Integrand size = 71, antiderivative size = 21 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=-x+\left (\log (\log (5))+\log \left (\log \left (-2 x+5 x^2\right )\right )\right )^2 \] Output:
-x+(ln(ln(5*x^2-2*x))+ln(ln(5)))^2
Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=-x+\log ^2(\log (5) \log (x (-2+5 x))) \] Input:
Integrate[((2*x - 5*x^2)*Log[-2*x + 5*x^2] + (-4 + 20*x)*Log[Log[5]] + (-4 + 20*x)*Log[Log[-2*x + 5*x^2]])/((-2*x + 5*x^2)*Log[-2*x + 5*x^2]),x]
Output:
-x + Log[Log[5]*Log[x*(-2 + 5*x)]]^2
Time = 1.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2026, 7292, 7239, 7293, 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.
\(\displaystyle \int \frac {(20 x-4) \log \left (\log \left (5 x^2-2 x\right )\right )+\left (2 x-5 x^2\right ) \log \left (5 x^2-2 x\right )+(20 x-4) \log (\log (5))}{\left (5 x^2-2 x\right ) \log \left (5 x^2-2 x\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {(20 x-4) \log \left (\log \left (5 x^2-2 x\right )\right )+\left (2 x-5 x^2\right ) \log \left (5 x^2-2 x\right )+(20 x-4) \log (\log (5))}{x (5 x-2) \log \left (5 x^2-2 x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(20 x-4) \left (-\log \left (\log \left (5 x^2-2 x\right )\right )\right )-\left (2 x-5 x^2\right ) \log \left (5 x^2-2 x\right )-(20 x-4) \log (\log (5))}{(2-5 x) x \log (x (5 x-2))}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 x-\frac {4 (5 x-1) \log (\log (5) \log (x (5 x-2)))}{x \log (x (5 x-2))}-2}{2-5 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 (5 x-1) \log (\log (5) \log (x (5 x-2)))}{x (5 x-2) \log (x (5 x-2))}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log ^2(\log (5) \log (-((2-5 x) x)))-x\) |
Input:
Int[((2*x - 5*x^2)*Log[-2*x + 5*x^2] + (-4 + 20*x)*Log[Log[5]] + (-4 + 20* x)*Log[Log[-2*x + 5*x^2]])/((-2*x + 5*x^2)*Log[-2*x + 5*x^2]),x]
Output:
-x + Log[Log[5]*Log[-((2 - 5*x)*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 = 0.44 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62
method | result | size |
default | \(-x +2 \ln \left (\ln \left (5 x^{2}-2 x \right )\right ) \ln \left (\ln \left (5\right )\right )+{\ln \left (\ln \left (5 x^{2}-2 x \right )\right )}^{2}\) | \(34\) |
norman | \(-x +2 \ln \left (\ln \left (5 x^{2}-2 x \right )\right ) \ln \left (\ln \left (5\right )\right )+{\ln \left (\ln \left (5 x^{2}-2 x \right )\right )}^{2}\) | \(34\) |
parts | \(-x +2 \ln \left (\ln \left (5 x^{2}-2 x \right )\right ) \ln \left (\ln \left (5\right )\right )+{\ln \left (\ln \left (5 x^{2}-2 x \right )\right )}^{2}\) | \(34\) |
Input:
int(((20*x-4)*ln(ln(5*x^2-2*x))+(20*x-4)*ln(ln(5))+(-5*x^2+2*x)*ln(5*x^2-2 *x))/(5*x^2-2*x)/ln(5*x^2-2*x),x,method=_RETURNVERBOSE)
Output:
-x+2*ln(ln(5*x^2-2*x))*ln(ln(5))+ln(ln(5*x^2-2*x))^2
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=2 \, \log \left (\log \left (5\right )\right ) \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right ) + \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right )^{2} - x \] Input:
integrate(((20*x-4)*log(log(5*x^2-2*x))+(20*x-4)*log(log(5))+(-5*x^2+2*x)* log(5*x^2-2*x))/(5*x^2-2*x)/log(5*x^2-2*x),x, algorithm="fricas")
Output:
2*log(log(5))*log(log(5*x^2 - 2*x)) + log(log(5*x^2 - 2*x))^2 - x
Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=- x + \log {\left (\log {\left (5 x^{2} - 2 x \right )} \right )}^{2} + 2 \log {\left (\log {\left (5 \right )} \right )} \log {\left (\log {\left (5 x^{2} - 2 x \right )} \right )} \] Input:
integrate(((20*x-4)*ln(ln(5*x**2-2*x))+(20*x-4)*ln(ln(5))+(-5*x**2+2*x)*ln (5*x**2-2*x))/(5*x**2-2*x)/ln(5*x**2-2*x),x)
Output:
-x + log(log(5*x**2 - 2*x))**2 + 2*log(log(5))*log(log(5*x**2 - 2*x))
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=\log \left (\log \left (5 \, x - 2\right ) + \log \left (x\right )\right )^{2} + 2 \, \log \left (\log \left (5 \, x - 2\right ) + \log \left (x\right )\right ) \log \left (\log \left (5\right )\right ) - x \] Input:
integrate(((20*x-4)*log(log(5*x^2-2*x))+(20*x-4)*log(log(5))+(-5*x^2+2*x)* log(5*x^2-2*x))/(5*x^2-2*x)/log(5*x^2-2*x),x, algorithm="maxima")
Output:
log(log(5*x - 2) + log(x))^2 + 2*log(log(5*x - 2) + log(x))*log(log(5)) - x
Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=2 \, \log \left (\log \left (5\right )\right ) \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right ) + \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right )^{2} - x \] Input:
integrate(((20*x-4)*log(log(5*x^2-2*x))+(20*x-4)*log(log(5))+(-5*x^2+2*x)* log(5*x^2-2*x))/(5*x^2-2*x)/log(5*x^2-2*x),x, algorithm="giac")
Output:
2*log(log(5))*log(log(5*x^2 - 2*x)) + log(log(5*x^2 - 2*x))^2 - x
Time = 3.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx={\ln \left (\ln \left (5\,x^2-2\,x\right )\right )}^2+2\,\ln \left (\ln \left (5\right )\right )\,\ln \left (\ln \left (5\,x^2-2\,x\right )\right )-x \] Input:
int(-(log(log(5))*(20*x - 4) + log(5*x^2 - 2*x)*(2*x - 5*x^2) + log(log(5* x^2 - 2*x))*(20*x - 4))/(log(5*x^2 - 2*x)*(2*x - 5*x^2)),x)
Output:
2*log(log(5*x^2 - 2*x))*log(log(5)) - x + log(log(5*x^2 - 2*x))^2
Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx={\mathrm {log}\left (\mathrm {log}\left (5 x^{2}-2 x \right )\right )}^{2}+2 \,\mathrm {log}\left (\mathrm {log}\left (5 x^{2}-2 x \right )\right ) \mathrm {log}\left (\mathrm {log}\left (5\right )\right )-x \] Input:
int(((20*x-4)*log(log(5*x^2-2*x))+(20*x-4)*log(log(5))+(-5*x^2+2*x)*log(5* x^2-2*x))/(5*x^2-2*x)/log(5*x^2-2*x),x)
Output:
log(log(5*x**2 - 2*x))**2 + 2*log(log(5*x**2 - 2*x))*log(log(5)) - x