3.53.41 \(\int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+((200-200 x) \log (x)+100 x \log ^2(x)) \log ^2(\log (4))}{(x-2 x^2+x^3) \log ^2(x)+(2 x-4 x^2+2 x^3) \log (x) \log ^2(\log (4))+(x-2 x^2+x^3) \log ^4(\log (4))} \, dx\)

Optimal. Leaf size=23 \[ \frac {100 \log ^2(x)}{(1-x) \left (\log (x)+\log ^2(\log (4))\right )} \]

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Rubi [F]  time = 1.43, 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 {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((100 - 100*x)*Log[x]^2 + 100*x*Log[x]^3 + ((200 - 200*x)*Log[x] + 100*x*Log[x]^2)*Log[Log[4]]^2)/((x - 2*
x^2 + x^3)*Log[x]^2 + (2*x - 4*x^2 + 2*x^3)*Log[x]*Log[Log[4]]^2 + (x - 2*x^2 + x^3)*Log[Log[4]]^4),x]

[Out]

100*Log[x] + (100*x*Log[x])/(1 - x) - (100*Log[Log[4]]^2)/(1 - x) + 100*Log[Log[4]]^4*Defer[Int][1/((-1 + x)*x
*(Log[x] + Log[Log[4]]^2)^2), x] + 100*Log[Log[4]]^4*Defer[Int][1/((-1 + x)^2*(Log[x] + Log[Log[4]]^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {100 \log (x) \left (\log (x)+x \log ^2(x)-2 (-1+x) \log ^2(\log (4))+x \log (x) \left (-1+\log ^2(\log (4))\right )\right )}{(1-x)^2 x \left (\log (x)+\log ^2(\log (4))\right )^2} \, dx\\ &=100 \int \frac {\log (x) \left (\log (x)+x \log ^2(x)-2 (-1+x) \log ^2(\log (4))+x \log (x) \left (-1+\log ^2(\log (4))\right )\right )}{(1-x)^2 x \left (\log (x)+\log ^2(\log (4))\right )^2} \, dx\\ &=100 \int \left (\frac {\log (x)}{(-1+x)^2}+\frac {\log ^4(\log (4))}{(-1+x) x \left (\log (x)+\log ^2(\log (4))\right )^2}+\frac {\log ^4(\log (4))}{(-1+x)^2 \left (\log (x)+\log ^2(\log (4))\right )}+\frac {1-x \left (1+\log ^2(\log (4))\right )}{(1-x)^2 x}\right ) \, dx\\ &=100 \int \frac {\log (x)}{(-1+x)^2} \, dx+100 \int \frac {1-x \left (1+\log ^2(\log (4))\right )}{(1-x)^2 x} \, dx+\left (100 \log ^4(\log (4))\right ) \int \frac {1}{(-1+x) x \left (\log (x)+\log ^2(\log (4))\right )^2} \, dx+\left (100 \log ^4(\log (4))\right ) \int \frac {1}{(-1+x)^2 \left (\log (x)+\log ^2(\log (4))\right )} \, dx\\ &=\frac {100 x \log (x)}{1-x}+100 \int \frac {1}{-1+x} \, dx+100 \int \left (\frac {1}{1-x}+\frac {1}{x}-\frac {\log ^2(\log (4))}{(-1+x)^2}\right ) \, dx+\left (100 \log ^4(\log (4))\right ) \int \frac {1}{(-1+x) x \left (\log (x)+\log ^2(\log (4))\right )^2} \, dx+\left (100 \log ^4(\log (4))\right ) \int \frac {1}{(-1+x)^2 \left (\log (x)+\log ^2(\log (4))\right )} \, dx\\ &=100 \log (x)+\frac {100 x \log (x)}{1-x}-\frac {100 \log ^2(\log (4))}{1-x}+\left (100 \log ^4(\log (4))\right ) \int \frac {1}{(-1+x) x \left (\log (x)+\log ^2(\log (4))\right )^2} \, dx+\left (100 \log ^4(\log (4))\right ) \int \frac {1}{(-1+x)^2 \left (\log (x)+\log ^2(\log (4))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 21, normalized size = 0.91 \begin {gather*} -\frac {100 \log ^2(x)}{(-1+x) \left (\log (x)+\log ^2(\log (4))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((100 - 100*x)*Log[x]^2 + 100*x*Log[x]^3 + ((200 - 200*x)*Log[x] + 100*x*Log[x]^2)*Log[Log[4]]^2)/((
x - 2*x^2 + x^3)*Log[x]^2 + (2*x - 4*x^2 + 2*x^3)*Log[x]*Log[Log[4]]^2 + (x - 2*x^2 + x^3)*Log[Log[4]]^4),x]

