3.7.94 \(\int \frac {8+(80-20 x) \log (\frac {324}{16-8 x+x^2})+(-4+21 x-5 x^2) \log ^2(\frac {324}{16-8 x+x^2})}{(-80+20 x) \log (\frac {324}{16-8 x+x^2})+(-20 x+5 x^2) \log ^2(\frac {324}{16-8 x+x^2})} \, dx\)

Optimal. Leaf size=26 \[ e^4-x+\frac {1}{5} \log \left (x+\frac {4}{\log \left (\frac {324}{(-4+x)^2}\right )}\right ) \]

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Rubi [F]  time = 1.31, 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+(80-20 x) \log \left (\frac {324}{16-8 x+x^2}\right )+\left (-4+21 x-5 x^2\right ) \log ^2\left (\frac {324}{16-8 x+x^2}\right )}{(-80+20 x) \log \left (\frac {324}{16-8 x+x^2}\right )+\left (-20 x+5 x^2\right ) \log ^2\left (\frac {324}{16-8 x+x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8 + (80 - 20*x)*Log[324/(16 - 8*x + x^2)] + (-4 + 21*x - 5*x^2)*Log[324/(16 - 8*x + x^2)]^2)/((-80 + 20*x
)*Log[324/(16 - 8*x + x^2)] + (-20*x + 5*x^2)*Log[324/(16 - 8*x + x^2)]^2),x]

[Out]

-x + Log[x]/5 - Log[Log[324/(4 - x)^2]]/5 - (2*Defer[Int][(4 + x*Log[324/(-4 + x)^2])^(-1), x])/5 - (8*Defer[I
nt][1/((-4 + x)*(4 + x*Log[324/(-4 + x)^2])), x])/5 - (4*Defer[Int][1/(x*(4 + x*Log[324/(-4 + x)^2])), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-8-(80-20 x) \log \left (\frac {324}{16-8 x+x^2}\right )-\left (-4+21 x-5 x^2\right ) \log ^2\left (\frac {324}{16-8 x+x^2}\right )}{5 (4-x) \log \left (\frac {324}{(-4+x)^2}\right ) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx\\ &=\frac {1}{5} \int \frac {-8-(80-20 x) \log \left (\frac {324}{16-8 x+x^2}\right )-\left (-4+21 x-5 x^2\right ) \log ^2\left (\frac {324}{16-8 x+x^2}\right )}{(4-x) \log \left (\frac {324}{(-4+x)^2}\right ) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx\\ &=\frac {1}{5} \int \left (-\frac {20}{4+x \log \left (\frac {324}{(-4+x)^2}\right )}+\frac {8}{(-4+x) \log \left (\frac {324}{(-4+x)^2}\right ) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )}-\frac {(-1+5 x) \log \left (\frac {324}{(-4+x)^2}\right )}{4+x \log \left (\frac {324}{(-4+x)^2}\right )}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {(-1+5 x) \log \left (\frac {324}{(-4+x)^2}\right )}{4+x \log \left (\frac {324}{(-4+x)^2}\right )} \, dx\right )+\frac {8}{5} \int \frac {1}{(-4+x) \log \left (\frac {324}{(-4+x)^2}\right ) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx-4 \int \frac {1}{4+x \log \left (\frac {324}{(-4+x)^2}\right )} \, dx\\ &=-\left (\frac {1}{5} \int \left (\frac {-1+5 x}{x}-\frac {4 (-1+5 x)}{x \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )}\right ) \, dx\right )+\frac {8}{5} \int \left (\frac {1}{4 (-4+x) \log \left (\frac {324}{(-4+x)^2}\right )}-\frac {x}{4 (-4+x) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )}\right ) \, dx-4 \int \frac {1}{4+x \log \left (\frac {324}{(-4+x)^2}\right )} \, dx\\ &=-\left (\frac {1}{5} \int \frac {-1+5 x}{x} \, dx\right )+\frac {2}{5} \int \frac {1}{(-4+x) \log \left (\frac {324}{(-4+x)^2}\right )} \, dx-\frac {2}{5} \int \frac {x}{(-4+x) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx+\frac {4}{5} \int \frac {-1+5 x}{x \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx-4 \int \frac {1}{4+x \log \left (\frac {324}{(-4+x)^2}\right )} \, dx\\ &=-\left (\frac {1}{5} \int \left (5-\frac {1}{x}\right ) \, dx\right )-\frac {2}{5} \int \left (\frac {1}{4+x \log \left (\frac {324}{(-4+x)^2}\right )}+\frac {4}{(-4+x) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )}\right ) \, dx+\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{x \log \left (\frac {324}{x^2}\right )} \, dx,x,-4+x\right )+\frac {4}{5} \int \left (\frac {5}{4+x \log \left (\frac {324}{(-4+x)^2}\right )}-\frac {1}{x \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )}\right ) \, dx-4 \int \frac {1}{4+x \log \left (\frac {324}{(-4+x)^2}\right )} \, dx\\ &=-x+\frac {\log (x)}{5}-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {324}{(-4+x)^2}\right )\right )-\frac {2}{5} \int \frac {1}{4+x \log \left (\frac {324}{(-4+x)^2}\right )} \, dx-\frac {4}{5} \int \frac {1}{x \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx-\frac {8}{5} \int \frac {1}{(-4+x) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx\\ &=-x+\frac {\log (x)}{5}-\frac {1}{5} \log \left (\log \left (\frac {324}{(4-x)^2}\right )\right )-\frac {2}{5} \int \frac {1}{4+x \log \left (\frac {324}{(-4+x)^2}\right )} \, dx-\frac {4}{5} \int \frac {1}{x \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx-\frac {8}{5} \int \frac {1}{(-4+x) \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 33, normalized size = 1.27 \begin {gather*} \frac {1}{5} \left (20-5 x-\log \left (\log \left (\frac {324}{(-4+x)^2}\right )\right )+\log \left (4+x \log \left (\frac {324}{(-4+x)^2}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + (80 - 20*x)*Log[324/(16 - 8*x + x^2)] + (-4 + 21*x - 5*x^2)*Log[324/(16 - 8*x + x^2)]^2)/((-80
+ 20*x)*Log[324/(16 - 8*x + x^2)] + (-20*x + 5*x^2)*Log[324/(16 - 8*x + x^2)]^2),x]

