\(\int \frac {e^3 (-16-4 x)+(-4 e^3 x+35 \log (3)) \log (\frac {-4 e^3 x+35 \log (3)}{\log (3)}) \log (\log (\frac {-4 e^3 x+35 \log (3)}{\log (3)}))}{(e^3 (-16 x-4 x^2)+(140+35 x) \log (3)) \log (\frac {-4 e^3 x+35 \log (3)}{\log (3)}) \log (\log (\frac {-4 e^3 x+35 \log (3)}{\log (3)}))} \, dx\) [53]

Optimal result

Integrand size = 121, antiderivative size = 24 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log (4+x)+\log \left (\log \left (\log \left (25-2 \left (-5+\frac {2 e^3 x}{\log (3)}\right )\right )\right )\right ) \]

[Out]

ln(4+x)+ln(ln(ln(35-4*exp(3)*x/ln(3))))

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6820, 6816} \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log (x+4)+\log \left (\log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )\right ) \]

[In]

Int[(E^3*(-16 - 4*x) + (-4*E^3*x + 35*Log[3])*Log[(-4*E^3*x + 35*Log[3])/Log[3]]*Log[Log[(-4*E^3*x + 35*Log[3]
)/Log[3]]])/((E^3*(-16*x - 4*x^2) + (140 + 35*x)*Log[3])*Log[(-4*E^3*x + 35*Log[3])/Log[3]]*Log[Log[(-4*E^3*x
+ 35*Log[3])/Log[3]]]),x]

[Out]

Log[4 + x] + Log[Log[Log[35 - (4*E^3*x)/Log[3]]]]

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 \left (\frac {1}{4+x}+\frac {4 e^3}{\left (4 e^3 x-35 \log (3)\right ) \log \left (35-\frac {4 e^3 x}{\log (3)}\right ) \log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )}\right ) \, dx \\ & = \log (4+x)+\left (4 e^3\right ) \int \frac {1}{\left (4 e^3 x-35 \log (3)\right ) \log \left (35-\frac {4 e^3 x}{\log (3)}\right ) \log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )} \, dx \\ & = \log (4+x)+\log \left (\log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log (4+x)+\log \left (\log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )\right ) \]

[In]

Integrate[(E^3*(-16 - 4*x) + (-4*E^3*x + 35*Log[3])*Log[(-4*E^3*x + 35*Log[3])/Log[3]]*Log[Log[(-4*E^3*x + 35*
Log[3])/Log[3]]])/((E^3*(-16*x - 4*x^2) + (140 + 35*x)*Log[3])*Log[(-4*E^3*x + 35*Log[3])/Log[3]]*Log[Log[(-4*
E^3*x + 35*Log[3])/Log[3]]]),x]

[Out]

Log[4 + x] + Log[Log[Log[35 - (4*E^3*x)/Log[3]]]]

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00

method	result	size
norman	\(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\)	\(24\)
risch	\(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\)	\(24\)
parts	\(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\)	\(24\)
default	\(\ln \left (4+x \right )+\ln \left (\ln \left (-\ln \left (\ln \left (3\right )\right )+\ln \left (35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}\right )\right )\right )\)	\(25\)
parallelrisch	\(\frac {\left (16 \ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right ) {\mathrm e}^{6}+16 \,{\mathrm e}^{6} \ln \left (4+x \right )\right ) {\mathrm e}^{-6}}{16}\)	\(42\)

[In]

int(((35*ln(3)-4*x*exp(3))*ln((35*ln(3)-4*x*exp(3))/ln(3))*ln(ln((35*ln(3)-4*x*exp(3))/ln(3)))+(-16-4*x)*exp(3
))/((35*x+140)*ln(3)+(-4*x^2-16*x)*exp(3))/ln((35*ln(3)-4*x*exp(3))/ln(3))/ln(ln((35*ln(3)-4*x*exp(3))/ln(3)))
,x,method=_RETURNVERBOSE)

[Out]

ln(ln(ln((35*ln(3)-4*x*exp(3))/ln(3))))+ln(4+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log \left (x + 4\right ) + \log \left (\log \left (\log \left (-\frac {4 \, x e^{3} - 35 \, \log \left (3\right )}{\log \left (3\right )}\right )\right )\right ) \]