[Out]

(-100*Log[x]^2)/((-1 + x)*(Log[x] + Log[Log[4]]^2))

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fricas [A]  time = 0.46, size = 26, normalized size = 1.13 \begin {gather*} -\frac {100 \, \log \relax (x)^{2}}{{\left (x - 1\right )} \log \left (2 \, \log \relax (2)\right )^{2} + {\left (x - 1\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x*log(x)^2+(-200*x+200)*log(x))*log(2*log(2))^2+100*x*log(x)^3+(-100*x+100)*log(x)^2)/((x^3-2*
x^2+x)*log(2*log(2))^4+(2*x^3-4*x^2+2*x)*log(x)*log(2*log(2))^2+(x^3-2*x^2+x)*log(x)^2),x, algorithm="fricas")

[Out]

-100*log(x)^2/((x - 1)*log(2*log(2))^2 + (x - 1)*log(x))

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giac [B]  time = 0.28, size = 127, normalized size = 5.52 \begin {gather*} -\frac {100 \, {\left (\log \relax (2)^{4} + 4 \, \log \relax (2)^{3} \log \left (\log \relax (2)\right ) + 6 \, \log \relax (2)^{2} \log \left (\log \relax (2)\right )^{2} + 4 \, \log \relax (2) \log \left (\log \relax (2)\right )^{3} + \log \left (\log \relax (2)\right )^{4}\right )}}{x \log \relax (2)^{2} + 2 \, x \log \relax (2) \log \left (\log \relax (2)\right ) + x \log \left (\log \relax (2)\right )^{2} - \log \relax (2)^{2} + x \log \relax (x) - 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) - \log \left (\log \relax (2)\right )^{2} - \log \relax (x)} + \frac {100 \, {\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )}}{x - 1} - \frac {100 \, \log \relax (x)}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x*log(x)^2+(-200*x+200)*log(x))*log(2*log(2))^2+100*x*log(x)^3+(-100*x+100)*log(x)^2)/((x^3-2*
x^2+x)*log(2*log(2))^4+(2*x^3-4*x^2+2*x)*log(x)*log(2*log(2))^2+(x^3-2*x^2+x)*log(x)^2),x, algorithm="giac")

[Out]

-100*(log(2)^4 + 4*log(2)^3*log(log(2)) + 6*log(2)^2*log(log(2))^2 + 4*log(2)*log(log(2))^3 + log(log(2))^4)/(
x*log(2)^2 + 2*x*log(2)*log(log(2)) + x*log(log(2))^2 - log(2)^2 + x*log(x) - 2*log(2)*log(log(2)) - log(log(2
))^2 - log(x)) + 100*(log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2)/(x - 1) - 100*log(x)/(x - 1)

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maple [A]  time = 0.42, size = 24, normalized size = 1.04

method	result	size
norman	\(-\frac {100 \ln \relax (x )^{2}}{\left (x -1\right ) \left (\ln \relax (x )+\ln \left (2 \ln \relax (2)\right )^{2}\right )}\)	\(24\)
risch	\(-\frac {100 \ln \relax (x )}{x -1}+\frac {100 \ln \relax (2)^{2}}{x -1}+\frac {200 \ln \relax (2) \ln \left (\ln \relax (2)\right )}{x -1}+\frac {100 \ln \left (\ln \relax (2)\right )^{2}}{x -1}-\frac {100 \left (\ln \relax (2)^{4}+4 \ln \relax (2)^{3} \ln \left (\ln \relax (2)\right )+6 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right )^{2}+4 \ln \relax (2) \ln \left (\ln \relax (2)\right )^{3}+\ln \left (\ln \relax (2)\right )^{4}\right )}{\left (x -1\right ) \left (\ln \relax (2)^{2}+2 \ln \relax (2) \ln \left (\ln \relax (2)\right )+\ln \left (\ln \relax (2)\right )^{2}+\ln \relax (x )\right )}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((100*x*ln(x)^2+(-200*x+200)*ln(x))*ln(2*ln(2))^2+100*x*ln(x)^3+(-100*x+100)*ln(x)^2)/((x^3-2*x^2+x)*ln(2*
ln(2))^4+(2*x^3-4*x^2+2*x)*ln(x)*ln(2*ln(2))^2+(x^3-2*x^2+x)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

