Integrand size = 76, antiderivative size = 18 \[ \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{\left (-4 x+x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx=5-2 x+\frac {10}{\log \left (x \log \left (-4+x^2\right )\right )} \] Output:
5-2*x+10/ln(x*ln(x^2-4))
Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{\left (-4 x+x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx=-2 x+\frac {10}{\log \left (x \log \left (-4+x^2\right )\right )} \] Input:
Integrate[(-20*x^2 + (40 - 10*x^2)*Log[-4 + x^2] + (8*x - 2*x^3)*Log[-4 + x^2]*Log[x*Log[-4 + x^2]]^2)/((-4*x + x^3)*Log[-4 + x^2]*Log[x*Log[-4 + x^ 2]]^2),x]
Output:
-2*x + 10/Log[x*Log[-4 + x^2]]
Time = 0.59 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2026, 7239, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (x^2-4\right )+\left (8 x-2 x^3\right ) \log \left (x^2-4\right ) \log ^2\left (x \log \left (x^2-4\right )\right )}{\left (x^3-4 x\right ) \log \left (x^2-4\right ) \log ^2\left (x \log \left (x^2-4\right )\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (x^2-4\right )+\left (8 x-2 x^3\right ) \log \left (x^2-4\right ) \log ^2\left (x \log \left (x^2-4\right )\right )}{x \left (x^2-4\right ) \log \left (x^2-4\right ) \log ^2\left (x \log \left (x^2-4\right )\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {-\frac {20 x}{\left (x^2-4\right ) \log \left (x^2-4\right )}-\frac {10}{x}}{\log ^2\left (x \log \left (x^2-4\right )\right )}-2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {10}{\log \left (x \log \left (x^2-4\right )\right )}-2 x\) |
Input:
Int[(-20*x^2 + (40 - 10*x^2)*Log[-4 + x^2] + (8*x - 2*x^3)*Log[-4 + x^2]*L og[x*Log[-4 + x^2]]^2)/((-4*x + x^3)*Log[-4 + x^2]*Log[x*Log[-4 + x^2]]^2) ,x]
Output:
-2*x + 10/Log[x*Log[-4 + x^2]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56
method | result | size |
parallelrisch | \(\frac {80-16 \ln \left (x \ln \left (x^{2}-4\right )\right ) x}{8 \ln \left (x \ln \left (x^{2}-4\right )\right )}\) | \(28\) |
risch | \(-2 x -\frac {20 i}{\pi \,\operatorname {csgn}\left (i \ln \left (x^{2}-4\right )\right ) {\operatorname {csgn}\left (i x \ln \left (x^{2}-4\right )\right )}^{2}-\pi \,\operatorname {csgn}\left (i \ln \left (x^{2}-4\right )\right ) \operatorname {csgn}\left (i x \ln \left (x^{2}-4\right )\right ) \operatorname {csgn}\left (i x \right )-\pi {\operatorname {csgn}\left (i x \ln \left (x^{2}-4\right )\right )}^{3}+\pi {\operatorname {csgn}\left (i x \ln \left (x^{2}-4\right )\right )}^{2} \operatorname {csgn}\left (i x \right )-2 i \ln \left (x \right )-2 i \ln \left (\ln \left (x^{2}-4\right )\right )}\) | \(116\) |
Input:
int(((-2*x^3+8*x)*ln(x^2-4)*ln(x*ln(x^2-4))^2+(-10*x^2+40)*ln(x^2-4)-20*x^ 2)/(x^3-4*x)/ln(x^2-4)/ln(x*ln(x^2-4))^2,x,method=_RETURNVERBOSE)
Output:
1/8*(80-16*ln(x*ln(x^2-4))*x)/ln(x*ln(x^2-4))
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{\left (-4 x+x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx=-\frac {2 \, {\left (x \log \left (x \log \left (x^{2} - 4\right )\right ) - 5\right )}}{\log \left (x \log \left (x^{2} - 4\right )\right )} \] Input:
integrate(((-2*x^3+8*x)*log(x^2-4)*log(x*log(x^2-4))^2+(-10*x^2+40)*log(x^ 2-4)-20*x^2)/(x^3-4*x)/log(x^2-4)/log(x*log(x^2-4))^2,x, algorithm="fricas ")
Output:
-2*(x*log(x*log(x^2 - 4)) - 5)/log(x*log(x^2 - 4))
Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{\left (-4 x+x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx=- 2 x + \frac {10}{\log {\left (x \log {\left (x^{2} - 4 \right )} \right )}} \] Input:
integrate(((-2*x**3+8*x)*ln(x**2-4)*ln(x*ln(x**2-4))**2+(-10*x**2+40)*ln(x **2-4)-20*x**2)/(x**3-4*x)/ln(x**2-4)/ln(x*ln(x**2-4))**2,x)
Output:
-2*x + 10/log(x*log(x**2 - 4))
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.94 \[ \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{\left (-4 x+x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx=-\frac {2 \, {\left (x \log \left (x\right ) + x \log \left (\log \left (x + 2\right ) + \log \left (x - 2\right )\right ) - 5\right )}}{\log \left (x\right ) + \log \left (\log \left (x + 2\right ) + \log \left (x - 2\right )\right )} \] Input:
integrate(((-2*x^3+8*x)*log(x^2-4)*log(x*log(x^2-4))^2+(-10*x^2+40)*log(x^ 2-4)-20*x^2)/(x^3-4*x)/log(x^2-4)/log(x*log(x^2-4))^2,x, algorithm="maxima ")
Output:
-2*(x*log(x) + x*log(log(x + 2) + log(x - 2)) - 5)/(log(x) + log(log(x + 2 ) + log(x - 2)))
Time = 0.60 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{\left (-4 x+x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx=-2 \, x + \frac {10}{\log \left (x \log \left (x^{2} - 4\right )\right )} \] Input:
integrate(((-2*x^3+8*x)*log(x^2-4)*log(x*log(x^2-4))^2+(-10*x^2+40)*log(x^ 2-4)-20*x^2)/(x^3-4*x)/log(x^2-4)/log(x*log(x^2-4))^2,x, algorithm="giac")
Output:
-2*x + 10/log(x*log(x^2 - 4))
Time = 3.74 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{\left (-4 x+x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx=\frac {10}{\ln \left (x\,\ln \left (x^2-4\right )\right )}-2\,x \] Input:
int((log(x^2 - 4)*(10*x^2 - 40) + 20*x^2 - log(x^2 - 4)*log(x*log(x^2 - 4) )^2*(8*x - 2*x^3))/(log(x^2 - 4)*log(x*log(x^2 - 4))^2*(4*x - x^3)),x)
Output:
10/log(x*log(x^2 - 4)) - 2*x
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{\left (-4 x+x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx=\frac {-2 \,\mathrm {log}\left (\mathrm {log}\left (x^{2}-4\right ) x \right ) x +10}{\mathrm {log}\left (\mathrm {log}\left (x^{2}-4\right ) x \right )} \] Input:
int(((-2*x^3+8*x)*log(x^2-4)*log(x*log(x^2-4))^2+(-10*x^2+40)*log(x^2-4)-2 0*x^2)/(x^3-4*x)/log(x^2-4)/log(x*log(x^2-4))^2,x)
Output:
(2*( - log(log(x**2 - 4)*x)*x + 5))/log(log(x**2 - 4)*x)