3.95.94 \(\int \frac {(8-2 x) \log (-4+x)+(x+(-4+x) \log (-4+x)) \log (x^2)+(-4+x) \log ^2(x^2)}{(-4+x) \log ^2(x^2)} \, dx\)

Optimal. Leaf size=14 \[ x+\frac {x \log (-4+x)}{\log \left (x^2\right )} \]

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Rubi [F]  time = 0.43, 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 {(8-2 x) \log (-4+x)+(x+(-4+x) \log (-4+x)) \log \left (x^2\right )+(-4+x) \log ^2\left (x^2\right )}{(-4+x) \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 14, normalized size = 1.00 \begin {gather*} x+\frac {x \log (-4+x)}{\log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (x*Log[-4 + x])/Log[x^2]

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fricas [A]  time = 0.72, size = 20, normalized size = 1.43 \begin {gather*} \frac {x \log \left (x^{2}\right ) + x \log \left (x - 4\right )}{\log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*log(x^2)^2+((x-4)*log(x-4)+x)*log(x^2)+(-2*x+8)*log(x-4))/(x-4)/log(x^2)^2,x, algorithm="fric
as")

[Out]

(x*log(x^2) + x*log(x - 4))/log(x^2)

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giac [A]  time = 0.29, size = 14, normalized size = 1.00 \begin {gather*} x + \frac {x \log \left (x - 4\right )}{\log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*log(x^2)^2+((x-4)*log(x-4)+x)*log(x^2)+(-2*x+8)*log(x-4))/(x-4)/log(x^2)^2,x, algorithm="giac
")

[Out]

x + x*log(x - 4)/log(x^2)

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maple [A]  time = 0.10, size = 21, normalized size = 1.50

method	result	size
norman	\(\frac {x \ln \left (x^{2}\right )+x \ln \left (x -4\right )}{\ln \left (x^{2}\right )}\)	\(21\)
risch	\(\frac {2 i x \ln \left (x -4\right )}{\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )}+x\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-4)*ln(x^2)^2+((x-4)*ln(x-4)+x)*ln(x^2)+(-2*x+8)*ln(x-4))/(x-4)/ln(x^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 18, normalized size = 1.29 \begin {gather*} \frac {x \log \left (x - 4\right ) + 2 \, x \log \relax (x)}{2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*log(x^2)^2+((x-4)*log(x-4)+x)*log(x^2)+(-2*x+8)*log(x-4))/(x-4)/log(x^2)^2,x, algorithm="maxi
ma")

[Out]

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

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mupad [B]  time = 6.33, size = 60, normalized size = 4.29 \begin {gather*} \frac {3\,x}{2}+\frac {x\,\ln \left (x-4\right )}{2}+\frac {8}{x-4}+\frac {x\,\ln \left (x-4\right )-\frac {x\,\ln \left (x^2\right )\,\left (x-4\,\ln \left (x-4\right )+x\,\ln \left (x-4\right )\right )}{2\,\left (x-4\right )}}{\ln \left (x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)^2*(x - 4) - log(x - 4)*(2*x - 8) + log(x^2)*(x + log(x - 4)*(x - 4)))/(log(x^2)^2*(x - 4)),x)

[Out]

(3*x)/2 + (x*log(x - 4))/2 + 8/(x - 4) + (x*log(x - 4) - (x*log(x^2)*(x - 4*log(x - 4) + x*log(x - 4)))/(2*(x
- 4)))/log(x^2)

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sympy [A]  time = 0.33, size = 12, normalized size = 0.86 \begin {gather*} x + \frac {x \log {\left (x - 4 \right )}}{\log {\left (x^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*ln(x**2)**2+((x-4)*ln(x-4)+x)*ln(x**2)+(-2*x+8)*ln(x-4))/(x-4)/ln(x**2)**2,x)

[Out]

x + x*log(x - 4)/log(x**2)

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