3.61.87 \(\int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx\)

Optimal. Leaf size=16 \[ 6-x-\frac {x}{(1-4 \log (5))^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {8} \begin {gather*} -\frac {2 x \left (1+8 \log ^2(5)-4 \log (5)\right )}{(1-\log (625))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + 8*Log[5] - 16*Log[5]^2)/(1 - 8*Log[5] + 16*Log[5]^2),x]

[Out]

(-2*x*(1 - 4*Log[5] + 8*Log[5]^2))/(1 - Log[625])^2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {2 x \left (1-4 \log (5)+8 \log ^2(5)\right )}{(1-\log (625))^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.00, size = 58, normalized size = 3.62 \begin {gather*} -\frac {2 x}{1-8 \log (5)+16 \log ^2(5)}+\frac {8 x \log (5)}{1-8 \log (5)+16 \log ^2(5)}-\frac {16 x \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 8*Log[5] - 16*Log[5]^2)/(1 - 8*Log[5] + 16*Log[5]^2),x]

[Out]

(-2*x)/(1 - 8*Log[5] + 16*Log[5]^2) + (8*x*Log[5])/(1 - 8*Log[5] + 16*Log[5]^2) - (16*x*Log[5]^2)/(1 - 8*Log[5
] + 16*Log[5]^2)

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fricas [A]  time = 0.51, size = 30, normalized size = 1.88 \begin {gather*} -\frac {2 \, {\left (8 \, x \log \relax (5)^{2} - 4 \, x \log \relax (5) + x\right )}}{16 \, \log \relax (5)^{2} - 8 \, \log \relax (5) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)^2+8*log(5)-2)/(16*log(5)^2-8*log(5)+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 29, normalized size = 1.81 \begin {gather*} -\frac {2 \, {\left (8 \, \log \relax (5)^{2} - 4 \, \log \relax (5) + 1\right )} x}{16 \, \log \relax (5)^{2} - 8 \, \log \relax (5) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)^2+8*log(5)-2)/(16*log(5)^2-8*log(5)+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 24, normalized size = 1.50

method	result	size
norman	\(-\frac {2 \left (8 \ln \relax (5)^{2}-4 \ln \relax (5)+1\right ) x}{\left (4 \ln \relax (5)-1\right )^{2}}\)	\(24\)
default	\(\frac {\left (-16 \ln \relax (5)^{2}+8 \ln \relax (5)-2\right ) x}{16 \ln \relax (5)^{2}-8 \ln \relax (5)+1}\)	\(29\)
risch	\(-\frac {16 x \ln \relax (5)^{2}}{16 \ln \relax (5)^{2}-8 \ln \relax (5)+1}+\frac {8 x \ln \relax (5)}{16 \ln \relax (5)^{2}-8 \ln \relax (5)+1}-\frac {2 x}{16 \ln \relax (5)^{2}-8 \ln \relax (5)+1}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*ln(5)^2+8*ln(5)-2)/(16*ln(5)^2-8*ln(5)+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 29, normalized size = 1.81 \begin {gather*} -\frac {2 \, {\left (8 \, \log \relax (5)^{2} - 4 \, \log \relax (5) + 1\right )} x}{16 \, \log \relax (5)^{2} - 8 \, \log \relax (5) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)^2+8*log(5)-2)/(16*log(5)^2-8*log(5)+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.00, size = 29, normalized size = 1.81 \begin {gather*} -\frac {x\,\left (16\,{\ln \relax (5)}^2-8\,\ln \relax (5)+2\right )}{16\,{\ln \relax (5)}^2-8\,\ln \relax (5)+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*log(5)^2 - 8*log(5) + 2)/(16*log(5)^2 - 8*log(5) + 1),x)

[Out]

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

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sympy [B]  time = 0.05, size = 27, normalized size = 1.69 \begin {gather*} \frac {x \left (- 16 \log {\relax (5 )}^{2} - 2 + 8 \log {\relax (5 )}\right )}{- 8 \log {\relax (5 )} + 1 + 16 \log {\relax (5 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*ln(5)**2+8*ln(5)-2)/(16*ln(5)**2-8*ln(5)+1),x)

[Out]

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

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