3.56.39 \(\int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx\)

Optimal. Leaf size=19 \[ \left (1-e^{-\frac {1250}{x \log (2 x)}}\right )^2 \]

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Rubi [A]  time = 1.63, antiderivative size = 29, normalized size of antiderivative = 1.53, number of steps used = 6, number of rules used = 4, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6741, 12, 6742, 6706} \begin {gather*} e^{-\frac {2500}{x \log (2 x)}}-2 e^{-\frac {1250}{x \log (2 x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-2500 - 2500*Log[2*x])/E^(1250/(x*Log[2*x])) + (2500 + 2500*Log[2*x])/E^(2500/(x*Log[2*x])))/(x^2*Log[2*
x]^2),x]

[Out]

E^(-2500/(x*Log[2*x])) - 2/E^(1250/(x*Log[2*x]))

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 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2500 e^{-\frac {2500}{x \log (2 x)}} \left (1-e^{\frac {1250}{x \log (2 x)}}\right ) (1+\log (2 x))}{x^2 \log ^2(2 x)} \, dx\\ &=2500 \int \frac {e^{-\frac {2500}{x \log (2 x)}} \left (1-e^{\frac {1250}{x \log (2 x)}}\right ) (1+\log (2 x))}{x^2 \log ^2(2 x)} \, dx\\ &=2500 \int \left (\frac {e^{-\frac {2500}{x \log (2 x)}} (1+\log (2 x))}{x^2 \log ^2(2 x)}-\frac {e^{-\frac {1250}{x \log (2 x)}} (1+\log (2 x))}{x^2 \log ^2(2 x)}\right ) \, dx\\ &=2500 \int \frac {e^{-\frac {2500}{x \log (2 x)}} (1+\log (2 x))}{x^2 \log ^2(2 x)} \, dx-2500 \int \frac {e^{-\frac {1250}{x \log (2 x)}} (1+\log (2 x))}{x^2 \log ^2(2 x)} \, dx\\ &=e^{-\frac {2500}{x \log (2 x)}}-2 e^{-\frac {1250}{x \log (2 x)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 37, normalized size = 1.95 \begin {gather*} -2500 \left (-\frac {e^{-\frac {2500}{x \log (2 x)}}}{2500}+\frac {e^{-\frac {1250}{x \log (2 x)}}}{1250}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2500 - 2500*Log[2*x])/E^(1250/(x*Log[2*x])) + (2500 + 2500*Log[2*x])/E^(2500/(x*Log[2*x])))/(x^2*
Log[2*x]^2),x]

[Out]

-2500*(-1/2500*1/E^(2500/(x*Log[2*x])) + 1/(1250*E^(1250/(x*Log[2*x]))))

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fricas [A]  time = 0.73, size = 27, normalized size = 1.42 \begin {gather*} -2 \, e^{\left (-\frac {1250}{x \log \left (2 \, x\right )}\right )} + e^{\left (-\frac {2500}{x \log \left (2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2500*log(2*x)+2500)*exp(-1250/x/log(2*x))^2+(-2500*log(2*x)-2500)*exp(-1250/x/log(2*x)))/x^2/log(2
*x)^2,x, algorithm="fricas")

[Out]

-2*e^(-1250/(x*log(2*x))) + e^(-2500/(x*log(2*x)))

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giac [B]  time = 0.16, size = 34, normalized size = 1.79 \begin {gather*} -{\left (2 \, e^{\left (\frac {1250}{x \log \relax (2) + x \log \relax (x)}\right )} - 1\right )} e^{\left (-\frac {2500}{x \log \relax (2) + x \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2500*log(2*x)+2500)*exp(-1250/x/log(2*x))^2+(-2500*log(2*x)-2500)*exp(-1250/x/log(2*x)))/x^2/log(2
*x)^2,x, algorithm="giac")

[Out]

-(2*e^(1250/(x*log(2) + x*log(x))) - 1)*e^(-2500/(x*log(2) + x*log(x)))

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maple [A]  time = 0.42, size = 28, normalized size = 1.47

method	result	size
risch	\({\mathrm e}^{-\frac {2500}{x \ln \left (2 x \right )}}-2 \,{\mathrm e}^{-\frac {1250}{x \ln \left (2 x \right )}}\)	\(28\)
default	\(\frac {-2 x \ln \relax (x ) {\mathrm e}^{-\frac {1250}{x \left (\ln \relax (2)+\ln \relax (x )\right )}}-2 \ln \relax (2) x \,{\mathrm e}^{-\frac {1250}{x \left (\ln \relax (2)+\ln \relax (x )\right )}}}{x \left (\ln \relax (2)+\ln \relax (x )\right )}+\frac {x \ln \relax (x ) {\mathrm e}^{-\frac {2500}{x \left (\ln \relax (2)+\ln \relax (x )\right )}}+\ln \relax (2) x \,{\mathrm e}^{-\frac {2500}{x \left (\ln \relax (2)+\ln \relax (x )\right )}}}{x \left (\ln \relax (2)+\ln \relax (x )\right )}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2500*ln(2*x)+2500)*exp(-1250/x/ln(2*x))^2+(-2500*ln(2*x)-2500)*exp(-1250/x/ln(2*x)))/x^2/ln(2*x)^2,x,met
hod=_RETURNVERBOSE)

[Out]

exp(-2500/x/ln(2*x))-2*exp(-1250/x/ln(2*x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2500*log(2*x)+2500)*exp(-1250/x/log(2*x))^2+(-2500*log(2*x)-2500)*exp(-1250/x/log(2*x)))/x^2/log(2
*x)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 3.61, size = 31, normalized size = 1.63 \begin {gather*} {\mathrm {e}}^{-\frac {2500}{x\,\ln \relax (2)+x\,\ln \relax (x)}}-2\,{\mathrm {e}}^{-\frac {1250}{x\,\ln \relax (2)+x\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1250/(x*log(2*x)))*(2500*log(2*x) + 2500) - exp(-2500/(x*log(2*x)))*(2500*log(2*x) + 2500))/(x^2*lo
g(2*x)^2),x)

[Out]

exp(-2500/(x*log(2) + x*log(x))) - 2*exp(-1250/(x*log(2) + x*log(x)))

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sympy [A]  time = 0.35, size = 22, normalized size = 1.16 \begin {gather*} - 2 e^{- \frac {1250}{x \log {\left (2 x \right )}}} + e^{- \frac {2500}{x \log {\left (2 x \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2500*ln(2*x)+2500)*exp(-1250/x/ln(2*x))**2+(-2500*ln(2*x)-2500)*exp(-1250/x/ln(2*x)))/x**2/ln(2*x)
**2,x)

[Out]

-2*exp(-1250/(x*log(2*x))) + exp(-2500/(x*log(2*x)))

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