\(\int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log (-\frac {5}{-3-x+\log (48)})}{-3 x-x^2+x \log (48)+(-3 x-x^2+x \log (48)) \log (-\frac {5}{-3-x+\log (48)})+(-3-x+\log (48)+(-3-x+\log (48)) \log (-\frac {5}{-3-x+\log (48)})) \log (1+\log (-\frac {5}{-3-x+\log (48)}))} \, dx\) [9047]

Optimal result

Integrand size = 112, antiderivative size = 18 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x+\log \left (1+\log \left (\frac {5}{3+x-\log (48)}\right )\right )\right ) \]

[Out]

ln(ln(1+ln(5/(-ln(48)+3+x)))+x)

Rubi [A] (verified)

Time = 0.16 (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, 6820, 6816} \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x+\log \left (\log \left (\frac {5}{x+3-\log (48)}\right )+1\right )\right ) \]

[In]

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])])*Lo
g[1 + Log[-5/(-3 - x + Log[48])]]),x]

[Out]

Log[x + Log[1 + Log[5/(3 + x - Log[48])]]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-x^2+x (-3+\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 \\ & = \int \frac {x+2 \left (1-\frac {\log (48)}{2}\right )+(3+x-\log (48)) \log \left (\frac {5}{3+x-\log (48)}\right )}{(3+x-\log (48)) \left (1+\log \left (\frac {5}{3+x-\log (48)}\right )\right ) \left (x+\log \left (1+\log \left (\frac {5}{3+x-\log (48)}\right )\right )\right )} \, dx \\ & = \log \left (x+\log \left (1+\log \left (\frac {5}{3+x-\log (48)}\right )\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (-x-\log \left (1+\log \left (\frac {5}{3+x-\log (48)}\right )\right )\right ) \]

[In]

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]

[Out]

Log[-x - Log[1 + Log[5/(3 + x - Log[48])]]]

Maple [A] (verified)

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	\(\ln \left (\ln \left (\ln \left (-\frac {5}{\ln \left (48\right )-3-x}\right )+1\right )+x \right )\)	\(19\)

[In]

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)

[Out]

ln(ln(ln(-5/(ln(48)-3-x))+1)+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x + \log \left (\log \left (\frac {5}{x - \log \left (48\right ) + 3}\right ) + 1\right )\right ) \]

[In]

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(48)-x^2-3*x)*log(-5/(log(48)-3-x))+x*log(48)-x^2-3*x),x, algorithm="frica
s")

[Out]

log(x + log(log(5/(x - log(48) + 3)) + 1))

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log {\left (x + \log {\left (\log {\left (- \frac {5}{- x - 3 + \log {\left (48 \right )}} \right )} + 1 \right )} \right )} \]

[In]

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)

[Out]

log(x + log(log(-5/(-x - 3 + log(48))) + 1))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x + \log \left (\log \left (5\right ) - \log \left (x - \log \left (3\right ) - 4 \, \log \left (2\right ) + 3\right ) + 1\right )\right ) \]

[In]

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(48)-x^2-3*x)*log(-5/(log(48)-3-x))+x*log(48)-x^2-3*x),x, algorithm="maxim
a")

[Out]

log(x + log(log(5) - log(x - log(3) - 4*log(2) + 3) + 1))

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\log \left (x + \log \left (2 i \, \pi + \log \left (5\right ) - \log \left (x - \log \left (48\right ) + 3\right ) + 1\right )\right ) \]

[In]

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(48)-x^2-3*x)*log(-5/(log(48)-3-x))+x*log(48)-x^2-3*x),x, algorithm="giac"
)

[Out]

log(x + log(2*I*pi + log(5) - log(x - log(48) + 3) + 1))

Mupad [F(-1)]

Timed out. \[ \int \frac {-2-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )}{-3 x-x^2+x \log (48)+\left (-3 x-x^2+x \log (48)\right ) \log \left (-\frac {5}{-3-x+\log (48)}\right )+\left (-3-x+\log (48)+(-3-x+\log (48)) \log \left (-\frac {5}{-3-x+\log (48)}\right )\right ) \log \left (1+\log \left (-\frac {5}{-3-x+\log (48)}\right )\right )} \, dx=\text {Hanged} \]

[In]

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 - log(48) + 3))*(x - log(48) + 3) + 3) + log(5/(x - log(48) + 3))*(3*x - x*l
og(48) + x^2) + x^2),x)

[Out]

\text{Hanged}