\(\int \frac {e^{\frac {1}{x+10 \log (25)+\log (\frac {\log (4)}{-16+x})}} (17-x)}{-16 x^2+x^3+(-320 x+20 x^2) \log (25)+(-1600+100 x) \log ^2(25)+(-32 x+2 x^2+(-320+20 x) \log (25)) \log (\frac {\log (4)}{-16+x})+(-16+x) \log ^2(\frac {\log (4)}{-16+x})} \, dx\) [258]

Optimal result

Integrand size = 100, antiderivative size = 19 \[ \int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (17-x)}{-16 x^2+x^3+\left (-320 x+20 x^2\right ) \log (25)+(-1600+100 x) \log ^2(25)+\left (-32 x+2 x^2+(-320+20 x) \log (25)\right ) \log \left (\frac {\log (4)}{-16+x}\right )+(-16+x) \log ^2\left (\frac {\log (4)}{-16+x}\right )} \, dx=e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6820, 6838} \[ \int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (17-x)}{-16 x^2+x^3+\left (-320 x+20 x^2\right ) \log (25)+(-1600+100 x) \log ^2(25)+\left (-32 x+2 x^2+(-320+20 x) \log (25)\right ) \log \left (\frac {\log (4)}{-16+x}\right )+(-16+x) \log ^2\left (\frac {\log (4)}{-16+x}\right )} \, dx=e^{\frac {1}{x+\log \left (-\frac {\log (4)}{16-x}\right )+10 \log (25)}} \]

[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 6820

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

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

Mathematica [A] (verified)

Time = 2.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (17-x)}{-16 x^2+x^3+\left (-320 x+20 x^2\right ) \log (25)+(-1600+100 x) \log ^2(25)+\left (-32 x+2 x^2+(-320+20 x) \log (25)\right ) \log \left (\frac {\log (4)}{-16+x}\right )+(-16+x) \log ^2\left (\frac {\log (4)}{-16+x}\right )} \, dx=e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} \]

[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)

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (17-x)}{-16 x^2+x^3+\left (-320 x+20 x^2\right ) \log (25)+(-1600+100 x) \log ^2(25)+\left (-32 x+2 x^2+(-320+20 x) \log (25)\right ) \log \left (\frac {\log (4)}{-16+x}\right )+(-16+x) \log ^2\left (\frac {\log (4)}{-16+x}\right )} \, dx=e^{\left (\frac {1}{x + 20 \, \log \left (5\right ) + \log \left (\frac {2 \, \log \left (2\right )}{x - 16}\right )}\right )} \]

[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))))

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (17-x)}{-16 x^2+x^3+\left (-320 x+20 x^2\right ) \log (25)+(-1600+100 x) \log ^2(25)+\left (-32 x+2 x^2+(-320+20 x) \log (25)\right ) \log \left (\frac {\log (4)}{-16+x}\right )+(-16+x) \log ^2\left (\frac {\log (4)}{-16+x}\right )} \, dx=e^{\frac {1}{x + \log {\left (\frac {2 \log {\left (2 \right )}}{x - 16} \right )} + 20 \log {\left (5 \right )}}} \]

[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)))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (19) = 38\).

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.89 \[ \int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (17-x)}{-16 x^2+x^3+\left (-320 x+20 x^2\right ) \log (25)+(-1600+100 x) \log ^2(25)+\left (-32 x+2 x^2+(-320+20 x) \log (25)\right ) \log \left (\frac {\log (4)}{-16+x}\right )+(-16+x) \log ^2\left (\frac {\log (4)}{-16+x}\right )} \, dx=\frac {x e^{\left (\frac {1}{x + 20 \, \log \left (5\right ) + \log \left (2\right ) - \log \left (x - 16\right ) + \log \left (\log \left (2\right )\right )}\right )}}{x - 17} - \frac {17 \, e^{\left (\frac {1}{x + 20 \, \log \left (5\right ) + \log \left (2\right ) - \log \left (x - 16\right ) + \log \left (\log \left (2\right )\right )}\right )}}{x - 17} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (17-x)}{-16 x^2+x^3+\left (-320 x+20 x^2\right ) \log (25)+(-1600+100 x) \log ^2(25)+\left (-32 x+2 x^2+(-320+20 x) \log (25)\right ) \log \left (\frac {\log (4)}{-16+x}\right )+(-16+x) \log ^2\left (\frac {\log (4)}{-16+x}\right )} \, dx=e^{\left (\frac {1}{x + 20 \, \log \left (5\right ) + \log \left (\frac {2 \, \log \left (2\right )}{x - 16}\right )}\right )} \]

[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))))

Mupad [B] (verification not implemented)

Time = 8.66 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {1}{x+10 \log (25)+\log \left (\frac {\log (4)}{-16+x}\right )}} (17-x)}{-16 x^2+x^3+\left (-320 x+20 x^2\right ) \log (25)+(-1600+100 x) \log ^2(25)+\left (-32 x+2 x^2+(-320+20 x) \log (25)\right ) \log \left (\frac {\log (4)}{-16+x}\right )+(-16+x) \log ^2\left (\frac {\log (4)}{-16+x}\right )} \, dx={\mathrm {e}}^{\frac {1}{x+\ln \left (\frac {190734863281250\,\ln \left (2\right )}{x-16}\right )}} \]

[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))))