3.93.55 \(\int -\frac {e^{5-\frac {1}{125} e^{1+2 \log (5) \log (x)}+2 \log (5) \log (x)} \log (5)}{500 x} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{8} e^{4-e^{1+\log (5) (-3+2 \log (x))}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2274, 6706} \begin {gather*} \frac {\log (5) e^{4-e 5^{2 \log (x)-3}}}{4 \log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1/500*(E^(5 - E^(1 + 2*Log[5]*Log[x])/125 + 2*Log[5]*Log[x])*Log[5])/x,x]

[Out]

(E^(4 - 5^(-3 + 2*Log[x])*E)*Log[5])/(4*Log[25])

Rule 12

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

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]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\frac {1}{500} \log (5) \int \frac {\exp \left (5-\frac {1}{125} e^{1+2 \log (5) \log (x)}+2 \log (5) \log (x)\right )}{x} \, dx\right )\\ &=-\left (\frac {1}{500} \log (5) \int e^{5-\frac {1}{125} e^{1+2 \log (5) \log (x)}} x^{-1+2 \log (5)} \, dx\right )\\ &=\frac {e^{4-5^{-3+2 \log (x)} e} \log (5)}{4 \log (25)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 19, normalized size = 0.83 \begin {gather*} \frac {1}{8} e^{4-5^{-3+2 \log (x)} e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1/500*(E^(5 - E^(1 + 2*Log[5]*Log[x])/125 + 2*Log[5]*Log[x])*Log[5])/x,x]

[Out]

E^(4 - 5^(-3 + 2*Log[x])*E)/8

________________________________________________________________________________________

fricas [A]  time = 1.26, size = 22, normalized size = 0.96 \begin {gather*} e^{\left (-e^{\left (2 \, \log \relax (5) \log \relax (x) - 3 \, \log \relax (5) + 1\right )} - 3 \, \log \relax (2) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*log(5)*exp(2*log(5)*log(x)-3*log(5)+1)*exp(-exp(2*log(5)*log(x)-3*log(5)+1)-3*log(2)+4)/x,x, algo
rithm="fricas")

[Out]

e^(-e^(2*log(5)*log(x) - 3*log(5) + 1) - 3*log(2) + 4)

________________________________________________________________________________________

giac [A]  time = 0.22, size = 24, normalized size = 1.04 \begin {gather*} \frac {1}{1000} \, e^{\left (-e^{\left (2 \, \log \relax (5) \log \relax (x) - 3 \, \log \relax (5) + 1\right )} + 3 \, \log \relax (5) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*log(5)*exp(2*log(5)*log(x)-3*log(5)+1)*exp(-exp(2*log(5)*log(x)-3*log(5)+1)-3*log(2)+4)/x,x, algo
rithm="giac")

[Out]

1/1000*e^(-e^(2*log(5)*log(x) - 3*log(5) + 1) + 3*log(5) + 4)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 16, normalized size = 0.70

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {x^{2 \ln \relax (5)} {\mathrm e}}{125}+4}}{8}\)	\(16\)
derivativedivides	\({\mathrm e}^{-{\mathrm e}^{2 \ln \relax (5) \ln \relax (x )-3 \ln \relax (5)+1}-3 \ln \relax (2)+4}\)	\(23\)
default	\({\mathrm e}^{-{\mathrm e}^{2 \ln \relax (5) \ln \relax (x )-3 \ln \relax (5)+1}-3 \ln \relax (2)+4}\)	\(23\)
norman	\({\mathrm e}^{-{\mathrm e}^{2 \ln \relax (5) \ln \relax (x )-3 \ln \relax (5)+1}-3 \ln \relax (2)+4}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*ln(5)*exp(2*ln(5)*ln(x)-3*ln(5)+1)*exp(-exp(2*ln(5)*ln(x)-3*ln(5)+1)-3*ln(2)+4)/x,x,method=_RETURNVERBO
SE)

[Out]

1/8*exp(-1/125*x^(2*ln(5))*exp(1)+4)

________________________________________________________________________________________

maxima [A]  time = 0.44, size = 16, normalized size = 0.70 \begin {gather*} \frac {1}{8} \, e^{\left (-\frac {1}{125} \, e^{\left (2 \, \log \relax (5) \log \relax (x) + 1\right )} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*log(5)*exp(2*log(5)*log(x)-3*log(5)+1)*exp(-exp(2*log(5)*log(x)-3*log(5)+1)-3*log(2)+4)/x,x, algo
rithm="maxima")

[Out]

1/8*e^(-1/125*e^(2*log(5)*log(x) + 1) + 4)

________________________________________________________________________________________

mupad [B]  time = 7.30, size = 15, normalized size = 0.65 \begin {gather*} \frac {{\mathrm {e}}^4\,{\mathrm {e}}^{-\frac {x^{2\,\ln \relax (5)}\,\mathrm {e}}{125}}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(4 - 3*log(2) - exp(2*log(5)*log(x) - 3*log(5) + 1))*exp(2*log(5)*log(x) - 3*log(5) + 1)*log(5))/x,
x)

[Out]

(exp(4)*exp(-(x^(2*log(5))*exp(1))/125))/8

________________________________________________________________________________________

sympy [A]  time = 6.60, size = 20, normalized size = 0.87 \begin {gather*} \frac {e^{4} e^{- \frac {e e^{2 \log {\relax (5 )} \log {\relax (x )}}}{125}}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*ln(5)*exp(2*ln(5)*ln(x)-3*ln(5)+1)*exp(-exp(2*ln(5)*ln(x)-3*ln(5)+1)-3*ln(2)+4)/x,x)

[Out]

exp(4)*exp(-E*exp(2*log(5)*log(x))/125)/8

________________________________________________________________________________________