\(\int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x))}{(4 x-x^2) \log ^2(x)+(-60 x+15 x^2) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} (-x \log ^2(x)+15 x \log ^4(x))} \, dx\) [6040]

Optimal result

Integrand size = 114, antiderivative size = 34 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=\log \left (\frac {x}{\left (4-e^{2-\frac {3}{\log (x)}}-x\right ) \left (3-\frac {1}{5 \log ^2(x)}\right )}\right ) \]

[Out]

ln(x/(3-1/5/ln(x)^2)/(4-exp(2-3/ln(x))-x))

Rubi [A] (verified)

Time = 3.38 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6873, 6874, 6816, 1816, 266} \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=-\log \left (1-15 \log ^2(x)\right )+\log (x)-\log \left (x e^{\frac {3}{\log (x)}}-4 e^{\frac {3}{\log (x)}}+e^2\right )+2 \log (\log (x))+\frac {3}{\log (x)} \]

[In]

Int[((8 - 2*x)*Log[x] + 4*Log[x]^2 - 60*Log[x]^4 + E^((-3 + 2*Log[x])/Log[x])*(3 - 2*Log[x] - 46*Log[x]^2 + 15
*Log[x]^4))/((4*x - x^2)*Log[x]^2 + (-60*x + 15*x^2)*Log[x]^4 + E^((-3 + 2*Log[x])/Log[x])*(-(x*Log[x]^2) + 15
*x*Log[x]^4)),x]

[Out]

3/Log[x] + Log[x] - Log[E^2 - 4*E^(3/Log[x]) + E^(3/Log[x])*x] + 2*Log[Log[x]] - Log[1 - 15*Log[x]^2]

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 6816

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

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {3}{\log (x)}} \left (-((8-2 x) \log (x))-4 \log ^2(x)+60 \log ^4(x)-e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )\right )}{x \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right ) \log ^2(x) \left (1-15 \log ^2(x)\right )} \, dx \\ & = \int \left (-\frac {e^{\frac {3}{\log (x)}} \left (12-3 x+x \log ^2(x)\right )}{x \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right ) \log ^2(x)}+\frac {3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)}{x \log ^2(x) \left (-1+15 \log ^2(x)\right )}\right ) \, dx \\ & = -\int \frac {e^{\frac {3}{\log (x)}} \left (12-3 x+x \log ^2(x)\right )}{x \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right ) \log ^2(x)} \, dx+\int \frac {3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)}{x \log ^2(x) \left (-1+15 \log ^2(x)\right )} \, dx \\ & = -\log \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right )+\text {Subst}\left (\int \frac {3-2 x-46 x^2+15 x^4}{x^2 \left (-1+15 x^2\right )} \, dx,x,\log (x)\right ) \\ & = -\log \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right )+\text {Subst}\left (\int \left (1-\frac {3}{x^2}+\frac {2}{x}-\frac {30 x}{-1+15 x^2}\right ) \, dx,x,\log (x)\right ) \\ & = \frac {3}{\log (x)}+\log (x)-\log \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right )+2 \log (\log (x))-30 \text {Subst}\left (\int \frac {x}{-1+15 x^2} \, dx,x,\log (x)\right ) \\ & = \frac {3}{\log (x)}+\log (x)-\log \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right )+2 \log (\log (x))-\log \left (1-15 \log ^2(x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=\frac {3}{\log (x)}+\log (x)-\log \left (e^2-4 e^{\frac {3}{\log (x)}}+e^{\frac {3}{\log (x)}} x\right )+2 \log (\log (x))-\log \left (1-15 \log ^2(x)\right ) \]

[In]

Integrate[((8 - 2*x)*Log[x] + 4*Log[x]^2 - 60*Log[x]^4 + E^((-3 + 2*Log[x])/Log[x])*(3 - 2*Log[x] - 46*Log[x]^
2 + 15*Log[x]^4))/((4*x - x^2)*Log[x]^2 + (-60*x + 15*x^2)*Log[x]^4 + E^((-3 + 2*Log[x])/Log[x])*(-(x*Log[x]^2
) + 15*x*Log[x]^4)),x]

[Out]

3/Log[x] + Log[x] - Log[E^2 - 4*E^(3/Log[x]) + E^(3/Log[x])*x] + 2*Log[Log[x]] - Log[1 - 15*Log[x]^2]

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(2 \ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (x \right )^{2}-\frac {1}{15}\right )-\ln \left (x +{\mathrm e}^{\frac {2 \ln \left (x \right )-3}{\ln \left (x \right )}}-4\right )+\ln \left (x \right )\)	\(36\)
risch	\(\ln \left (x \right )+\frac {3}{\ln \left (x \right )}-\ln \left (\ln \left (x \right )^{2}-\frac {1}{15}\right )+2 \ln \left (\ln \left (x \right )\right )+\frac {2 \ln \left (x \right )-3}{\ln \left (x \right )}-\ln \left (x +{\mathrm e}^{\frac {2 \ln \left (x \right )-3}{\ln \left (x \right )}}-4\right )\)	\(53\)

[In]

int(((15*ln(x)^4-46*ln(x)^2-2*ln(x)+3)*exp((2*ln(x)-3)/ln(x))-60*ln(x)^4+4*ln(x)^2+(-2*x+8)*ln(x))/((15*x*ln(x
)^4-x*ln(x)^2)*exp((2*ln(x)-3)/ln(x))+(15*x^2-60*x)*ln(x)^4+(-x^2+4*x)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

