\(\int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx\) [6087]

Optimal result

Integrand size = 27, antiderivative size = 16 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=6-x-\frac {x}{(1-4 \log (5))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {8} \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 x \left (1+8 \log ^2(5)-4 \log (5)\right )}{(1-\log (625))^2} \]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(16)=32\).

Time = 0.00 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.62 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 x}{1-8 \log (5)+16 \log ^2(5)}+\frac {8 x \log (5)}{1-8 \log (5)+16 \log ^2(5)}-\frac {16 x \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \]

[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)

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50

method	result	size
norman	\(-\frac {2 \left (8 \ln \left (5\right )^{2}-4 \ln \left (5\right )+1\right ) x}{\left (4 \ln \left (5\right )-1\right )^{2}}\)	\(24\)
parallelrisch	\(\frac {\left (-16 \ln \left (5\right )^{2}+8 \ln \left (5\right )-2\right ) x}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}\)	\(29\)
default	\(\frac {2 \left (-8 \ln \left (5\right )^{2}+4 \ln \left (5\right )-1\right ) x}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}\)	\(30\)
risch	\(-\frac {16 x \ln \left (5\right )^{2}}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}+\frac {8 x \ln \left (5\right )}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}-\frac {2 x}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}\)	\(59\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.88 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 \, {\left (8 \, x \log \left (5\right )^{2} - 4 \, x \log \left (5\right ) + x\right )}}{16 \, \log \left (5\right )^{2} - 8 \, \log \left (5\right ) + 1} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\).

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=\frac {x \left (- 16 \log {\left (5 \right )}^{2} - 2 + 8 \log {\left (5 \right )}\right )}{- 8 \log {\left (5 \right )} + 1 + 16 \log {\left (5 \right )}^{2}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 \, {\left (8 \, \log \left (5\right )^{2} - 4 \, \log \left (5\right ) + 1\right )} x}{16 \, \log \left (5\right )^{2} - 8 \, \log \left (5\right ) + 1} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 \, {\left (8 \, \log \left (5\right )^{2} - 4 \, \log \left (5\right ) + 1\right )} x}{16 \, \log \left (5\right )^{2} - 8 \, \log \left (5\right ) + 1} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {x\,\left (16\,{\ln \left (5\right )}^2-8\,\ln \left (5\right )+2\right )}{16\,{\ln \left (5\right )}^2-8\,\ln \left (5\right )+1} \]

[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)