Integrand size = 65, antiderivative size = 21 \[ \int \frac {-1-8 x^2+\log (x)}{\left (-16 x^2-8 x^3-x \log (x)+4 x^2 \log (\log (5))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx=\log \left (\log \left (2 \left (-4-2 x-\frac {\log (x)}{4 x}+\log (\log (5))\right )\right )\right ) \] Output:
ln(ln(-1/2*ln(x)/x-8-4*x+2*ln(ln(5))))
Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-1-8 x^2+\log (x)}{\left (-16 x^2-8 x^3-x \log (x)+4 x^2 \log (\log (5))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx=\log \left (\log \left (-8-4 x-\frac {\log (x)}{2 x}+2 \log (\log (5))\right )\right ) \] Input:
Integrate[(-1 - 8*x^2 + Log[x])/((-16*x^2 - 8*x^3 - x*Log[x] + 4*x^2*Log[L og[5]])*Log[(-16*x - 8*x^2 - Log[x] + 4*x*Log[Log[5]])/(2*x)]),x]
Output:
Log[Log[-8 - 4*x - Log[x]/(2*x) + 2*Log[Log[5]]]]
Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6, 7235}
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.
\(\displaystyle \int \frac {-8 x^2+\log (x)-1}{\left (-8 x^3-16 x^2+4 x^2 \log (\log (5))-x \log (x)\right ) \log \left (\frac {-8 x^2-16 x+4 x \log (\log (5))-\log (x)}{2 x}\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-8 x^2+\log (x)-1}{\left (-8 x^3+x^2 (4 \log (\log (5))-16)-x \log (x)\right ) \log \left (\frac {-8 x^2-16 x+4 x \log (\log (5))-\log (x)}{2 x}\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \log \left (\log \left (-\frac {8 x^2+16 x-4 x \log (\log (5))+\log (x)}{2 x}\right )\right )\) |
Input:
Int[(-1 - 8*x^2 + Log[x])/((-16*x^2 - 8*x^3 - x*Log[x] + 4*x^2*Log[Log[5]] )*Log[(-16*x - 8*x^2 - Log[x] + 4*x*Log[Log[5]])/(2*x)]),x]
Output:
Log[Log[-1/2*(16*x + 8*x^2 + Log[x] - 4*x*Log[Log[5]])/x]]
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 0.79 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \(\ln \left (\ln \left (\frac {4 x \ln \left (\ln \left (5\right )\right )-\ln \left (x \right )-8 x^{2}-16 x}{2 x}\right )\right )\) | \(27\) |
default | \(\ln \left (\ln \left (2\right )-\ln \left (\frac {4 x \ln \left (\ln \left (5\right )\right )-\ln \left (x \right )-8 x^{2}-16 x}{x}\right )\right )\) | \(31\) |
risch | \(\ln \left (\ln \left (2\right )-\ln \left (x \right )+\ln \left (x \ln \left (\ln \left (5\right )\right )-2 x^{2}-4 x -\frac {\ln \left (x \right )}{4}\right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (-x \ln \left (\ln \left (5\right )\right )+2 x^{2}+4 x +\frac {\ln \left (x \right )}{4}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-x \ln \left (\ln \left (5\right )\right )+2 x^{2}+4 x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (-x \ln \left (\ln \left (5\right )\right )+2 x^{2}+4 x +\frac {\ln \left (x \right )}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-x \ln \left (\ln \left (5\right )\right )+2 x^{2}+4 x +\frac {\ln \left (x \right )}{4}\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (-x \ln \left (\ln \left (5\right )\right )+2 x^{2}+4 x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (-x \ln \left (\ln \left (5\right )\right )+2 x^{2}+4 x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )}^{3}}{2}\right )\) | \(214\) |
Input:
int((ln(x)-8*x^2-1)/(4*ln(ln(5))*x^2-x*ln(x)-8*x^3-16*x^2)/ln(1/2*(4*x*ln( ln(5))-ln(x)-8*x^2-16*x)/x),x,method=_RETURNVERBOSE)
Output:
ln(ln(1/2*(4*x*ln(ln(5))-ln(x)-8*x^2-16*x)/x))
