\(\int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+(-40 x^2+4 x^3+12 x^2 \log (x)) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx\) [7471]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 66, antiderivative size = 19 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {x^3 \log (5 x)}{-3+\frac {x}{2}+\log (x)} \]

[Out]

x^3*ln(5*x)/(1/2*x-3+ln(x))

Rubi [F]

\[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx \]

[In]

Int[(-12*x^2 + 2*x^3 + 4*x^2*Log[x] + (-40*x^2 + 4*x^3 + 12*x^2*Log[x])*Log[5*x])/(36 - 12*x + x^2 + (-24 + 4*
x)*Log[x] + 4*Log[x]^2),x]

[Out]

2*Defer[Int][x^2/(-6 + x + 2*Log[x]), x] - 40*Defer[Int][(x^2*Log[5*x])/(-6 + x + 2*Log[x])^2, x] + 4*Defer[In
t][(x^3*Log[5*x])/(-6 + x + 2*Log[x])^2, x] + 12*Defer[Int][(x^2*Log[x]*Log[5*x])/(-6 + x + 2*Log[x])^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x^2 (-6+x+2 (-10+x) \log (5 x)+\log (x) (2+6 \log (5 x)))}{(6-x-2 \log (x))^2} \, dx \\ & = 2 \int \frac {x^2 (-6+x+2 (-10+x) \log (5 x)+\log (x) (2+6 \log (5 x)))}{(6-x-2 \log (x))^2} \, dx \\ & = 2 \int \left (\frac {x^2}{-6+x+2 \log (x)}+\frac {2 x^2 (-10+x+3 \log (x)) \log (5 x)}{(-6+x+2 \log (x))^2}\right ) \, dx \\ & = 2 \int \frac {x^2}{-6+x+2 \log (x)} \, dx+4 \int \frac {x^2 (-10+x+3 \log (x)) \log (5 x)}{(-6+x+2 \log (x))^2} \, dx \\ & = 2 \int \frac {x^2}{-6+x+2 \log (x)} \, dx+4 \int \left (-\frac {10 x^2 \log (5 x)}{(-6+x+2 \log (x))^2}+\frac {x^3 \log (5 x)}{(-6+x+2 \log (x))^2}+\frac {3 x^2 \log (x) \log (5 x)}{(-6+x+2 \log (x))^2}\right ) \, dx \\ & = 2 \int \frac {x^2}{-6+x+2 \log (x)} \, dx+4 \int \frac {x^3 \log (5 x)}{(-6+x+2 \log (x))^2} \, dx+12 \int \frac {x^2 \log (x) \log (5 x)}{(-6+x+2 \log (x))^2} \, dx-40 \int \frac {x^2 \log (5 x)}{(-6+x+2 \log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {2 x^3 \log (5 x)}{-6+x+2 \log (x)} \]

[In]

Integrate[(-12*x^2 + 2*x^3 + 4*x^2*Log[x] + (-40*x^2 + 4*x^3 + 12*x^2*Log[x])*Log[5*x])/(36 - 12*x + x^2 + (-2
4 + 4*x)*Log[x] + 4*Log[x]^2),x]

[Out]

(2*x^3*Log[5*x])/(-6 + x + 2*Log[x])

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

method result size
parallelrisch \(\frac {2 x^{3} \ln \left (5 x \right )}{-6+2 \ln \left (x \right )+x}\) \(19\)
risch \(x^{3}-\frac {\left (-6-2 \ln \left (5\right )+x \right ) x^{3}}{-6+2 \ln \left (x \right )+x}\) \(26\)
default \(\frac {2 x^{3} \ln \left (x \right )}{-6+2 \ln \left (x \right )+x}+\frac {2 \ln \left (5\right ) x^{3}}{-6+2 \ln \left (x \right )+x}\) \(34\)

[In]

int(((12*x^2*ln(x)+4*x^3-40*x^2)*ln(5*x)+4*x^2*ln(x)+2*x^3-12*x^2)/(4*ln(x)^2+(4*x-24)*ln(x)+x^2-12*x+36),x,me
thod=_RETURNVERBOSE)

[Out]

2*x^3*ln(5*x)/(-6+2*ln(x)+x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {2 \, {\left (x^{3} \log \left (5\right ) + x^{3} \log \left (x\right )\right )}}{x + 2 \, \log \left (x\right ) - 6} \]

[In]

integrate(((12*x^2*log(x)+4*x^3-40*x^2)*log(5*x)+4*x^2*log(x)+2*x^3-12*x^2)/(4*log(x)^2+(4*x-24)*log(x)+x^2-12
*x+36),x, algorithm="fricas")

[Out]

2*(x^3*log(5) + x^3*log(x))/(x + 2*log(x) - 6)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=x^{3} + \frac {- x^{4} + 2 x^{3} \log {\left (5 \right )} + 6 x^{3}}{x + 2 \log {\left (x \right )} - 6} \]

[In]

integrate(((12*x**2*ln(x)+4*x**3-40*x**2)*ln(5*x)+4*x**2*ln(x)+2*x**3-12*x**2)/(4*ln(x)**2+(4*x-24)*ln(x)+x**2
-12*x+36),x)

[Out]

x**3 + (-x**4 + 2*x**3*log(5) + 6*x**3)/(x + 2*log(x) - 6)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {2 \, {\left (x^{3} \log \left (5\right ) + x^{3} \log \left (x\right )\right )}}{x + 2 \, \log \left (x\right ) - 6} \]

[In]

integrate(((12*x^2*log(x)+4*x^3-40*x^2)*log(5*x)+4*x^2*log(x)+2*x^3-12*x^2)/(4*log(x)^2+(4*x-24)*log(x)+x^2-12
*x+36),x, algorithm="maxima")

[Out]

2*(x^3*log(5) + x^3*log(x))/(x + 2*log(x) - 6)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=x^{3} - \frac {x^{4} - 2 \, x^{3} \log \left (5\right ) - 6 \, x^{3}}{x + 2 \, \log \left (x\right ) - 6} \]

[In]

integrate(((12*x^2*log(x)+4*x^3-40*x^2)*log(5*x)+4*x^2*log(x)+2*x^3-12*x^2)/(4*log(x)^2+(4*x-24)*log(x)+x^2-12
*x+36),x, algorithm="giac")

[Out]

x^3 - (x^4 - 2*x^3*log(5) - 6*x^3)/(x + 2*log(x) - 6)

Mupad [B] (verification not implemented)

Time = 12.87 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-12 x^2+2 x^3+4 x^2 \log (x)+\left (-40 x^2+4 x^3+12 x^2 \log (x)\right ) \log (5 x)}{36-12 x+x^2+(-24+4 x) \log (x)+4 \log ^2(x)} \, dx=\frac {2\,x^3\,\left (\ln \left (5\right )+\ln \left (x\right )\right )}{x+2\,\ln \left (x\right )-6} \]

[In]

int((4*x^2*log(x) + log(5*x)*(12*x^2*log(x) - 40*x^2 + 4*x^3) - 12*x^2 + 2*x^3)/(4*log(x)^2 - 12*x + log(x)*(4
*x - 24) + x^2 + 36),x)

[Out]

(2*x^3*(log(5) + log(x)))/(x + 2*log(x) - 6)