3.68.9 \(\int \frac {256-1280 x^2+(-256-2560 x) \log (x)-1280 \log ^2(x)}{(x^2+5 x^3+(x+10 x^2) \log (x)+5 x \log ^2(x)) \log ^2(\frac {x+5 x^2+5 x \log (x)}{2 x+2 \log (x)})} \, dx\)

Optimal. Leaf size=25 \[ 2+\log (2)+\frac {256}{\log \left (\frac {1}{2} \left (5 x+\frac {x}{x+\log (x)}\right )\right )} \]

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Rubi [F]  time = 1.85, 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 {256-1280 x^2+(-256-2560 x) \log (x)-1280 \log ^2(x)}{\left (x^2+5 x^3+\left (x+10 x^2\right ) \log (x)+5 x \log ^2(x)\right ) \log ^2\left (\frac {x+5 x^2+5 x \log (x)}{2 x+2 \log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(256 - 1280*x^2 + (-256 - 2560*x)*Log[x] - 1280*Log[x]^2)/((x^2 + 5*x^3 + (x + 10*x^2)*Log[x] + 5*x*Log[x]
^2)*Log[(x + 5*x^2 + 5*x*Log[x])/(2*x + 2*Log[x])]^2),x]

[Out]

256*Defer[Int][1/(x*(x + 5*x^2 + Log[x] + 10*x*Log[x] + 5*Log[x]^2)*Log[(x*(1 + 5*x + 5*Log[x]))/(2*(x + Log[x
]))]^2), x] - 1280*Defer[Int][x/((x + 5*x^2 + Log[x] + 10*x*Log[x] + 5*Log[x]^2)*Log[(x*(1 + 5*x + 5*Log[x]))/
(2*(x + Log[x]))]^2), x] - 2560*Defer[Int][Log[x]/((x + 5*x^2 + Log[x] + 10*x*Log[x] + 5*Log[x]^2)*Log[(x*(1 +
5*x + 5*Log[x]))/(2*(x + Log[x]))]^2), x] - 256*Defer[Int][Log[x]/(x*(x + 5*x^2 + Log[x] + 10*x*Log[x] + 5*Lo
g[x]^2)*Log[(x*(1 + 5*x + 5*Log[x]))/(2*(x + Log[x]))]^2), x] - 1280*Defer[Int][Log[x]^2/(x*(x + 5*x^2 + Log[x
] + 10*x*Log[x] + 5*Log[x]^2)*Log[(x*(1 + 5*x + 5*Log[x]))/(2*(x + Log[x]))]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {256 \left (1-5 x^2-\log (x)-10 x \log (x)-5 \log ^2(x)\right )}{\left (x^2+5 x^3+\left (x+10 x^2\right ) \log (x)+5 x \log ^2(x)\right ) \log ^2\left (\frac {x+5 x^2+5 x \log (x)}{2 (x+\log (x))}\right )} \, dx\\ &=256 \int \frac {1-5 x^2-\log (x)-10 x \log (x)-5 \log ^2(x)}{\left (x^2+5 x^3+\left (x+10 x^2\right ) \log (x)+5 x \log ^2(x)\right ) \log ^2\left (\frac {x+5 x^2+5 x \log (x)}{2 (x+\log (x))}\right )} \, dx\\ &=256 \int \left (\frac {1}{x \left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )}-\frac {5 x}{\left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )}-\frac {10 \log (x)}{\left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )}-\frac {\log (x)}{x \left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )}-\frac {5 \log ^2(x)}{x \left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )}\right ) \, dx\\ &=256 \int \frac {1}{x \left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )} \, dx-256 \int \frac {\log (x)}{x \left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )} \, dx-1280 \int \frac {x}{\left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )} \, dx-1280 \int \frac {\log ^2(x)}{x \left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )} \, dx-2560 \int \frac {\log (x)}{\left (x+5 x^2+\log (x)+10 x \log (x)+5 \log ^2(x)\right ) \log ^2\left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.37, size = 25, normalized size = 1.00 \begin {gather*} \frac {256}{\log \left (\frac {x (1+5 x+5 \log (x))}{2 (x+\log (x))}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(256 - 1280*x^2 + (-256 - 2560*x)*Log[x] - 1280*Log[x]^2)/((x^2 + 5*x^3 + (x + 10*x^2)*Log[x] + 5*x*
Log[x]^2)*Log[(x + 5*x^2 + 5*x*Log[x])/(2*x + 2*Log[x])]^2),x]

[Out]

256/Log[(x*(1 + 5*x + 5*Log[x]))/(2*(x + Log[x]))]

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fricas [A]  time = 0.62, size = 25, normalized size = 1.00 \begin {gather*} \frac {256}{\log \left (\frac {5 \, x^{2} + 5 \, x \log \relax (x) + x}{2 \, {\left (x + \log \relax (x)\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1280*log(x)^2+(-2560*x-256)*log(x)-1280*x^2+256)/(5*x*log(x)^2+(10*x^2+x)*log(x)+5*x^3+x^2)/log((5
*x*log(x)+5*x^2+x)/(2*x+2*log(x)))^2,x, algorithm="fricas")