[In]

integrate(((35*log(3)-4*x*exp(3))*log((35*log(3)-4*x*exp(3))/log(3))*log(log((35*log(3)-4*x*exp(3))/log(3)))+(
-16-4*x)*exp(3))/((35*x+140)*log(3)+(-4*x^2-16*x)*exp(3))/log((35*log(3)-4*x*exp(3))/log(3))/log(log((35*log(3
)-4*x*exp(3))/log(3))),x, algorithm="fricas")

[Out]

log(x + 4) + log(log(log(-(4*x*e^3 - 35*log(3))/log(3))))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log {\left (x + 4 \right )} + \log {\left (\log {\left (\log {\left (\frac {- 4 x e^{3} + 35 \log {\left (3 \right )}}{\log {\left (3 \right )}} \right )} \right )} \right )} \]

[In]

integrate(((35*ln(3)-4*x*exp(3))*ln((35*ln(3)-4*x*exp(3))/ln(3))*ln(ln((35*ln(3)-4*x*exp(3))/ln(3)))+(-16-4*x)
*exp(3))/((35*x+140)*ln(3)+(-4*x**2-16*x)*exp(3))/ln((35*ln(3)-4*x*exp(3))/ln(3))/ln(ln((35*ln(3)-4*x*exp(3))/
ln(3))),x)

[Out]

log(x + 4) + log(log(log((-4*x*exp(3) + 35*log(3))/log(3))))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log \left (x + 4\right ) + \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right ) \]

[In]

integrate(((35*log(3)-4*x*exp(3))*log((35*log(3)-4*x*exp(3))/log(3))*log(log((35*log(3)-4*x*exp(3))/log(3)))+(
-16-4*x)*exp(3))/((35*x+140)*log(3)+(-4*x^2-16*x)*exp(3))/log((35*log(3)-4*x*exp(3))/log(3))/log(log((35*log(3
)-4*x*exp(3))/log(3))),x, algorithm="maxima")

[Out]

log(x + 4) + log(log(log(-4*x*e^3 + 35*log(3)) - log(log(3))))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (19) = 38\).

Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.38 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\frac {35 \, \log \left (3\right ) \log \left (4 \, x e^{3} - 35 \, \log \left (3\right )\right ) - 35 \, \log \left (3\right ) \log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) + 16 \, e^{3} \log \left (x + 4\right ) + 35 \, \log \left (3\right ) \log \left (x + 4\right ) + 16 \, e^{3} \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right ) + 35 \, \log \left (3\right ) \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right )}{16 \, e^{3} + 35 \, \log \left (3\right )} \]

[In]

integrate(((35*log(3)-4*x*exp(3))*log((35*log(3)-4*x*exp(3))/log(3))*log(log((35*log(3)-4*x*exp(3))/log(3)))+(
-16-4*x)*exp(3))/((35*x+140)*log(3)+(-4*x^2-16*x)*exp(3))/log((35*log(3)-4*x*exp(3))/log(3))/log(log((35*log(3
)-4*x*exp(3))/log(3))),x, algorithm="giac")

[Out]

(35*log(3)*log(4*x*e^3 - 35*log(3)) - 35*log(3)*log(-4*x*e^3 + 35*log(3)) + 16*e^3*log(x + 4) + 35*log(3)*log(
x + 4) + 16*e^3*log(log(log(-4*x*e^3 + 35*log(3)) - log(log(3)))) + 35*log(3)*log(log(log(-4*x*e^3 + 35*log(3)
) - log(log(3)))))/(16*e^3 + 35*log(3))

Mupad [B] (verification not implemented)

Time = 10.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\ln \left (x+4\right )+\ln \left (\ln \left (\ln \left (35\,\ln \left (3\right )-4\,x\,{\mathrm {e}}^3\right )-\ln \left (\ln \left (3\right )\right )\right )\right ) \]

[In]

int(-(exp(3)*(4*x + 16) - log(log((35*log(3) - 4*x*exp(3))/log(3)))*log((35*log(3) - 4*x*exp(3))/log(3))*(35*l
og(3) - 4*x*exp(3)))/(log(log((35*log(3) - 4*x*exp(3))/log(3)))*log((35*log(3) - 4*x*exp(3))/log(3))*(log(3)*(
35*x + 140) - exp(3)*(16*x + 4*x^2))),x)

[Out]

log(x + 4) + log(log(log(35*log(3) - 4*x*exp(3)) - log(log(3))))