\(\int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx\) [5539]

Optimal result

Integrand size = 55, antiderivative size = 19 \[ \int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx=\left (1-e^{-\frac {1250}{x \log (2 x)}}\right )^2 \]

[Out]

(1-exp(-1250/x/ln(2*x)))^2

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6873, 12, 6874, 6838} \[ \int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx=e^{-\frac {2500}{x \log (2 x)}}-2 e^{-\frac {1250}{x \log (2 x)}} \]

[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 6838

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 6873

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

Rule 6874

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.95 \[ \int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx=-2500 \left (-\frac {e^{-\frac {2500}{x \log (2 x)}}}{2500}+\frac {e^{-\frac {1250}{x \log (2 x)}}}{1250}\right ) \]

[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]))))

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 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	\(\frac {-2 x \ln \left (2\right ) {\mathrm e}^{-\frac {1250}{\left (\ln \left (2\right )+\ln \left (x \right )\right ) x}}-2 \ln \left (x \right ) x \,{\mathrm e}^{-\frac {1250}{\left (\ln \left (2\right )+\ln \left (x \right )\right ) x}}}{x \left (\ln \left (2\right )+\ln \left (x \right )\right )}+{\mathrm e}^{-\frac {2500}{\left (\ln \left (2\right )+\ln \left (x \right )\right ) x}}\)	\(63\)

[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))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx=-2 \, e^{\left (-\frac {1250}{x \log \left (2 \, x\right )}\right )} + e^{\left (-\frac {2500}{x \log \left (2 \, x\right )}\right )} \]

[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)))

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx=\text {Exception raised: TypeError} \]

[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]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx=\text {Exception raised: RuntimeError} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (16) = 32\).

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.79 \[ \int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx=-{\left (2 \, e^{\left (\frac {1250}{x \log \left (2\right ) + x \log \left (x\right )}\right )} - 1\right )} e^{\left (-\frac {2500}{x \log \left (2\right ) + x \log \left (x\right )}\right )} \]

[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)))

Mupad [B] (verification not implemented)

Time = 11.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {e^{-\frac {1250}{x \log (2 x)}} (-2500-2500 \log (2 x))+e^{-\frac {2500}{x \log (2 x)}} (2500+2500 \log (2 x))}{x^2 \log ^2(2 x)} \, dx={\mathrm {e}}^{-\frac {2500}{x\,\ln \left (2\right )+x\,\ln \left (x\right )}}-2\,{\mathrm {e}}^{-\frac {1250}{x\,\ln \left (2\right )+x\,\ln \left (x\right )}} \]

[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)))