[Out]

(20 - 5*x - Log[Log[324/(-4 + x)^2]] + Log[4 + x*Log[324/(-4 + x)^2]])/5

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fricas [B]  time = 0.79, size = 48, normalized size = 1.85 \begin {gather*} -x + \frac {1}{5} \, \log \relax (x) + \frac {1}{5} \, \log \left (\frac {x \log \left (\frac {324}{x^{2} - 8 \, x + 16}\right ) + 4}{x}\right ) - \frac {1}{5} \, \log \left (\log \left (\frac {324}{x^{2} - 8 \, x + 16}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+21*x-4)*log(324/(x^2-8*x+16))^2+(-20*x+80)*log(324/(x^2-8*x+16))+8)/((5*x^2-20*x)*log(324/(
x^2-8*x+16))^2+(20*x-80)*log(324/(x^2-8*x+16))),x, algorithm="fricas")

[Out]

-x + 1/5*log(x) + 1/5*log((x*log(324/(x^2 - 8*x + 16)) + 4)/x) - 1/5*log(log(324/(x^2 - 8*x + 16)))

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giac [A]  time = 0.69, size = 40, normalized size = 1.54 \begin {gather*} -x + \frac {1}{5} \, \log \left (x \log \left (\frac {324}{x^{2} - 8 \, x + 16}\right ) + 4\right ) - \frac {1}{5} \, \log \left (\log \left (\frac {324}{x^{2} - 8 \, x + 16}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+21*x-4)*log(324/(x^2-8*x+16))^2+(-20*x+80)*log(324/(x^2-8*x+16))+8)/((5*x^2-20*x)*log(324/(
x^2-8*x+16))^2+(20*x-80)*log(324/(x^2-8*x+16))),x, algorithm="giac")

[Out]

