3.18.52 \(\int \frac {-4+18 x^2+x^2 \log (x^2)}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log (x^2)} \, dx\)

Optimal. Leaf size=24 \[ \log \left (2+\frac {1}{2 x}+2 x+\log (4)+\frac {1}{8} x \log \left (x^2\right )\right ) \]

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Rubi [F]  time = 0.62, 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 {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4 + 18*x^2 + x^2*Log[x^2])/(4*x + 16*x^2 + 16*x^3 + 8*x^2*Log[4] + x^3*Log[x^2]),x]

[Out]

Log[x] + 8*Defer[Int][1/(x*(-4 - 16*x^2 - 16*x*(1 + Log[2]) - x^2*Log[x^2])), x] - 8*(2 + Log[4])*Defer[Int][(
4 + 16*x^2 + 16*x*(1 + Log[2]) + x^2*Log[x^2])^(-1), x] + 2*Defer[Int][x/(4 + 16*x^2 + 16*x*(1 + Log[2]) + x^2
*Log[x^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^3+x^2 (16+8 \log (4))+x^3 \log \left (x^2\right )} \, dx\\ &=\int \left (\frac {1}{x}+\frac {2 \left (-4+x^2-4 x (2+\log (4))\right )}{x \left (4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )\right )}\right ) \, dx\\ &=\log (x)+2 \int \frac {-4+x^2-4 x (2+\log (4))}{x \left (4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )\right )} \, dx\\ &=\log (x)+2 \int \left (\frac {4}{x \left (-4-16 x^2-16 x (1+\log (2))-x^2 \log \left (x^2\right )\right )}+\frac {x}{4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )}+\frac {4 (-2-\log (4))}{4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )}\right ) \, dx\\ &=\log (x)+2 \int \frac {x}{4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )} \, dx+8 \int \frac {1}{x \left (-4-16 x^2-16 x (1+\log (2))-x^2 \log \left (x^2\right )\right )} \, dx-(8 (2+\log (4))) \int \frac {1}{4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 29, normalized size = 1.21 \begin {gather*} -\log (x)+\log \left (4+16 x+16 x^2+8 x \log (4)+x^2 \log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 18*x^2 + x^2*Log[x^2])/(4*x + 16*x^2 + 16*x^3 + 8*x^2*Log[4] + x^3*Log[x^2]),x]

[Out]

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

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fricas [A]  time = 0.80, size = 35, normalized size = 1.46 \begin {gather*} \frac {1}{2} \, \log \left (x^{2}\right ) + \log \left (\frac {x^{2} \log \left (x^{2}\right ) + 16 \, x^{2} + 16 \, x \log \relax (2) + 16 \, x + 4}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x^2)+18*x^2-4)/(x^3*log(x^2)+16*x^2*log(2)+16*x^3+16*x^2+4*x),x, algorithm="fricas")

[Out]

1/2*log(x^2) + log((x^2*log(x^2) + 16*x^2 + 16*x*log(2) + 16*x + 4)/x^2)

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giac [A]  time = 0.19, size = 29, normalized size = 1.21 \begin {gather*} \log \left (x^{2} \log \left (x^{2}\right ) + 16 \, x^{2} + 16 \, x \log \relax (2) + 16 \, x + 4\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x^2)+18*x^2-4)/(x^3*log(x^2)+16*x^2*log(2)+16*x^3+16*x^2+4*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 30, normalized size = 1.25

method	result	size
risch	\(\ln \relax (x )+\ln \left (\ln \left (x^{2}\right )+\frac {16 x \ln \relax (2)+16 x^{2}+16 x +4}{x^{2}}\right )\)	\(30\)
norman	\(-\frac {\ln \left (x^{2}\right )}{2}+\ln \left (x^{2} \ln \left (x^{2}\right )+16 x \ln \relax (2)+16 x^{2}+16 x +4\right )\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(x^2)+18*x^2-4)/(x^3*ln(x^2)+16*x^2*ln(2)+16*x^3+16*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(ln(x^2)+4*(4*x*ln(2)+4*x^2+4*x+1)/x^2)

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maxima [A]  time = 0.56, size = 28, normalized size = 1.17 \begin {gather*} \log \relax (x) + \log \left (\frac {x^{2} \log \relax (x) + 8 \, x^{2} + 8 \, x {\left (\log \relax (2) + 1\right )} + 2}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x^2)+18*x^2-4)/(x^3*log(x^2)+16*x^2*log(2)+16*x^3+16*x^2+4*x),x, algorithm="maxima")

[Out]

log(x) + log((x^2*log(x) + 8*x^2 + 8*x*(log(2) + 1) + 2)/x^2)

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mupad [B]  time = 1.34, size = 31, normalized size = 1.29 \begin {gather*} \ln \left (16\,x+16\,x\,\ln \relax (2)+x^2\,\ln \left (x^2\right )+16\,x^2+4\right )-\frac {\ln \left (x^2\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*log(x^2) + 18*x^2 - 4)/(4*x + x^3*log(x^2) + 16*x^2*log(2) + 16*x^2 + 16*x^3),x)

[Out]

log(16*x + 16*x*log(2) + x^2*log(x^2) + 16*x^2 + 4) - log(x^2)/2

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sympy [A]  time = 0.22, size = 29, normalized size = 1.21 \begin {gather*} \log {\relax (x )} + \log {\left (\log {\left (x^{2} \right )} + \frac {16 x^{2} + 16 x \log {\relax (2 )} + 16 x + 4}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(x**2)+18*x**2-4)/(x**3*ln(x**2)+16*x**2*ln(2)+16*x**3+16*x**2+4*x),x)

[Out]

log(x) + log(log(x**2) + (16*x**2 + 16*x*log(2) + 16*x + 4)/x**2)

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