[Out]

256/log(1/2*(5*x^2 + 5*x*log(x) + x)/(x + log(x)))

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giac [A]  time = 0.26, size = 28, normalized size = 1.12 \begin {gather*} \frac {256}{\log \left (5 \, x + 5 \, \log \relax (x) + 1\right ) - \log \left (2 \, x + 2 \, \log \relax (x)\right ) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1280*log(x)^2+(-2560*x-256)*log(x)-1280*x^2+256)/(5*x*log(x)^2+(10*x^2+x)*log(x)+5*x^3+x^2)/log((5
*x*log(x)+5*x^2+x)/(2*x+2*log(x)))^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.09, size = 273, normalized size = 10.92

method	result	size
risch	\(\frac {512 i}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{x +\ln \relax (x )}\right ) \mathrm {csgn}\left (i \left (x +\ln \relax (x )+\frac {1}{5}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x +\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x +\ln \relax (x )+\frac {1}{5}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i x \left (x +\ln \relax (x )+\frac {1}{5}\right )}{x +\ln \relax (x )}\right )^{3}-2 i \ln \relax (2)+2 i \ln \relax (5)+2 i \ln \left (x +\ln \relax (x )+\frac {1}{5}\right )+2 i \ln \relax (x )-2 i \ln \left (x +\ln \relax (x )\right )}\)	\(273\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1280*ln(x)^2+(-2560*x-256)*ln(x)-1280*x^2+256)/(5*x*ln(x)^2+(10*x^2+x)*ln(x)+5*x^3+x^2)/ln((5*x*ln(x)+5*
x^2+x)/(2*x+2*ln(x)))^2,x,method=_RETURNVERBOSE)

[Out]

512*I/(Pi*csgn(I*x)*csgn(I/(x+ln(x))*(x+ln(x)+1/5))*csgn(I*x/(x+ln(x))*(x+ln(x)+1/5))-Pi*csgn(I*x)*csgn(I*x/(x
+ln(x))*(x+ln(x)+1/5))^2+Pi*csgn(I/(x+ln(x)))*csgn(I*(x+ln(x)+1/5))*csgn(I/(x+ln(x))*(x+ln(x)+1/5))-Pi*csgn(I/
(x+ln(x)))*csgn(I/(x+ln(x))*(x+ln(x)+1/5))^2-Pi*csgn(I*(x+ln(x)+1/5))*csgn(I/(x+ln(x))*(x+ln(x)+1/5))^2+Pi*csg
n(I/(x+ln(x))*(x+ln(x)+1/5))^3-Pi*csgn(I/(x+ln(x))*(x+ln(x)+1/5))*csgn(I*x/(x+ln(x))*(x+ln(x)+1/5))^2+Pi*csgn(
I*x/(x+ln(x))*(x+ln(x)+1/5))^3-2*I*ln(2)+2*I*ln(5)+2*I*ln(x+ln(x)+1/5)+2*I*ln(x)-2*I*ln(x+ln(x)))

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maxima [A]  time = 0.51, size = 28, normalized size = 1.12 \begin {gather*} -\frac {256}{\log \relax (2) - \log \left (5 \, x + 5 \, \log \relax (x) + 1\right ) + \log \left (x + \log \relax (x)\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1280*log(x)^2+(-2560*x-256)*log(x)-1280*x^2+256)/(5*x*log(x)^2+(10*x^2+x)*log(x)+5*x^3+x^2)/log((5
*x*log(x)+5*x^2+x)/(2*x+2*log(x)))^2,x, algorithm="maxima")

[Out]

-256/(log(2) - log(5*x + 5*log(x) + 1) + log(x + log(x)) - log(x))

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mupad [B]  time = 5.16, size = 28, normalized size = 1.12 \begin {gather*} \frac {256}{\ln \left (\frac {x+5\,x\,\ln \relax (x)+5\,x^2}{2\,x+2\,\ln \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(1280*log(x)^2 + log(x)*(2560*x + 256) + 1280*x^2 - 256)/(log((x + 5*x*log(x) + 5*x^2)/(2*x + 2*log(x)))^
2*(5*x*log(x)^2 + log(x)*(x + 10*x^2) + x^2 + 5*x^3)),x)

[Out]

256/log((x + 5*x*log(x) + 5*x^2)/(2*x + 2*log(x)))

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sympy [A]  time = 0.41, size = 24, normalized size = 0.96 \begin {gather*} \frac {256}{\log {\left (\frac {5 x^{2} + 5 x \log {\relax (x )} + x}{2 x + 2 \log {\relax (x )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1280*ln(x)**2+(-2560*x-256)*ln(x)-1280*x**2+256)/(5*x*ln(x)**2+(10*x**2+x)*ln(x)+5*x**3+x**2)/ln((
5*x*ln(x)+5*x**2+x)/(2*x+2*ln(x)))**2,x)

[Out]

256/log((5*x**2 + 5*x*log(x) + x)/(2*x + 2*log(x)))

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