3.2.46 \(\int -\frac {e^{\frac {5+(5-\log ^2(9)) \log (x)}{-5+\log ^2(9)}}}{x} \, dx\)

Optimal. Leaf size=16 \[ \frac {e^{\frac {5}{-5+\log ^2(9)}}}{x} \]

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Rubi [A]  time = 0.06, antiderivative size = 18, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2274, 7, 12, 30} \begin {gather*} \frac {e^{-\frac {5}{5-\log ^2(9)}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-(E^((5 + (5 - Log[9]^2)*Log[x])/(-5 + Log[9]^2))/x),x]

[Out]

1/(E^(5/(5 - Log[9]^2))*x)

Rule 7

Int[(u_.)*(Px_)^(p_), x_Symbol] :> Int[u*Px^Simplify[p], x] /; PolyQ[Px, x] &&  !RationalQ[p] && FreeQ[p, x] &
& RationalQ[Simplify[p]]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\int e^{\frac {5}{-5+\log ^2(9)}} x^{-1+\frac {5-\log ^2(9)}{-5+\log ^2(9)}} \, dx\\ &=-\int \frac {e^{\frac {5}{-5+\log ^2(9)}}}{x^2} \, dx\\ &=-\left (e^{-\frac {5}{5-\log ^2(9)}} \int \frac {1}{x^2} \, dx\right )\\ &=\frac {e^{-\frac {5}{5-\log ^2(9)}}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {5}{-5+\log ^2(9)}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-(E^((5 + (5 - Log[9]^2)*Log[x])/(-5 + Log[9]^2))/x),x]

[Out]

E^(5/(-5 + Log[9]^2))/x

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fricas [A]  time = 0.82, size = 17, normalized size = 1.06 \begin {gather*} \frac {e^{\left (\frac {5}{4 \, \log \relax (3)^{2} - 5}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(((-4*log(3)^2+5)*log(x)+5)/(4*log(3)^2-5))/x,x, algorithm="fricas")

[Out]

e^(5/(4*log(3)^2 - 5))/x

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giac [B]  time = 0.65, size = 46, normalized size = 2.88 \begin {gather*} e^{\left (-\frac {4 \, \log \relax (3)^{2} \log \relax (x)}{4 \, \log \relax (3)^{2} - 5} + \frac {5 \, \log \relax (x)}{4 \, \log \relax (3)^{2} - 5} + \frac {5}{4 \, \log \relax (3)^{2} - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(((-4*log(3)^2+5)*log(x)+5)/(4*log(3)^2-5))/x,x, algorithm="giac")

[Out]

e^(-4*log(3)^2*log(x)/(4*log(3)^2 - 5) + 5*log(x)/(4*log(3)^2 - 5) + 5/(4*log(3)^2 - 5))

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maple [A]  time = 0.04, size = 18, normalized size = 1.12

method	result	size
norman	\(\frac {{\mathrm e}^{\frac {5}{4 \ln \relax (3)^{2}-5}}}{x}\)	\(18\)
risch	\(\frac {{\mathrm e}^{\frac {5}{4 \ln \relax (3)^{2}-5}}}{x}\)	\(18\)
gosper	\({\mathrm e}^{-\frac {4 \ln \relax (3)^{2} \ln \relax (x )-5 \ln \relax (x )-5}{4 \ln \relax (3)^{2}-5}}\)	\(28\)
derivativedivides	\(-\frac {{\mathrm e}^{\frac {\left (-4 \ln \relax (3)^{2}+5\right ) \ln \relax (x )+5}{4 \ln \relax (3)^{2}-5}} \left (4 \ln \relax (3)^{2}-5\right )}{-4 \ln \relax (3)^{2}+5}\)	\(46\)
default	\(-\frac {{\mathrm e}^{\frac {\left (-4 \ln \relax (3)^{2}+5\right ) \ln \relax (x )+5}{4 \ln \relax (3)^{2}-5}} \left (4 \ln \relax (3)^{2}-5\right )}{-4 \ln \relax (3)^{2}+5}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(((-4*ln(3)^2+5)*ln(x)+5)/(4*ln(3)^2-5))/x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 26, normalized size = 1.62 \begin {gather*} e^{\left (-\frac {{\left (4 \, \log \relax (3)^{2} - 5\right )} \log \relax (x) - 5}{4 \, \log \relax (3)^{2} - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(((-4*log(3)^2+5)*log(x)+5)/(4*log(3)^2-5))/x,x, algorithm="maxima")

[Out]

e^(-((4*log(3)^2 - 5)*log(x) - 5)/(4*log(3)^2 - 5))

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mupad [B]  time = 0.25, size = 17, normalized size = 1.06 \begin {gather*} \frac {{\mathrm {e}}^{\frac {5}{4\,{\ln \relax (3)}^2-5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-(log(x)*(4*log(3)^2 - 5) - 5)/(4*log(3)^2 - 5))/x,x)

[Out]

exp(5/(4*log(3)^2 - 5))/x

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sympy [A]  time = 0.22, size = 15, normalized size = 0.94 \begin {gather*} \frac {1}{x e^{- \frac {5}{-5 + 4 \log {\relax (3 )}^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(((-4*ln(3)**2+5)*ln(x)+5)/(4*ln(3)**2-5))/x,x)

[Out]

exp(5/(-5 + 4*log(3)**2))/x

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