3.3.58 \(\int \frac {e^{\frac {1}{x+10 \log (25)+\log (\frac {\log (4)}{-16+x})}} (17-x)}{-16 x^2+x^3+(-320 x+20 x^2) \log (25)+(-1600+100 x) \log ^2(25)+(-32 x+2 x^2+(-320+20 x) \log (25)) \log (\frac {\log (4)}{-16+x})+(-16+x) \log ^2(\frac {\log (4)}{-16+x})} \, dx\)

Optimal. Leaf size=19 \[ e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} \]

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Rubi [A]  time = 0.56, antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 2, number of rules used = 2, integrand size = 100, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6688, 6706} \begin {gather*} e^{\frac {1}{x+\log \left (-\frac {\log (4)}{16-x}\right )+10 \log (25)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x + 10*Log[25] + Log[Log[4]/(-16 + x)])^(-1)*(17 - x))/(-16*x^2 + x^3 + (-320*x + 20*x^2)*Log[25] + (-
1600 + 100*x)*Log[25]^2 + (-32*x + 2*x^2 + (-320 + 20*x)*Log[25])*Log[Log[4]/(-16 + x)] + (-16 + x)*Log[Log[4]
/(-16 + x)]^2),x]

[Out]

E^(x + 10*Log[25] + Log[-(Log[4]/(16 - x))])^(-1)

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (-17+x)}{(16-x) \left (x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )\right )^2} \, dx\\ &=e^{\frac {1}{x+10 \log (25)+\log \left (-\frac {\log (4)}{16-x}\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.68, size = 19, normalized size = 1.00 \begin {gather*} e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x + 10*Log[25] + Log[Log[4]/(-16 + x)])^(-1)*(17 - x))/(-16*x^2 + x^3 + (-320*x + 20*x^2)*Log[25
] + (-1600 + 100*x)*Log[25]^2 + (-32*x + 2*x^2 + (-320 + 20*x)*Log[25])*Log[Log[4]/(-16 + x)] + (-16 + x)*Log[
Log[4]/(-16 + x)]^2),x]

[Out]

E^(x + 10*Log[25] + Log[Log[4]/(-16 + x)])^(-1)

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fricas [A]  time = 0.94, size = 19, normalized size = 1.00 \begin {gather*} e^{\left (\frac {1}{x + 20 \, \log \relax (5) + \log \left (\frac {2 \, \log \relax (2)}{x - 16}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x+17)*exp(1/(log(2*log(2)/(x-16))+20*log(5)+x))/((x-16)*log(2*log(2)/(x-16))^2+(2*(20*x-320)*log(5
)+2*x^2-32*x)*log(2*log(2)/(x-16))+4*(100*x-1600)*log(5)^2+2*(20*x^2-320*x)*log(5)+x^3-16*x^2),x, algorithm="f
ricas")

[Out]

e^(1/(x + 20*log(5) + log(2*log(2)/(x - 16))))

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giac [A]  time = 0.33, size = 19, normalized size = 1.00 \begin {gather*} e^{\left (\frac {1}{x + 20 \, \log \relax (5) + \log \left (\frac {2 \, \log \relax (2)}{x - 16}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x+17)*exp(1/(log(2*log(2)/(x-16))+20*log(5)+x))/((x-16)*log(2*log(2)/(x-16))^2+(2*(20*x-320)*log(5
)+2*x^2-32*x)*log(2*log(2)/(x-16))+4*(100*x-1600)*log(5)^2+2*(20*x^2-320*x)*log(5)+x^3-16*x^2),x, algorithm="g
iac")

[Out]

e^(1/(x + 20*log(5) + log(2*log(2)/(x - 16))))

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maple [A]  time = 0.26, size = 20, normalized size = 1.05

method	result	size
risch	\({\mathrm e}^{\frac {1}{\ln \left (\frac {2 \ln \relax (2)}{x -16}\right )+20 \ln \relax (5)+x}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x+17)*exp(1/(ln(2*ln(2)/(x-16))+20*ln(5)+x))/((x-16)*ln(2*ln(2)/(x-16))^2+(2*(20*x-320)*ln(5)+2*x^2-32*x
)*ln(2*ln(2)/(x-16))+4*(100*x-1600)*ln(5)^2+2*(20*x^2-320*x)*ln(5)+x^3-16*x^2),x,method=_RETURNVERBOSE)

[Out]

exp(1/(ln(2*ln(2)/(x-16))+20*ln(5)+x))

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maxima [B]  time = 0.61, size = 55, normalized size = 2.89 \begin {gather*} \frac {x e^{\left (\frac {1}{x + 20 \, \log \relax (5) + \log \relax (2) - \log \left (x - 16\right ) + \log \left (\log \relax (2)\right )}\right )}}{x - 17} - \frac {17 \, e^{\left (\frac {1}{x + 20 \, \log \relax (5) + \log \relax (2) - \log \left (x - 16\right ) + \log \left (\log \relax (2)\right )}\right )}}{x - 17} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x+17)*exp(1/(log(2*log(2)/(x-16))+20*log(5)+x))/((x-16)*log(2*log(2)/(x-16))^2+(2*(20*x-320)*log(5
)+2*x^2-32*x)*log(2*log(2)/(x-16))+4*(100*x-1600)*log(5)^2+2*(20*x^2-320*x)*log(5)+x^3-16*x^2),x, algorithm="m
axima")

[Out]

x*e^(1/(x + 20*log(5) + log(2) - log(x - 16) + log(log(2))))/(x - 17) - 17*e^(1/(x + 20*log(5) + log(2) - log(
x - 16) + log(log(2))))/(x - 17)

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mupad [B]  time = 0.85, size = 15, normalized size = 0.79 \begin {gather*} {\mathrm {e}}^{\frac {1}{x+\ln \left (\frac {190734863281250\,\ln \relax (2)}{x-16}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1/(x + 20*log(5) + log((2*log(2))/(x - 16))))*(x - 17))/(log((2*log(2))/(x - 16))^2*(x - 16) - 2*log
(5)*(320*x - 20*x^2) + log((2*log(2))/(x - 16))*(2*log(5)*(20*x - 320) - 32*x + 2*x^2) + 4*log(5)^2*(100*x - 1
600) - 16*x^2 + x^3),x)

[Out]

exp(1/(x + log((190734863281250*log(2))/(x - 16))))

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sympy [A]  time = 0.62, size = 19, normalized size = 1.00 \begin {gather*} e^{\frac {1}{x + \log {\left (\frac {2 \log {\relax (2 )}}{x - 16} \right )} + 20 \log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x+17)*exp(1/(ln(2*ln(2)/(x-16))+20*ln(5)+x))/((x-16)*ln(2*ln(2)/(x-16))**2+(2*(20*x-320)*ln(5)+2*x
**2-32*x)*ln(2*ln(2)/(x-16))+4*(100*x-1600)*ln(5)**2+2*(20*x**2-320*x)*ln(5)+x**3-16*x**2),x)

[Out]

exp(1/(x + log(2*log(2)/(x - 16)) + 20*log(5)))

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