Integrand size = 112, antiderivative size = 18 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x+\log \left (1+\log \left (\frac {5}{3+x-\log (48)}\right )\right )\right ) \] Output:
ln(ln(1+ln(5/(-ln(48)+3+x)))+x)
Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (-x-\log \left (1+\log \left (\frac {5}{3+x-\log (48)}\right )\right )\right ) \] Input:
Integrate[(-2 - x + Log[48] + (-3 - x + Log[48])*Log[-5/(-3 - x + Log[48]) ])/(-3*x - x^2 + x*Log[48] + (-3*x - x^2 + x*Log[48])*Log[-5/(-3 - x + Log [48])] + (-3 - x + Log[48] + (-3 - x + Log[48])*Log[-5/(-3 - x + Log[48])] )*Log[1 + Log[-5/(-3 - x + Log[48])]]),x]
Output:
Log[-x - Log[1 + Log[5/(3 + x - Log[48])]]]
Time = 0.54 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6, 7292, 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.
\(\displaystyle \int \frac {-x+(-x-3+\log (48)) \log \left (-\frac {5}{-x-3+\log (48)}\right )-2+\log (48)}{-x^2+\left (-x^2-3 x+x \log (48)\right ) \log \left (-\frac {5}{-x-3+\log (48)}\right )-3 x+x \log (48)+\left (-x+(-x-3+\log (48)) \log \left (-\frac {5}{-x-3+\log (48)}\right )-3+\log (48)\right ) \log \left (\log \left (-\frac {5}{-x-3+\log (48)}\right )+1\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x+(-x-3+\log (48)) \log \left (-\frac {5}{-x-3+\log (48)}\right )-2+\log (48)}{-x^2+\left (-x^2-3 x+x \log (48)\right ) \log \left (-\frac {5}{-x-3+\log (48)}\right )+x (\log (48)-3)+\left (-x+(-x-3+\log (48)) \log \left (-\frac {5}{-x-3+\log (48)}\right )-3+\log (48)\right ) \log \left (\log \left (-\frac {5}{-x-3+\log (48)}\right )+1\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x-(-x-3+\log (48)) \log \left (-\frac {5}{-x-3+\log (48)}\right )+2 \left (1-\frac {\log (48)}{2}\right )}{(x+3-\log (48)) \left (\log \left (\frac {5}{x+3-\log (48)}\right )+1\right ) \left (x+\log \left (\log \left (\frac {5}{x+3-\log (48)}\right )+1\right )\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \log \left (x+\log \left (\log \left (\frac {5}{x+3-\log (48)}\right )+1\right )\right )\) |
Input:
Int[(-2 - x + Log[48] + (-3 - x + Log[48])*Log[-5/(-3 - x + Log[48])])/(-3 *x - x^2 + x*Log[48] + (-3*x - x^2 + x*Log[48])*Log[-5/(-3 - x + Log[48])] + (-3 - x + Log[48] + (-3 - x + Log[48])*Log[-5/(-3 - x + Log[48])])*Log[ 1 + Log[-5/(-3 - x + Log[48])]]),x]
Output:
Log[x + Log[1 + Log[5/(3 + x - Log[48])]]]
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 = 5.82 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06
method | result | size |
norman | \(\ln \left (x +\ln \left (\ln \left (-\frac {5}{\ln \left (48\right )-3-x}\right )+1\right )\right )\) | \(19\) |
parallelrisch | \(\ln \left (x +\ln \left (\ln \left (-\frac {5}{\ln \left (48\right )-3-x}\right )+1\right )\right )\) | \(19\) |
Input:
int(((ln(48)-3-x)*ln(-5/(ln(48)-3-x))+ln(48)-x-2)/(((ln(48)-3-x)*ln(-5/(ln (48)-3-x))+ln(48)-3-x)*ln(ln(-5/(ln(48)-3-x))+1)+(x*ln(48)-x^2-3*x)*ln(-5/ (ln(48)-3-x))+x*ln(48)-x^2-3*x),x,method=_RETURNVERBOSE)
