\(\int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx\) [6275]

Optimal result

Integrand size = 39, antiderivative size = 17 \[ \int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx=1+e^5-\frac {x}{4+\log (x)+\log (\log (3))} \]

[Out]

exp(5)+1-x/(ln(x)+ln(ln(3))+4)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 5.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6820, 2334, 2336, 2209, 2408, 6617} \[ \int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx=-\frac {(\log (x \log (3))+3) \operatorname {ExpIntegralEi}(\log (x)+\log (\log (3))+4)}{e^4 \log (3)}+\frac {(\log (x \log (3))+4) \operatorname {ExpIntegralEi}(\log (x \log (3))+4)}{e^4 \log (3)}-\frac {\operatorname {ExpIntegralEi}(\log (x \log (3))+4)}{e^4 \log (3)}-x+\frac {x (\log (x \log (3))+3)}{\log (x)+4+\log (\log (3))} \]

[In]

Int[(-3 - Log[x] - Log[Log[3]])/(16 + 8*Log[x] + Log[x]^2 + (8 + 2*Log[x])*Log[Log[3]] + Log[Log[3]]^2),x]

[Out]

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2336

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2408

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 6617

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(a + b*x)*(ExpIntegralEi[a + b*x]/b), x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, x]

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 {-3-\log (x \log (3))}{\left (\log (x)+4 \left (1+\frac {1}{4} \log (\log (3))\right )\right )^2} \, dx \\ & = -\frac {\text {Ei}(4+\log (x)+\log (\log (3))) (3+\log (x \log (3)))}{e^4 \log (3)}+\frac {x (3+\log (x \log (3)))}{4+\log (x)+\log (\log (3))}+\int \left (\frac {\text {Ei}(4+\log (x \log (3)))}{e^4 x \log (3)}-\frac {1}{4+\log (x \log (3))}\right ) \, dx \\ & = -\frac {\text {Ei}(4+\log (x)+\log (\log (3))) (3+\log (x \log (3)))}{e^4 \log (3)}+\frac {x (3+\log (x \log (3)))}{4+\log (x)+\log (\log (3))}+\frac {\int \frac {\text {Ei}(4+\log (x \log (3)))}{x} \, dx}{e^4 \log (3)}-\int \frac {1}{4+\log (x \log (3))} \, dx \\ & = -\frac {\text {Ei}(4+\log (x)+\log (\log (3))) (3+\log (x \log (3)))}{e^4 \log (3)}+\frac {x (3+\log (x \log (3)))}{4+\log (x)+\log (\log (3))}-\frac {\text {Subst}\left (\int \frac {e^x}{4+x} \, dx,x,\log (x \log (3))\right )}{\log (3)}+\frac {\text {Subst}(\int \text {Ei}(4+x) \, dx,x,\log (x \log (3)))}{e^4 \log (3)} \\ & = -x-\frac {\text {Ei}(4+\log (x \log (3)))}{e^4 \log (3)}-\frac {\text {Ei}(4+\log (x)+\log (\log (3))) (3+\log (x \log (3)))}{e^4 \log (3)}+\frac {x (3+\log (x \log (3)))}{4+\log (x)+\log (\log (3))}+\frac {\text {Ei}(4+\log (x \log (3))) (4+\log (x \log (3)))}{e^4 \log (3)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx=-\frac {x}{4+\log (x \log (3))} \]

[In]

Integrate[(-3 - Log[x] - Log[Log[3]])/(16 + 8*Log[x] + Log[x]^2 + (8 + 2*Log[x])*Log[Log[3]] + Log[Log[3]]^2),
x]

[Out]

-(x/(4 + Log[x*Log[3]]))

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {x}{\ln \left (x \right )+\ln \left (\ln \left (3\right )\right )+4}\)	\(13\)
norman	\(-\frac {x}{\ln \left (x \right )+\ln \left (\ln \left (3\right )\right )+4}\)	\(13\)
risch	\(-\frac {x}{\ln \left (x \right )+\ln \left (\ln \left (3\right )\right )+4}\)	\(13\)
parallelrisch	\(-\frac {x}{\ln \left (x \right )+\ln \left (\ln \left (3\right )\right )+4}\)	\(13\)

[In]

int((-ln(ln(3))-3-ln(x))/(ln(ln(3))^2+(2*ln(x)+8)*ln(ln(3))+ln(x)^2+8*ln(x)+16),x,method=_RETURNVERBOSE)

[Out]

-x/(ln(x)+ln(ln(3))+4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx=-\frac {x}{\log \left (x\right ) + \log \left (\log \left (3\right )\right ) + 4} \]

[In]

integrate((-log(log(3))-3-log(x))/(log(log(3))^2+(2*log(x)+8)*log(log(3))+log(x)^2+8*log(x)+16),x, algorithm="
fricas")

[Out]

-x/(log(x) + log(log(3)) + 4)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx=- \frac {x}{\log {\left (x \right )} + \log {\left (\log {\left (3 \right )} \right )} + 4} \]

[In]

integrate((-ln(ln(3))-3-ln(x))/(ln(ln(3))**2+(2*ln(x)+8)*ln(ln(3))+ln(x)**2+8*ln(x)+16),x)

[Out]

-x/(log(x) + log(log(3)) + 4)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx=-\frac {x}{\log \left (x\right ) + \log \left (\log \left (3\right )\right ) + 4} \]

[In]

integrate((-log(log(3))-3-log(x))/(log(log(3))^2+(2*log(x)+8)*log(log(3))+log(x)^2+8*log(x)+16),x, algorithm="
maxima")

[Out]

-x/(log(x) + log(log(3)) + 4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx=-\frac {x}{\log \left (x\right ) + \log \left (\log \left (3\right )\right ) + 4} \]

[In]

integrate((-log(log(3))-3-log(x))/(log(log(3))^2+(2*log(x)+8)*log(log(3))+log(x)^2+8*log(x)+16),x, algorithm="
giac")

[Out]

-x/(log(x) + log(log(3)) + 4)

Mupad [B] (verification not implemented)

Time = 12.31 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {-3-\log (x)-\log (\log (3))}{16+8 \log (x)+\log ^2(x)+(8+2 \log (x)) \log (\log (3))+\log ^2(\log (3))} \, dx=-\frac {x}{\ln \left (\ln \left (3\right )\right )+\ln \left (x\right )+4} \]

[In]

int(-(log(log(3)) + log(x) + 3)/(8*log(x) + log(log(3))^2 + log(x)^2 + log(log(3))*(2*log(x) + 8) + 16),x)

[Out]

-x/(log(log(3)) + log(x) + 4)