\(\int \frac {2 e^{\frac {e^{39/16}+\log (3) \log (\frac {x^2}{9})}{\log (3)}}}{x} \, dx\) [1299]

Optimal result

Integrand size = 29, antiderivative size = 19 \[ \int \frac {2 e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx=\frac {1}{9} e^{\frac {e^{39/16}}{\log (3)}} x^2 \]

[Out]

exp(exp(39/16)/ln(3)+ln(1/9*x^2))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 1600, 30} \[ \int \frac {2 e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx=\frac {1}{9} x^2 e^{\frac {e^{39/16}}{\log (3)}} \]

[In]

Int[(2*E^((E^(39/16) + Log[3]*Log[x^2/9])/Log[3]))/x,x]

[Out]

(E^(E^(39/16)/Log[3])*x^2)/9

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 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx \\ & = 2 \int \frac {1}{9} e^{\frac {e^{39/16}}{\log (3)}} x \, dx \\ & = \frac {1}{9} \left (2 e^{\frac {e^{39/16}}{\log (3)}}\right ) \int x \, dx \\ & = \frac {1}{9} e^{\frac {e^{39/16}}{\log (3)}} x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx=\frac {1}{9} e^{\frac {e^{39/16}}{\log (3)}} x^2 \]

[In]

Integrate[(2*E^((E^(39/16) + Log[3]*Log[x^2/9])/Log[3]))/x,x]

[Out]

(E^(E^(39/16)/Log[3])*x^2)/9

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
norman	\(\frac {{\mathrm e}^{\frac {{\mathrm e}^{\frac {39}{16}}}{\ln \left (3\right )}} x^{2}}{9}\)	\(14\)
risch	\(\frac {{\mathrm e}^{\frac {{\mathrm e}^{\frac {39}{16}}}{\ln \left (3\right )}} x^{2}}{9}\)	\(14\)
gosper	\({\mathrm e}^{\frac {\ln \left (3\right ) \ln \left (\frac {x^{2}}{9}\right )+{\mathrm e}^{\frac {39}{16}}}{\ln \left (3\right )}}\)	\(19\)
derivativedivides	\({\mathrm e}^{\frac {\ln \left (3\right ) \ln \left (\frac {x^{2}}{9}\right )+{\mathrm e}^{\frac {39}{16}}}{\ln \left (3\right )}}\)	\(19\)
default	\({\mathrm e}^{\frac {\ln \left (3\right ) \ln \left (\frac {x^{2}}{9}\right )+{\mathrm e}^{\frac {39}{16}}}{\ln \left (3\right )}}\)	\(19\)
parallelrisch	\({\mathrm e}^{\frac {\ln \left (3\right ) \ln \left (\frac {x^{2}}{9}\right )+{\mathrm e}^{\frac {39}{16}}}{\ln \left (3\right )}}\)	\(19\)

[In]

int(2*exp((ln(3)*ln(1/9*x^2)+exp(39/16))/ln(3))/x,x,method=_RETURNVERBOSE)

[Out]

1/9*exp(exp(39/16)/ln(3))*x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {2 e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx=\frac {1}{9} \, x^{2} e^{\left (\frac {e^{\frac {39}{16}}}{\log \left (3\right )}\right )} \]

[In]

integrate(2*exp((log(3)*log(1/9*x^2)+exp(39/16))/log(3))/x,x, algorithm="fricas")

[Out]

1/9*x^2*e^(e^(39/16)/log(3))

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {2 e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx=\frac {x^{2} e^{\frac {e^{\frac {39}{16}}}{\log {\left (3 \right )}}}}{9} \]

[In]

integrate(2*exp((ln(3)*ln(1/9*x**2)+exp(39/16))/ln(3))/x,x)

[Out]

x**2*exp(exp(39/16)/log(3))/9

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {2 e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx=e^{\left (\frac {\log \left (3\right ) \log \left (\frac {1}{9} \, x^{2}\right ) + e^{\frac {39}{16}}}{\log \left (3\right )}\right )} \]

[In]

integrate(2*exp((log(3)*log(1/9*x^2)+exp(39/16))/log(3))/x,x, algorithm="maxima")

[Out]

e^((log(3)*log(1/9*x^2) + e^(39/16))/log(3))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {2 e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx=\frac {1}{9} \, x^{2} e^{\left (\frac {e^{\frac {39}{16}}}{\log \left (3\right )}\right )} \]

[In]

integrate(2*exp((log(3)*log(1/9*x^2)+exp(39/16))/log(3))/x,x, algorithm="giac")

[Out]

1/9*x^2*e^(e^(39/16)/log(3))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {2 e^{\frac {e^{39/16}+\log (3) \log \left (\frac {x^2}{9}\right )}{\log (3)}}}{x} \, dx=\frac {x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{39/16}}{\ln \left (3\right )}}}{9} \]

[In]

int((2*exp((exp(39/16) + log(3)*log(x^2/9))/log(3)))/x,x)

[Out]

(x^2*exp(exp(39/16)/log(3)))/9