3.42.26 \(\int -\frac {e^{\frac {1}{3+\log ^2(3+\log (x))}} \log (3+\log (x))}{54 x+18 x \log (x)+(36 x+12 x \log (x)) \log ^2(3+\log (x))+(6 x+2 x \log (x)) \log ^4(3+\log (x))} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{4} \left (7+e^{\frac {x}{3 x+x \log ^2(3+\log (x))}}\right ) \]

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Rubi [A]  time = 0.77, antiderivative size = 17, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 4, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 6742, 6725, 6706} \begin {gather*} \frac {1}{4} e^{\frac {1}{\log ^2(\log (x)+3)+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-((E^(3 + Log[3 + Log[x]]^2)^(-1)*Log[3 + Log[x]])/(54*x + 18*x*Log[x] + (36*x + 12*x*Log[x])*Log[3 + Log[
x]]^2 + (6*x + 2*x*Log[x])*Log[3 + Log[x]]^4)),x]

[Out]

E^(3 + Log[3 + Log[x]]^2)^(-1)/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\operatorname {Subst}\left (\int \frac {e^{\frac {1}{3+\log ^2(3+x)}} \log (3+x)}{2 (3+x) \left (3+\log ^2(3+x)\right )^2} \, dx,x,\log (x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^{\frac {1}{3+\log ^2(3+x)}} \log (3+x)}{(3+x) \left (3+\log ^2(3+x)\right )^2} \, dx,x,\log (x)\right )\right )\\ &=\frac {1}{4} e^{\frac {1}{3+\log ^2(3+\log (x))}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 17, normalized size = 0.68 \begin {gather*} \frac {1}{4} e^{\frac {1}{3+\log ^2(3+\log (x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-((E^(3 + Log[3 + Log[x]]^2)^(-1)*Log[3 + Log[x]])/(54*x + 18*x*Log[x] + (36*x + 12*x*Log[x])*Log[3
+ Log[x]]^2 + (6*x + 2*x*Log[x])*Log[3 + Log[x]]^4)),x]

[Out]

E^(3 + Log[3 + Log[x]]^2)^(-1)/4

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fricas [A]  time = 0.61, size = 14, normalized size = 0.56 \begin {gather*} \frac {1}{4} \, e^{\left (\frac {1}{\log \left (\log \relax (x) + 3\right )^{2} + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(3+log(x))*exp(1/(log(3+log(x))^2+3))/((2*x*log(x)+6*x)*log(3+log(x))^4+(12*x*log(x)+36*x)*log(3
+log(x))^2+18*x*log(x)+54*x),x, algorithm="fricas")

[Out]

1/4*e^(1/(log(log(x) + 3)^2 + 3))

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giac [A]  time = 0.22, size = 14, normalized size = 0.56 \begin {gather*} \frac {1}{4} \, e^{\left (\frac {1}{\log \left (\log \relax (x) + 3\right )^{2} + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(3+log(x))*exp(1/(log(3+log(x))^2+3))/((2*x*log(x)+6*x)*log(3+log(x))^4+(12*x*log(x)+36*x)*log(3
+log(x))^2+18*x*log(x)+54*x),x, algorithm="giac")

[Out]

1/4*e^(1/(log(log(x) + 3)^2 + 3))

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maple [A]  time = 0.02, size = 15, normalized size = 0.60

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-ln(3+ln(x))*exp(1/(ln(3+ln(x))^2+3))/((2*x*ln(x)+6*x)*ln(3+ln(x))^4+(12*x*ln(x)+36*x)*ln(3+ln(x))^2+18*x*
ln(x)+54*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.49, size = 14, normalized size = 0.56 \begin {gather*} \frac {1}{4} \, e^{\left (\frac {1}{\log \left (\log \relax (x) + 3\right )^{2} + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(3+log(x))*exp(1/(log(3+log(x))^2+3))/((2*x*log(x)+6*x)*log(3+log(x))^4+(12*x*log(x)+36*x)*log(3
+log(x))^2+18*x*log(x)+54*x),x, algorithm="maxima")

[Out]

1/4*e^(1/(log(log(x) + 3)^2 + 3))

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mupad [B]  time = 3.75, size = 14, normalized size = 0.56 \begin {gather*} \frac {{\mathrm {e}}^{\frac {1}{{\ln \left (\ln \relax (x)+3\right )}^2+3}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1/(log(log(x) + 3)^2 + 3))*log(log(x) + 3))/(54*x + log(log(x) + 3)^4*(6*x + 2*x*log(x)) + log(log(x
) + 3)^2*(36*x + 12*x*log(x)) + 18*x*log(x)),x)

[Out]

exp(1/(log(log(x) + 3)^2 + 3))/4

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sympy [A]  time = 0.69, size = 14, normalized size = 0.56 \begin {gather*} \frac {e^{\frac {1}{\log {\left (\log {\relax (x )} + 3 \right )}^{2} + 3}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-ln(3+ln(x))*exp(1/(ln(3+ln(x))**2+3))/((2*x*ln(x)+6*x)*ln(3+ln(x))**4+(12*x*ln(x)+36*x)*ln(3+ln(x))
**2+18*x*ln(x)+54*x),x)

[Out]

exp(1/(log(log(x) + 3)**2 + 3))/4

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