Output:
ln(x+ln(ln(-5/(ln(48)-3-x))+1))
Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x + \log \left (\log \left (\frac {5}{x - \log \left (48\right ) + 3}\right ) + 1\right )\right ) \] Input:
integrate(((log(48)-3-x)*log(-5/(log(48)-3-x))+log(48)-x-2)/(((log(48)-3-x )*log(-5/(log(48)-3-x))+log(48)-3-x)*log(log(-5/(log(48)-3-x))+1)+(x*log(4 8)-x^2-3*x)*log(-5/(log(48)-3-x))+x*log(48)-x^2-3*x),x, algorithm="fricas" )
Output:
log(x + log(log(5/(x - log(48) + 3)) + 1))
Time = 0.35 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log {\left (x + \log {\left (\log {\left (- \frac {5}{- x - 3 + \log {\left (48 \right )}} \right )} + 1 \right )} \right )} \] Input:
integrate(((ln(48)-3-x)*ln(-5/(ln(48)-3-x))+ln(48)-x-2)/(((ln(48)-3-x)*ln( -5/(ln(48)-3-x))+ln(48)-3-x)*ln(ln(-5/(ln(48)-3-x))+1)+(x*ln(48)-x**2-3*x) *ln(-5/(ln(48)-3-x))+x*ln(48)-x**2-3*x),x)
Output:
log(x + log(log(-5/(-x - 3 + log(48))) + 1))
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x + \log \left (\log \left (5\right ) - \log \left (x - \log \left (3\right ) - 4 \, \log \left (2\right ) + 3\right ) + 1\right )\right ) \] Input:
integrate(((log(48)-3-x)*log(-5/(log(48)-3-x))+log(48)-x-2)/(((log(48)-3-x )*log(-5/(log(48)-3-x))+log(48)-3-x)*log(log(-5/(log(48)-3-x))+1)+(x*log(4 8)-x^2-3*x)*log(-5/(log(48)-3-x))+x*log(48)-x^2-3*x),x, algorithm="maxima" )
Output:
log(x + log(log(5) - log(x - log(3) - 4*log(2) + 3) + 1))
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x + \log \left (2 i \, \pi + \log \left (5\right ) - \log \left (x - \log \left (48\right ) + 3\right ) + 1\right )\right ) \] Input:
integrate(((log(48)-3-x)*log(-5/(log(48)-3-x))+log(48)-x-2)/(((log(48)-3-x )*log(-5/(log(48)-3-x))+log(48)-3-x)*log(log(-5/(log(48)-3-x))+1)+(x*log(4 8)-x^2-3*x)*log(-5/(log(48)-3-x))+x*log(48)-x^2-3*x),x, algorithm="giac")
Output:
log(x + log(2*I*pi + log(5) - log(x - log(48) + 3) + 1))
Time = 116.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\ln \left (x+\ln \left (\ln \left (\frac {5}{x-\ln \left (48\right )+3}\right )+1\right )\right ) \] Input:
int((x - log(48) + log(5/(x - log(48) + 3))*(x - log(48) + 3) + 2)/(3*x - x*log(48) + log(log(5/(x - log(48) + 3)) + 1)*(x - log(48) + log(5/(x - lo g(48) + 3))*(x - log(48) + 3) + 3) + log(5/(x - log(48) + 3))*(3*x - x*log (48) + x^2) + x^2),x)
Output:
log(x + log(log(5/(x - log(48) + 3)) + 1))
Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (-\frac {5}{\mathrm {log}\left (48\right )-x -3}\right )+1\right )+x \right ) \] Input:
int(((log(48)-3-x)*log(-5/(log(48)-3-x))+log(48)-x-2)/(((log(48)-3-x)*log( -5/(log(48)-3-x))+log(48)-3-x)*log(log(-5/(log(48)-3-x))+1)+(x*log(48)-x^2 -3*x)*log(-5/(log(48)-3-x))+x*log(48)-x^2-3*x),x)
Output:
log(log(log(( - 5)/(log(48) - x - 3)) + 1) + x)