Integrand size = 112, antiderivative size = 18 \begin {dmath*} \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 ) \end {dmath*}
Time = 0.58 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \begin {dmath*} \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 ) \end {dmath*}
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]
Time = 0.53 (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 |
\(\displaystyle \log \left (x+\log \left (\log \left (\frac {5}{x+3-\log (48)}\right )+1\right )\right )\) |
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]
3.1.47.3.1 Defintions of rubi rules used
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 = 9.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06
method | result | size |
norman | \(\ln \left (\ln \left (\ln \left (-\frac {5}{\ln \left (48\right )-3-x}\right )+1\right )+x \right )\) | \(19\) |
parallelrisch | \(\ln \left (\ln \left (\ln \left (-\frac {5}{\ln \left (48\right )-3-x}\right )+1\right )+x \right )\) | \(19\) |
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)
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \begin {dmath*} \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 ) \end {dmath*}
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=\
Time = 0.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \begin {dmath*} \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 )} \end {dmath*}
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)
Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \begin {dmath*} \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 ) \end {dmath*}
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=\
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \begin {dmath*} \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 ) \end {dmath*}
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=\
Timed out. \begin {dmath*} \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=\text {Hanged} \end {dmath*}