2*ln(ln(x))-ln(ln(x)^2-1/15)-ln(x+exp((2*ln(x)-3)/ln(x))-4)+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=-\log \left (15 \, \log \left (x\right )^{2} - 1\right ) - \log \left (x + e^{\left (\frac {2 \, \log \left (x\right ) - 3}{\log \left (x\right )}\right )} - 4\right ) + \log \left (x\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((15*log(x)^4-46*log(x)^2-2*log(x)+3)*exp((2*log(x)-3)/log(x))-60*log(x)^4+4*log(x)^2+(-2*x+8)*log(x
))/((15*x*log(x)^4-x*log(x)^2)*exp((2*log(x)-3)/log(x))+(15*x^2-60*x)*log(x)^4+(-x^2+4*x)*log(x)^2),x, algorit
hm="fricas")

[Out]

-log(15*log(x)^2 - 1) - log(x + e^((2*log(x) - 3)/log(x)) - 4) + log(x) + 2*log(log(x))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=\log {\left (x \right )} - \log {\left (\log {\left (x \right )}^{2} - \frac {1}{15} \right )} - \log {\left (x + e^{\frac {2 \log {\left (x \right )} - 3}{\log {\left (x \right )}}} - 4 \right )} + 2 \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((15*ln(x)**4-46*ln(x)**2-2*ln(x)+3)*exp((2*ln(x)-3)/ln(x))-60*ln(x)**4+4*ln(x)**2+(-2*x+8)*ln(x))/(
(15*x*ln(x)**4-x*ln(x)**2)*exp((2*ln(x)-3)/ln(x))+(15*x**2-60*x)*ln(x)**4+(-x**2+4*x)*ln(x)**2),x)

[Out]

log(x) - log(log(x)**2 - 1/15) - log(x + exp((2*log(x) - 3)/log(x)) - 4) + 2*log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=\frac {3}{\log \left (x\right )} - \log \left (\log \left (x\right )^{2} - \frac {1}{15}\right ) - \log \left (x - 4\right ) + \log \left (x\right ) - \log \left (\frac {{\left (x - 4\right )} e^{\frac {3}{\log \left (x\right )}} + e^{2}}{x - 4}\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((15*log(x)^4-46*log(x)^2-2*log(x)+3)*exp((2*log(x)-3)/log(x))-60*log(x)^4+4*log(x)^2+(-2*x+8)*log(x
))/((15*x*log(x)^4-x*log(x)^2)*exp((2*log(x)-3)/log(x))+(15*x^2-60*x)*log(x)^4+(-x^2+4*x)*log(x)^2),x, algorit
hm="maxima")

[Out]

3/log(x) - log(log(x)^2 - 1/15) - log(x - 4) + log(x) - log(((x - 4)*e^(3/log(x)) + e^2)/(x - 4)) + 2*log(log(
x))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=-\log \left (15 \, \log \left (x\right )^{2} - 1\right ) - \log \left (x + e^{\left (\frac {2 \, \log \left (x\right ) - 3}{\log \left (x\right )}\right )} - 4\right ) + \log \left (x\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((15*log(x)^4-46*log(x)^2-2*log(x)+3)*exp((2*log(x)-3)/log(x))-60*log(x)^4+4*log(x)^2+(-2*x+8)*log(x
))/((15*x*log(x)^4-x*log(x)^2)*exp((2*log(x)-3)/log(x))+(15*x^2-60*x)*log(x)^4+(-x^2+4*x)*log(x)^2),x, algorit
hm="giac")

[Out]

-log(15*log(x)^2 - 1) - log(x + e^((2*log(x) - 3)/log(x)) - 4) + log(x) + 2*log(log(x))

Mupad [B] (verification not implemented)

Time = 10.88 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38 \[ \int \frac {(8-2 x) \log (x)+4 \log ^2(x)-60 \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (3-2 \log (x)-46 \log ^2(x)+15 \log ^4(x)\right )}{\left (4 x-x^2\right ) \log ^2(x)+\left (-60 x+15 x^2\right ) \log ^4(x)+e^{\frac {-3+2 \log (x)}{\log (x)}} \left (-x \log ^2(x)+15 x \log ^4(x)\right )} \, dx=2\,\ln \left (\frac {1}{x^2}\right )-\ln \left (x+{\mathrm {e}}^{-\frac {3}{\ln \left (x\right )}}\,{\mathrm {e}}^2-4\right )+2\,\ln \left (\ln \left (x\right )\right )-\ln \left (\frac {15\,{\ln \left (x\right )}^2-1}{x}\right )+4\,\ln \left (x\right ) \]

[In]

int((60*log(x)^4 - 4*log(x)^2 + log(x)*(2*x - 8) + exp((2*log(x) - 3)/log(x))*(2*log(x) + 46*log(x)^2 - 15*log
(x)^4 - 3))/(log(x)^4*(60*x - 15*x^2) - log(x)^2*(4*x - x^2) + exp((2*log(x) - 3)/log(x))*(x*log(x)^2 - 15*x*l
og(x)^4)),x)

[Out]

2*log(1/x^2) - log(x + exp(-3/log(x))*exp(2) - 4) + 2*log(log(x)) - log((15*log(x)^2 - 1)/x) + 4*log(x)