\(\int \frac {-60-30 x+(-30+15 x+6 x^2-3 x^3) \log (\frac {20-4 x^2}{x^2})}{(-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5) \log ^2(\frac {20-4 x^2}{x^2})} \, dx\) [3428]

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

Optimal result

Integrand size = 75, antiderivative size = 27 \[ \int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx=\frac {3 x}{4 (2+x)^2 \log \left (\frac {4 \left (\frac {5}{x}-x\right )}{x}\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx=\int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx \]

[In]

Int[(-60 - 30*x + (-30 + 15*x + 6*x^2 - 3*x^3)*Log[(20 - 4*x^2)/x^2])/((-160 - 240*x - 88*x^2 + 28*x^3 + 24*x^
4 + 4*x^5)*Log[(20 - 4*x^2)/x^2]^2),x]

[Out]

(15*Defer[Int][1/((2 + x)^2*Log[-4 + 20/x^2]^2), x])/2 - 30*Defer[Int][1/((2 + x)*Log[-4 + 20/x^2]^2), x] - (1
35*Defer[Int][1/((-5 + x^2)*Log[-4 + 20/x^2]^2), x])/2 + 30*Defer[Int][x/((-5 + x^2)*Log[-4 + 20/x^2]^2), x] +
 3*Defer[Int][1/((2 + x)^3*Log[-4 + 20/x^2]), x] - (3*Defer[Int][1/((2 + x)^2*Log[-4 + 20/x^2]), x])/4

Rubi steps \begin{align*} \text {integral}& = \int \frac {3 \left (10 (2+x)+\left (10-5 x-2 x^2+x^3\right ) \log \left (-4+\frac {20}{x^2}\right )\right )}{4 (2+x)^3 \left (5-x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx \\ & = \frac {3}{4} \int \frac {10 (2+x)+\left (10-5 x-2 x^2+x^3\right ) \log \left (-4+\frac {20}{x^2}\right )}{(2+x)^3 \left (5-x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx \\ & = \frac {3}{4} \int \left (-\frac {10}{(2+x)^2 \left (-5+x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )}+\frac {2-x}{(2+x)^3 \log \left (-4+\frac {20}{x^2}\right )}\right ) \, dx \\ & = \frac {3}{4} \int \frac {2-x}{(2+x)^3 \log \left (-4+\frac {20}{x^2}\right )} \, dx-\frac {15}{2} \int \frac {1}{(2+x)^2 \left (-5+x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx \\ & = \frac {3}{4} \int \left (\frac {4}{(2+x)^3 \log \left (-4+\frac {20}{x^2}\right )}-\frac {1}{(2+x)^2 \log \left (-4+\frac {20}{x^2}\right )}\right ) \, dx-\frac {15}{2} \int \left (-\frac {1}{(2+x)^2 \log ^2\left (-4+\frac {20}{x^2}\right )}+\frac {4}{(2+x) \log ^2\left (-4+\frac {20}{x^2}\right )}+\frac {9-4 x}{\left (-5+x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )}\right ) \, dx \\ & = -\left (\frac {3}{4} \int \frac {1}{(2+x)^2 \log \left (-4+\frac {20}{x^2}\right )} \, dx\right )+3 \int \frac {1}{(2+x)^3 \log \left (-4+\frac {20}{x^2}\right )} \, dx+\frac {15}{2} \int \frac {1}{(2+x)^2 \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx-\frac {15}{2} \int \frac {9-4 x}{\left (-5+x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx-30 \int \frac {1}{(2+x) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx \\ & = -\left (\frac {3}{4} \int \frac {1}{(2+x)^2 \log \left (-4+\frac {20}{x^2}\right )} \, dx\right )+3 \int \frac {1}{(2+x)^3 \log \left (-4+\frac {20}{x^2}\right )} \, dx-\frac {15}{2} \int \left (\frac {9}{\left (-5+x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )}-\frac {4 x}{\left (-5+x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )}\right ) \, dx+\frac {15}{2} \int \frac {1}{(2+x)^2 \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx-30 \int \frac {1}{(2+x) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx \\ & = -\left (\frac {3}{4} \int \frac {1}{(2+x)^2 \log \left (-4+\frac {20}{x^2}\right )} \, dx\right )+3 \int \frac {1}{(2+x)^3 \log \left (-4+\frac {20}{x^2}\right )} \, dx+\frac {15}{2} \int \frac {1}{(2+x)^2 \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx-30 \int \frac {1}{(2+x) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx+30 \int \frac {x}{\left (-5+x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx-\frac {135}{2} \int \frac {1}{\left (-5+x^2\right ) \log ^2\left (-4+\frac {20}{x^2}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx=\frac {3 x}{4 (2+x)^2 \log \left (-4+\frac {20}{x^2}\right )} \]

[In]

Integrate[(-60 - 30*x + (-30 + 15*x + 6*x^2 - 3*x^3)*Log[(20 - 4*x^2)/x^2])/((-160 - 240*x - 88*x^2 + 28*x^3 +
 24*x^4 + 4*x^5)*Log[(20 - 4*x^2)/x^2]^2),x]

