\(\int \frac {-48+180 x+96 x^2+12 x^3+12 x \log (\frac {x}{5})}{80 x^3+40 x^4+5 x^5+(-40 x^2-10 x^3) \log (\frac {x}{5})+5 x \log ^2(\frac {x}{5})} \, dx\) [1791]

Optimal result

Integrand size = 72, antiderivative size = 22 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {12}{5 \left (-x+\frac {\log \left (\frac {x}{5}\right )}{4+x}\right )} \]

[Out]

12/5/(ln(1/5*x)/(4+x)-x)

Rubi [F]

\[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=\int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx \]

[In]

Int[(-48 + 180*x + 96*x^2 + 12*x^3 + 12*x*Log[x/5])/(80*x^3 + 40*x^4 + 5*x^5 + (-40*x^2 - 10*x^3)*Log[x/5] + 5
*x*Log[x/5]^2),x]

[Out]

36*Defer[Int][(4*x + x^2 + Log[5] - Log[x])^(-2), x] - (48*Defer[Int][1/(x*(4*x + x^2 + Log[5] - Log[x])^2), x
])/5 + (96*Defer[Int][x/(4*x + x^2 + Log[5] - Log[x])^2, x])/5 + (12*Defer[Int][x^2/(4*x + x^2 + Log[5] - Log[
x])^2, x])/5 + 12*Defer[Subst][Defer[Int][Log[x]/(20*x + 25*x^2 + Log[5] - Log[5*x])^2, x], x, x/5]

Rubi steps \begin{align*} \text {integral}& = \int \frac {12 \left (-4+15 x+8 x^2+x^3+x \log \left (\frac {x}{5}\right )\right )}{5 x \left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx \\ & = \frac {12}{5} \int \frac {-4+15 x+8 x^2+x^3+x \log \left (\frac {x}{5}\right )}{x \left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx \\ & = \frac {12}{5} \int \left (\frac {15}{\left (4 x+x^2+\log (5)-\log (x)\right )^2}-\frac {4}{x \left (4 x+x^2+\log (5)-\log (x)\right )^2}+\frac {8 x}{\left (4 x+x^2+\log (5)-\log (x)\right )^2}+\frac {x^2}{\left (4 x+x^2+\log (5)-\log (x)\right )^2}+\frac {\log \left (\frac {x}{5}\right )}{\left (4 x+x^2+\log (5)-\log (x)\right )^2}\right ) \, dx \\ & = \frac {12}{5} \int \frac {x^2}{\left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx+\frac {12}{5} \int \frac {\log \left (\frac {x}{5}\right )}{\left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx-\frac {48}{5} \int \frac {1}{x \left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx+\frac {96}{5} \int \frac {x}{\left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx+36 \int \frac {1}{\left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx \\ & = \frac {12}{5} \int \frac {x^2}{\left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx-\frac {48}{5} \int \frac {1}{x \left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx+12 \text {Subst}\left (\int \frac {\log (x)}{\left (20 x+25 x^2+\log (5)-\log (5 x)\right )^2} \, dx,x,\frac {x}{5}\right )+\frac {96}{5} \int \frac {x}{\left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx+36 \int \frac {1}{\left (4 x+x^2+\log (5)-\log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {12 (4+x)}{5 \left (4 x+x^2+\log (5)-\log (x)\right )} \]

[In]

Integrate[(-48 + 180*x + 96*x^2 + 12*x^3 + 12*x*Log[x/5])/(80*x^3 + 40*x^4 + 5*x^5 + (-40*x^2 - 10*x^3)*Log[x/
5] + 5*x*Log[x/5]^2),x]

[Out]

(-12*(4 + x))/(5*(4*x + x^2 + Log[5] - Log[x]))

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {48}{5}+\frac {12 x}{5}}{-x^{2}+\ln \left (\frac {x}{5}\right )-4 x}\)	\(21\)
default	\(\frac {\frac {48}{5}+\frac {12 x}{5}}{-x^{2}+\ln \left (\frac {x}{5}\right )-4 x}\)	\(21\)
risch	\(-\frac {12 \left (4+x \right )}{5 \left (x^{2}+4 x -\ln \left (\frac {x}{5}\right )\right )}\)	\(21\)
norman	\(\frac {-\frac {48}{5}-\frac {12 x}{5}}{x^{2}+4 x -\ln \left (\frac {x}{5}\right )}\)	\(22\)
parallelrisch	\(\frac {-48-12 x}{5 x^{2}+20 x -5 \ln \left (\frac {x}{5}\right )}\)	\(23\)

[In]

int((12*x*ln(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*ln(1/5*x)^2+(-10*x^3-40*x^2)*ln(1/5*x)+5*x^5+40*x^4+80*x^3),x
,method=_RETURNVERBOSE)

[Out]

12/5*(4+x)/(-x^2+ln(1/5*x)-4*x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x - \log \left (\frac {1}{5} \, x\right )\right )}} \]

[In]

integrate((12*x*log(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*log(1/5*x)^2+(-10*x^3-40*x^2)*log(1/5*x)+5*x^5+40*x^4+
80*x^3),x, algorithm="fricas")

[Out]

-12/5*(x + 4)/(x^2 + 4*x - log(1/5*x))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=\frac {12 x + 48}{- 5 x^{2} - 20 x + 5 \log {\left (\frac {x}{5} \right )}} \]

[In]

integrate((12*x*ln(1/5*x)+12*x**3+96*x**2+180*x-48)/(5*x*ln(1/5*x)**2+(-10*x**3-40*x**2)*ln(1/5*x)+5*x**5+40*x
**4+80*x**3),x)

[Out]

(12*x + 48)/(-5*x**2 - 20*x + 5*log(x/5))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x + \log \left (5\right ) - \log \left (x\right )\right )}} \]

[In]

integrate((12*x*log(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*log(1/5*x)^2+(-10*x^3-40*x^2)*log(1/5*x)+5*x^5+40*x^4+
80*x^3),x, algorithm="maxima")

[Out]

-12/5*(x + 4)/(x^2 + 4*x + log(5) - log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x - \log \left (\frac {1}{5} \, x\right )\right )}} \]

[In]

integrate((12*x*log(1/5*x)+12*x^3+96*x^2+180*x-48)/(5*x*log(1/5*x)^2+(-10*x^3-40*x^2)*log(1/5*x)+5*x^5+40*x^4+
80*x^3),x, algorithm="giac")

[Out]

-12/5*(x + 4)/(x^2 + 4*x - log(1/5*x))

Mupad [B] (verification not implemented)

Time = 9.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-48+180 x+96 x^2+12 x^3+12 x \log \left (\frac {x}{5}\right )}{80 x^3+40 x^4+5 x^5+\left (-40 x^2-10 x^3\right ) \log \left (\frac {x}{5}\right )+5 x \log ^2\left (\frac {x}{5}\right )} \, dx=-\frac {\frac {12\,x}{5}+\frac {48}{5}}{4\,x-\ln \left (\frac {x}{5}\right )+x^2} \]

[In]

int((180*x + 12*x*log(x/5) + 96*x^2 + 12*x^3 - 48)/(5*x*log(x/5)^2 - log(x/5)*(40*x^2 + 10*x^3) + 80*x^3 + 40*
x^4 + 5*x^5),x)

[Out]

-((12*x)/5 + 48/5)/(4*x - log(x/5) + x^2)