-100*ln(x)^2/(x-1)/(ln(x)+ln(2*ln(2))^2)

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maxima [B]  time = 0.50, size = 54, normalized size = 2.35 \begin {gather*} -\frac {100 \, \log \relax (x)^{2}}{{\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )} x - \log \relax (2)^{2} + {\left (x - 1\right )} \log \relax (x) - 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) - \log \left (\log \relax (2)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x*log(x)^2+(-200*x+200)*log(x))*log(2*log(2))^2+100*x*log(x)^3+(-100*x+100)*log(x)^2)/((x^3-2*
x^2+x)*log(2*log(2))^4+(2*x^3-4*x^2+2*x)*log(x)*log(2*log(2))^2+(x^3-2*x^2+x)*log(x)^2),x, algorithm="maxima")

[Out]

-100*log(x)^2/((log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2)*x - log(2)^2 + (x - 1)*log(x) - 2*log(2)*log(
log(2)) - log(log(2))^2)

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mupad [B]  time = 3.84, size = 21, normalized size = 0.91 \begin {gather*} -\frac {100\,{\ln \relax (x)}^2}{\left (\ln \relax (x)+{\ln \left (\ln \relax (4)\right )}^2\right )\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*x*log(x)^3 - log(x)^2*(100*x - 100) + log(2*log(2))^2*(100*x*log(x)^2 - log(x)*(200*x - 200)))/(log(x
)^2*(x - 2*x^2 + x^3) + log(2*log(2))^4*(x - 2*x^2 + x^3) + log(2*log(2))^2*log(x)*(2*x - 4*x^2 + 2*x^3)),x)

[Out]

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

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sympy [B]  time = 0.31, size = 148, normalized size = 6.43 \begin {gather*} \frac {- 600 \log {\relax (2 )}^{2} \log {\left (\log {\relax (2 )} \right )}^{2} - 100 \log {\relax (2 )}^{4} - 100 \log {\left (\log {\relax (2 )} \right )}^{4} - 400 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )}^{3} - 400 \log {\relax (2 )}^{3} \log {\left (\log {\relax (2 )} \right )}}{2 x \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )} + x \log {\left (\log {\relax (2 )} \right )}^{2} + x \log {\relax (2 )}^{2} + \left (x - 1\right ) \log {\relax (x )} - \log {\relax (2 )}^{2} - \log {\left (\log {\relax (2 )} \right )}^{2} - 2 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )}} - \frac {100 \log {\relax (x )}}{x - 1} - \frac {- 100 \log {\relax (2 )}^{2} - 100 \log {\left (\log {\relax (2 )} \right )}^{2} - 200 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x*ln(x)**2+(-200*x+200)*ln(x))*ln(2*ln(2))**2+100*x*ln(x)**3+(-100*x+100)*ln(x)**2)/((x**3-2*x
**2+x)*ln(2*ln(2))**4+(2*x**3-4*x**2+2*x)*ln(x)*ln(2*ln(2))**2+(x**3-2*x**2+x)*ln(x)**2),x)

[Out]

(-600*log(2)**2*log(log(2))**2 - 100*log(2)**4 - 100*log(log(2))**4 - 400*log(2)*log(log(2))**3 - 400*log(2)**
3*log(log(2)))/(2*x*log(2)*log(log(2)) + x*log(log(2))**2 + x*log(2)**2 + (x - 1)*log(x) - log(2)**2 - log(log
(2))**2 - 2*log(2)*log(log(2))) - 100*log(x)/(x - 1) - (-100*log(2)**2 - 100*log(log(2))**2 - 200*log(2)*log(l
og(2)))/(x - 1)

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