[Out]

(3*x)/(4*(2 + x)^2*Log[-4 + 20/x^2])

Maple [A] (verified)

Time = 6.73 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method result size
norman \(\frac {3 x}{4 \left (2+x \right )^{2} \ln \left (\frac {-4 x^{2}+20}{x^{2}}\right )}\) \(23\)
parallelrisch \(\frac {3 x}{4 \left (x^{2}+4 x +4\right ) \ln \left (-\frac {4 \left (x^{2}-5\right )}{x^{2}}\right )}\) \(27\)
risch \(\frac {3 x}{4 \left (x^{2}+4 x +4\right ) \ln \left (\frac {-4 x^{2}+20}{x^{2}}\right )}\) \(28\)

[In]

int(((-3*x^3+6*x^2+15*x-30)*ln((-4*x^2+20)/x^2)-30*x-60)/(4*x^5+24*x^4+28*x^3-88*x^2-240*x-160)/ln((-4*x^2+20)
/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

3/4*x/(2+x)^2/ln((-4*x^2+20)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx=\frac {3 \, x}{4 \, {\left (x^{2} + 4 \, x + 4\right )} \log \left (-\frac {4 \, {\left (x^{2} - 5\right )}}{x^{2}}\right )} \]

[In]

integrate(((-3*x^3+6*x^2+15*x-30)*log((-4*x^2+20)/x^2)-30*x-60)/(4*x^5+24*x^4+28*x^3-88*x^2-240*x-160)/log((-4
*x^2+20)/x^2)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx=\frac {3 x}{\left (4 x^{2} + 16 x + 16\right ) \log {\left (\frac {20 - 4 x^{2}}{x^{2}} \right )}} \]

[In]

integrate(((-3*x**3+6*x**2+15*x-30)*ln((-4*x**2+20)/x**2)-30*x-60)/(4*x**5+24*x**4+28*x**3-88*x**2-240*x-160)/
ln((-4*x**2+20)/x**2)**2,x)

[Out]

3*x/((4*x**2 + 16*x + 16)*log((20 - 4*x**2)/x**2))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.33 \[ \int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx=\frac {3 \, x}{4 \, {\left (4 i \, \pi + {\left (i \, \pi + 2 \, \log \left (2\right )\right )} x^{2} - 4 \, {\left (-i \, \pi - 2 \, \log \left (2\right )\right )} x + {\left (x^{2} + 4 \, x + 4\right )} \log \left (x^{2} - 5\right ) - 2 \, {\left (x^{2} + 4 \, x + 4\right )} \log \left (x\right ) + 8 \, \log \left (2\right )\right )}} \]

[In]

integrate(((-3*x^3+6*x^2+15*x-30)*log((-4*x^2+20)/x^2)-30*x-60)/(4*x^5+24*x^4+28*x^3-88*x^2-240*x-160)/log((-4
*x^2+20)/x^2)^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (23) = 46\).

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx=\frac {3 \, x}{4 \, {\left (x^{2} \log \left (-\frac {4 \, {\left (x^{2} - 5\right )}}{x^{2}}\right ) + 4 \, x \log \left (-\frac {4 \, {\left (x^{2} - 5\right )}}{x^{2}}\right ) + 4 \, \log \left (-\frac {4 \, {\left (x^{2} - 5\right )}}{x^{2}}\right )\right )}} \]

[In]

integrate(((-3*x^3+6*x^2+15*x-30)*log((-4*x^2+20)/x^2)-30*x-60)/(4*x^5+24*x^4+28*x^3-88*x^2-240*x-160)/log((-4
*x^2+20)/x^2)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.75 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {-60-30 x+\left (-30+15 x+6 x^2-3 x^3\right ) \log \left (\frac {20-4 x^2}{x^2}\right )}{\left (-160-240 x-88 x^2+28 x^3+24 x^4+4 x^5\right ) \log ^2\left (\frac {20-4 x^2}{x^2}\right )} \, dx=\frac {3\,x}{4\,\ln \left (-\frac {4\,x^2-20}{x^2}\right )\,{\left (x+2\right )}^2}-\frac {12}{5\,{\left (x+2\right )}^2}-\frac {3\,x^2}{5\,{\left (x+2\right )}^2}-\frac {12\,x}{5\,{\left (x+2\right )}^2} \]

[In]

int((30*x - log(-(4*x^2 - 20)/x^2)*(15*x + 6*x^2 - 3*x^3 - 30) + 60)/(log(-(4*x^2 - 20)/x^2)^2*(240*x + 88*x^2
 - 28*x^3 - 24*x^4 - 4*x^5 + 160)),x)

[Out]

(3*x)/(4*log(-(4*x^2 - 20)/x^2)*(x + 2)^2) - 12/(5*(x + 2)^2) - (3*x^2)/(5*(x + 2)^2) - (12*x)/(5*(x + 2)^2)