-x + 1/5*log(x*log(324/(x^2 - 8*x + 16)) + 4) - 1/5*log(log(324/(x^2 - 8*x + 16)))

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maple [A]  time = 0.16, size = 41, normalized size = 1.58

method	result	size
norman	\(-x -\frac {\ln \left (\ln \left (\frac {324}{x^{2}-8 x +16}\right )\right )}{5}+\frac {\ln \left (\ln \left (\frac {324}{x^{2}-8 x +16}\right ) x +4\right )}{5}\)	\(41\)
risch	\(-x +\frac {\ln \relax (x )}{5}-\frac {\ln \left (\ln \left (\frac {324}{x^{2}-8 x +16}\right )\right )}{5}+\frac {\ln \left (\ln \left (\frac {324}{x^{2}-8 x +16}\right )+\frac {4}{x}\right )}{5}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^2+21*x-4)*ln(324/(x^2-8*x+16))^2+(-20*x+80)*ln(324/(x^2-8*x+16))+8)/((5*x^2-20*x)*ln(324/(x^2-8*x+1
6))^2+(20*x-80)*ln(324/(x^2-8*x+16))),x,method=_RETURNVERBOSE)

[Out]

-x-1/5*ln(ln(324/(x^2-8*x+16)))+1/5*ln(ln(324/(x^2-8*x+16))*x+4)

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maxima [B]  time = 0.67, size = 50, normalized size = 1.92 \begin {gather*} -x + \frac {1}{5} \, \log \relax (x) + \frac {1}{5} \, \log \left (-\frac {x {\left (2 \, \log \relax (3) + \log \relax (2)\right )} - x \log \left (x - 4\right ) + 2}{x}\right ) - \frac {1}{5} \, \log \left (-2 \, \log \relax (3) - \log \relax (2) + \log \left (x - 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+21*x-4)*log(324/(x^2-8*x+16))^2+(-20*x+80)*log(324/(x^2-8*x+16))+8)/((5*x^2-20*x)*log(324/(
x^2-8*x+16))^2+(20*x-80)*log(324/(x^2-8*x+16))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.59, size = 104, normalized size = 4.00 \begin {gather*} \frac {\ln \left (x^2+2\,x-8\right )}{5}-\frac {\ln \left (\frac {16\,x\,\ln \left (18\right )-32\,\ln \left (\frac {1}{{\left (x-4\right )}^2}\right )-64\,\ln \left (18\right )+8\,x\,\ln \left (\frac {1}{{\left (x-4\right )}^2}\right )+8\,x^2\,\ln \left (18\right )+4\,x^2\,\ln \left (\frac {1}{{\left (x-4\right )}^2}\right )}{x^2\,\left (x-4\right )}\right )}{5}-x-\frac {\ln \relax (x)}{5}+\frac {\ln \left (\frac {8\,x\,\ln \left (18\right )+4\,x\,\ln \left (\frac {1}{{\left (x-4\right )}^2}\right )+16}{x\,\left (x-4\right )}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(324/(x^2 - 8*x + 16))^2*(5*x^2 - 21*x + 4) + log(324/(x^2 - 8*x + 16))*(20*x - 80) - 8)/(log(324/(x^2
- 8*x + 16))^2*(20*x - 5*x^2) - log(324/(x^2 - 8*x + 16))*(20*x - 80)),x)

[Out]

log(2*x + x^2 - 8)/5 - log((16*x*log(18) - 32*log(1/(x - 4)^2) - 64*log(18) + 8*x*log(1/(x - 4)^2) + 8*x^2*log
(18) + 4*x^2*log(1/(x - 4)^2))/(x^2*(x - 4)))/5 - x - log(x)/5 + log((8*x*log(18) + 4*x*log(1/(x - 4)^2) + 16)
/(x*(x - 4)))/5

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**2+21*x-4)*ln(324/(x**2-8*x+16))**2+(-20*x+80)*ln(324/(x**2-8*x+16))+8)/((5*x**2-20*x)*ln(324
/(x**2-8*x+16))**2+(20*x-80)*ln(324/(x**2-8*x+16))),x)

[Out]

Exception raised: PolynomialError

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