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-1-8 x^2+\log (x)}{\left (-16 x^2-8 x^3-x \log (x)+4 x^2 \log (\log (5))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx=\log \left (\log \left (-\frac {8 \, x^{2} - 4 \, x \log \left (\log \left (5\right )\right ) + 16 \, x + \log \left (x\right )}{2 \, x}\right )\right ) \] Input:
integrate((log(x)-8*x^2-1)/(4*log(log(5))*x^2-x*log(x)-8*x^3-16*x^2)/log(1 /2*(4*x*log(log(5))-log(x)-8*x^2-16*x)/x),x, algorithm="fricas")
Output:
log(log(-1/2*(8*x^2 - 4*x*log(log(5)) + 16*x + log(x))/x))
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-1-8 x^2+\log (x)}{\left (-16 x^2-8 x^3-x \log (x)+4 x^2 \log (\log (5))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx=\log {\left (\log {\left (\frac {- 4 x^{2} - 8 x + 2 x \log {\left (\log {\left (5 \right )} \right )} - \frac {\log {\left (x \right )}}{2}}{x} \right )} \right )} \] Input:
integrate((ln(x)-8*x**2-1)/(4*ln(ln(5))*x**2-x*ln(x)-8*x**3-16*x**2)/ln(1/ 2*(4*x*ln(ln(5))-ln(x)-8*x**2-16*x)/x),x)
Output:
log(log((-4*x**2 - 8*x + 2*x*log(log(5)) - log(x)/2)/x))
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-1-8 x^2+\log (x)}{\left (-16 x^2-8 x^3-x \log (x)+4 x^2 \log (\log (5))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx=\log \left (-\log \left (2\right ) + \log \left (-8 \, x^{2} + 4 \, x {\left (\log \left (\log \left (5\right )\right ) - 4\right )} - \log \left (x\right )\right ) - \log \left (x\right )\right ) \] Input:
integrate((log(x)-8*x^2-1)/(4*log(log(5))*x^2-x*log(x)-8*x^3-16*x^2)/log(1 /2*(4*x*log(log(5))-log(x)-8*x^2-16*x)/x),x, algorithm="maxima")
Output:
log(-log(2) + log(-8*x^2 + 4*x*(log(log(5)) - 4) - log(x)) - log(x))
Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-1-8 x^2+\log (x)}{\left (-16 x^2-8 x^3-x \log (x)+4 x^2 \log (\log (5))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx=\log \left (-\log \left (2\right ) + \log \left (-8 \, x^{2} + 4 \, x \log \left (\log \left (5\right )\right ) - 16 \, x - \log \left (x\right )\right ) - \log \left (x\right )\right ) \] Input:
integrate((log(x)-8*x^2-1)/(4*log(log(5))*x^2-x*log(x)-8*x^3-16*x^2)/log(1 /2*(4*x*log(log(5))-log(x)-8*x^2-16*x)/x),x, algorithm="giac")
Output:
log(-log(2) + log(-8*x^2 + 4*x*log(log(5)) - 16*x - log(x)) - log(x))
Time = 9.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-1-8 x^2+\log (x)}{\left (-16 x^2-8 x^3-x \log (x)+4 x^2 \log (\log (5))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx=\ln \left (\ln \left (-\frac {16\,x+\ln \left (x\right )-4\,x\,\ln \left (\ln \left (5\right )\right )+8\,x^2}{2\,x}\right )\right ) \] Input:
int((8*x^2 - log(x) + 1)/(log(-(8*x + log(x)/2 - 2*x*log(log(5)) + 4*x^2)/ x)*(x*log(x) - 4*x^2*log(log(5)) + 16*x^2 + 8*x^3)),x)
Output:
log(log(-(16*x + log(x) - 4*x*log(log(5)) + 8*x^2)/(2*x)))
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-1-8 x^2+\log (x)}{\left (-16 x^2-8 x^3-x \log (x)+4 x^2 \log (\log (5))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {4 \,\mathrm {log}\left (\mathrm {log}\left (5\right )\right ) x -\mathrm {log}\left (x \right )-8 x^{2}-16 x}{2 x}\right )\right ) \] Input:
int((log(x)-8*x^2-1)/(4*log(log(5))*x^2-x*log(x)-8*x^3-16*x^2)/log(1/2*(4* x*log(log(5))-log(x)-8*x^2-16*x)/x),x)
Output:
log(log((4*log(log(5))*x - log(x) - 8*x**2 - 16*x)/(2*x)))