \relax (x)}^2\,\left (32\,\ln \relax (x)-32\,\ln \left (5\,x\right )-x+x\,\left (\ln \left (5\,x\right )-\ln \relax (x)\right )+16\right )}{x^2}-\frac {\ln \relax (x)\,\left (32\,\ln \relax (x)-32\,\ln \left (5\,x\right )-x-2\,x\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2+64\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2+2\,x\,\left (\ln \left (5\,x\right )-\ln \relax (x)\right )+8\right )}{x^2}}{\ln \left (5\,x\right )}-\frac {\frac {2\,\left (x-4\right )}{x^2\,\ln \left (5\,x\right )}+\frac {2\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (x-8\,\ln \left (5\,x\right )+x\,\ln \relax (x)+x\,\left (\ln \left (5\,x\right )-\ln \relax (x)\right )-4\right )}{x^2\,{\ln \left (5\,x\right )}^2}}{\ln \left (\ln \relax (x)\right )}-\frac {32\,\ln \left (5\,x\right )-32\,\ln \relax (x)-32\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2+x\,\left (2\,\ln \relax (x)-2\,\ln \left (5\,x\right )+{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2\right )}{x^2}-\frac {\frac {4\,\ln \relax (x)-4\,\ln \left (5\,x\right )+x\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2+x\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^3-8\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2-16\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^3+x\,\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}{x^2}+\frac {{\ln \relax (x)}^2\,\left (16\,\ln \relax (x)-16\,\ln \left (5\,x\right )-x+x\,\left (\ln \left (5\,x\right )-\ln \relax (x)\right )+8\right )}{x^2}-\frac {\ln \relax (x)\,\left (x-2\,x\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2+32\,{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2-4\right )}{x^2}}{2\,\ln \relax (x)\,\left (\ln \left (5\,x\right )-\ln \relax (x)\right )+{\ln \relax (x)}^2+{\left (\ln \left (5\,x\right )-\ln \relax (x)\right )}^2}+\frac {\ln \relax (x)\,\left (32\,\ln \left (5\,x\right )-32\,\ln \relax (x)+x\,\left (\ln \relax (x)-\ln \left (5\,x\right )+1\right )-16\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x))*(log(x)*(2*x - 8) + log(5*x)*log(x)*(2*x - 16)) + log(5*x)*(2*x - 8))/(x^3*log(5*x)^2*log(log
(x))^2*log(x)),x)

[Out]

((x - 8*log(5*x) + 8*log(x) - x*(log(5*x) - log(x))^2 + x*(log(5*x) - log(x))^3 + 16*(log(5*x) - log(x))^2 - 3
2*(log(5*x) - log(x))^3 + x*(log(5*x) - log(x)) - 4)/x^2 + (log(x)^2*(32*log(x) - 32*log(5*x) - x + x*(log(5*x
) - log(x)) + 16))/x^2 - (log(x)*(32*log(x) - 32*log(5*x) - x - 2*x*(log(5*x) - log(x))^2 + 64*(log(5*x) - log
(x))^2 + 2*x*(log(5*x) - log(x)) + 8))/x^2)/log(5*x) - ((2*(x - 4))/(x^2*log(5*x)) + (2*log(log(x))*log(x)*(x
- 8*log(5*x) + x*log(x) + x*(log(5*x) - log(x)) - 4))/(x^2*log(5*x)^2))/log(log(x)) - (32*log(5*x) - 32*log(x)
- 32*(log(5*x) - log(x))^2 + x*(2*log(x) - 2*log(5*x) + (log(5*x) - log(x))^2))/x^2 - ((4*log(x) - 4*log(5*x)
+ x*(log(5*x) - log(x))^2 + x*(log(5*x) - log(x))^3 - 8*(log(5*x) - log(x))^2 - 16*(log(5*x) - log(x))^3 + x*
(log(5*x) - log(x)))/x^2 + (log(x)^2*(16*log(x) - 16*log(5*x) - x + x*(log(5*x) - log(x)) + 8))/x^2 - (log(x)*
(x - 2*x*(log(5*x) - log(x))^2 + 32*(log(5*x) - log(x))^2 - 4))/x^2)/(2*log(x)*(log(5*x) - log(x)) + log(x)^2
+ (log(5*x) - log(x))^2) + (log(x)*(32*log(5*x) - 32*log(x) + x*(log(x) - log(5*x) + 1) - 16))/x^2

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sympy [A]  time = 0.32, size = 22, normalized size = 1.05 \begin {gather*} \frac {8 - 2 x}{\left (x^{2} \log {\relax (x )} + x^{2} \log {\relax (5 )}\right ) \log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-16)*ln(x)*ln(5*x)+(2*x-8)*ln(x))*ln(ln(x))+(2*x-8)*ln(5*x))/x**3/ln(x)/ln(5*x)**2/ln(ln(x))**
2,x)

[Out]

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

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