\(\int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+(-10 x+64 x^3-16 x^4) \log (5)+x^2 \log ^2(5)} \, dx\) [4186]

Optimal result

Integrand size = 82, antiderivative size = 31 \[ \int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+\left (-10 x+64 x^3-16 x^4\right ) \log (5)+x^2 \log ^2(5)} \, dx=5+\frac {-4+\frac {5}{x}}{8 (-4+x)-\frac {-\frac {5}{x}+\log (5)}{x}} \]

[Out]

5+(5/x-4)/(8*x-32-(-5/x+ln(5))/x)

Rubi [F(-1)]

Timed out. \[ \int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+\left (-10 x+64 x^3-16 x^4\right ) \log (5)+x^2 \log ^2(5)} \, dx=\text {\$Aborted} \]

[In]

Int[(25 - 40*x + 160*x^2 - 80*x^3 + 32*x^4 + 4*x^2*Log[5])/(25 - 320*x^2 + 80*x^3 + 1024*x^4 - 512*x^5 + 64*x^
6 + (-10*x + 64*x^3 - 16*x^4)*Log[5] + x^2*Log[5]^2),x]

[Out]

$Aborted

Rubi steps Aborted

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+\left (-10 x+64 x^3-16 x^4\right ) \log (5)+x^2 \log ^2(5)} \, dx=\frac {(5-4 x) x}{5-32 x^2+8 x^3-x \log (5)} \]

[In]

Integrate[(25 - 40*x + 160*x^2 - 80*x^3 + 32*x^4 + 4*x^2*Log[5])/(25 - 320*x^2 + 80*x^3 + 1024*x^4 - 512*x^5 +
64*x^6 + (-10*x + 64*x^3 - 16*x^4)*Log[5] + x^2*Log[5]^2),x]

[Out]

((5 - 4*x)*x)/(5 - 32*x^2 + 8*x^3 - x*Log[5])

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {x \left (-5+4 x \right )}{-8 x^{3}+x \ln \left (5\right )+32 x^{2}-5}\)	\(26\)
default	\(\frac {4 x^{2}-5 x}{-8 x^{3}+x \ln \left (5\right )+32 x^{2}-5}\)	\(29\)
norman	\(\frac {4 x^{2}-5 x}{-8 x^{3}+x \ln \left (5\right )+32 x^{2}-5}\)	\(29\)
risch	\(\frac {4 x^{2}-5 x}{-8 x^{3}+x \ln \left (5\right )+32 x^{2}-5}\)	\(29\)
parallelrisch	\(-\frac {-32 x^{2}+40 x}{8 \left (-8 x^{3}+x \ln \left (5\right )+32 x^{2}-5\right )}\)	\(30\)

[In]

int((4*x^2*ln(5)+32*x^4-80*x^3+160*x^2-40*x+25)/(x^2*ln(5)^2+(-16*x^4+64*x^3-10*x)*ln(5)+64*x^6-512*x^5+1024*x
^4+80*x^3-320*x^2+25),x,method=_RETURNVERBOSE)

[Out]

x*(-5+4*x)/(-8*x^3+x*ln(5)+32*x^2-5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+\left (-10 x+64 x^3-16 x^4\right ) \log (5)+x^2 \log ^2(5)} \, dx=-\frac {4 \, x^{2} - 5 \, x}{8 \, x^{3} - 32 \, x^{2} - x \log \left (5\right ) + 5} \]

[In]

integrate((4*x^2*log(5)+32*x^4-80*x^3+160*x^2-40*x+25)/(x^2*log(5)^2+(-16*x^4+64*x^3-10*x)*log(5)+64*x^6-512*x
^5+1024*x^4+80*x^3-320*x^2+25),x, algorithm="fricas")

[Out]

-(4*x^2 - 5*x)/(8*x^3 - 32*x^2 - x*log(5) + 5)

Sympy [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+\left (-10 x+64 x^3-16 x^4\right ) \log (5)+x^2 \log ^2(5)} \, dx=\frac {- 4 x^{2} + 5 x}{8 x^{3} - 32 x^{2} - x \log {\left (5 \right )} + 5} \]

[In]

integrate((4*x**2*ln(5)+32*x**4-80*x**3+160*x**2-40*x+25)/(x**2*ln(5)**2+(-16*x**4+64*x**3-10*x)*ln(5)+64*x**6
-512*x**5+1024*x**4+80*x**3-320*x**2+25),x)

[Out]

(-4*x**2 + 5*x)/(8*x**3 - 32*x**2 - x*log(5) + 5)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+\left (-10 x+64 x^3-16 x^4\right ) \log (5)+x^2 \log ^2(5)} \, dx=-\frac {4 \, x^{2} - 5 \, x}{8 \, x^{3} - 32 \, x^{2} - x \log \left (5\right ) + 5} \]

[In]

integrate((4*x^2*log(5)+32*x^4-80*x^3+160*x^2-40*x+25)/(x^2*log(5)^2+(-16*x^4+64*x^3-10*x)*log(5)+64*x^6-512*x
^5+1024*x^4+80*x^3-320*x^2+25),x, algorithm="maxima")

[Out]

-(4*x^2 - 5*x)/(8*x^3 - 32*x^2 - x*log(5) + 5)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+\left (-10 x+64 x^3-16 x^4\right ) \log (5)+x^2 \log ^2(5)} \, dx=-\frac {4 \, x^{2} - 5 \, x}{8 \, x^{3} - 32 \, x^{2} - x \log \left (5\right ) + 5} \]

[In]

integrate((4*x^2*log(5)+32*x^4-80*x^3+160*x^2-40*x+25)/(x^2*log(5)^2+(-16*x^4+64*x^3-10*x)*log(5)+64*x^6-512*x
^5+1024*x^4+80*x^3-320*x^2+25),x, algorithm="giac")

[Out]

-(4*x^2 - 5*x)/(8*x^3 - 32*x^2 - x*log(5) + 5)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {25-40 x+160 x^2-80 x^3+32 x^4+4 x^2 \log (5)}{25-320 x^2+80 x^3+1024 x^4-512 x^5+64 x^6+\left (-10 x+64 x^3-16 x^4\right ) \log (5)+x^2 \log ^2(5)} \, dx=\frac {x\,\left (4\,x-5\right )}{-8\,x^3+32\,x^2+\ln \left (5\right )\,x-5} \]

[In]

int((4*x^2*log(5) - 40*x + 160*x^2 - 80*x^3 + 32*x^4 + 25)/(x^2*log(5)^2 - log(5)*(10*x - 64*x^3 + 16*x^4) - 3
20*x^2 + 80*x^3 + 1024*x^4 - 512*x^5 + 64*x^6 + 25),x)

[Out]

(x*(4*x - 5))/(x*log(5) + 32*x^2 - 8*x^3 - 5)