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

Optimal. Leaf size=17 \[ 1+\frac {4 x}{2-\log (4)+x \log (x)} \]

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Rubi [F]  time = 0.19, 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 {8-4 x-4 \log (4)}{4-4 \log (4)+\log ^2(4)+(4 x-2 x \log (4)) \log (x)+x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 15, normalized size = 0.88 \begin {gather*} \frac {4 x}{2-\log (4)+x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x)/(2 - Log[4] + x*Log[x])

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fricas [A]  time = 0.75, size = 15, normalized size = 0.88 \begin {gather*} \frac {4 \, x}{x \log \relax (x) - 2 \, \log \relax (2) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x/(x*log(x) - 2*log(2) + 2)

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giac [A]  time = 0.13, size = 15, normalized size = 0.88 \begin {gather*} \frac {4 \, x}{x \log \relax (x) - 2 \, \log \relax (2) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x/(x*log(x) - 2*log(2) + 2)

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maple [A]  time = 0.07, size = 17, normalized size = 1.00

method	result	size
norman	\(-\frac {4 x}{-x \ln \relax (x )+2 \ln \relax (2)-2}\)	\(17\)
risch	\(-\frac {4 x}{-x \ln \relax (x )+2 \ln \relax (2)-2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 15, normalized size = 0.88 \begin {gather*} \frac {4 \, x}{x \log \relax (x) - 2 \, \log \relax (2) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x/(x*log(x) - 2*log(2) + 2)

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mupad [B]  time = 7.47, size = 15, normalized size = 0.88 \begin {gather*} \frac {4\,x}{x\,\ln \relax (x)-\ln \relax (4)+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x)/(x*log(x) - log(4) + 2)

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sympy [A]  time = 0.13, size = 14, normalized size = 0.82 \begin {gather*} \frac {4 x}{x \log {\relax (x )} - 2 \log {\relax (2 )} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x/(x*log(x) - 2*log(2) + 2)

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