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

Optimal. Leaf size=22 \[ \frac {12}{5 \left (-x+\frac {\log \left (\frac {x}{5}\right )}{4+x}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 1.01, 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 {-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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 \operatorname {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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.52, size = 22, normalized size = 1.00 \begin {gather*} -\frac {12 (4+x)}{5 \left (4 x+x^2+\log (5)-\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[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]))

________________________________________________________________________________________

fricas [A]  time = 0.78, size = 20, normalized size = 0.91 \begin {gather*} -\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x - \log \left (\frac {1}{5} \, x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

giac [A]  time = 0.19, size = 20, normalized size = 0.91 \begin {gather*} -\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x - \log \left (\frac {1}{5} \, x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 21, normalized size = 0.95

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

Verification of antiderivative is not currently implemented for this CAS.

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

________________________________________________________________________________________

maxima [A]  time = 0.85, size = 20, normalized size = 0.91 \begin {gather*} -\frac {12 \, {\left (x + 4\right )}}{5 \, {\left (x^{2} + 4 \, x + \log \relax (5) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

mupad [B]  time = 1.31, size = 22, normalized size = 1.00 \begin {gather*} -\frac {\frac {12\,x}{5}+\frac {48}{5}}{4\,x-\ln \left (\frac {x}{5}\right )+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 19, normalized size = 0.86 \begin {gather*} \frac {12 x + 48}{- 5 x^{2} - 20 x + 5 \log {\left (\frac {x}{5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________