\(\int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x))}{25 x-10 x \log (x)+x \log ^2(x)} \, dx\) [7761]

Optimal result

Integrand size = 76, antiderivative size = 21 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)} \]

[Out]

exp((-2-2*x)*exp(-exp(16)^2/(ln(x)-5)))

Rubi [F]

\[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=\int \frac {\exp \left (e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right ) \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx \]

[In]

Int[(E^((-2 - 2*x)/E^(E^32/(-5 + Log[x])) - E^32/(-5 + Log[x]))*(E^32*(-2 - 2*x) - 50*x + 20*x*Log[x] - 2*x*Lo
g[x]^2))/(25*x - 10*x*Log[x] + x*Log[x]^2),x]

[Out]

-2*Defer[Int][E^((-2 - 2*x)/E^(E^32/(-5 + Log[x])) - E^32/(-5 + Log[x])), x] - 2*Defer[Int][E^(32 + (-2 - 2*x)
/E^(E^32/(-5 + Log[x])) - E^32/(-5 + Log[x]))/(-5 + Log[x])^2, x] - 2*Defer[Int][E^(32 + (-2 - 2*x)/E^(E^32/(-
5 + Log[x])) - E^32/(-5 + Log[x]))/(x*(-5 + Log[x])^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \exp \left (e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right ) \left (-e^{32}-25 \left (1+\frac {e^{32}}{25}\right ) x+10 x \log (x)-x \log ^2(x)\right )}{x (5-\log (x))^2} \, dx \\ & = 2 \int \frac {\exp \left (e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right ) \left (-e^{32}-25 \left (1+\frac {e^{32}}{25}\right ) x+10 x \log (x)-x \log ^2(x)\right )}{x (5-\log (x))^2} \, dx \\ & = 2 \int \left (-\exp \left (e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right )-\frac {\exp \left (32+e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right ) (1+x)}{x (-5+\log (x))^2}\right ) \, dx \\ & = -\left (2 \int \exp \left (e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right ) \, dx\right )-2 \int \frac {\exp \left (32+e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right ) (1+x)}{x (-5+\log (x))^2} \, dx \\ & = -\left (2 \int \exp \left (e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right ) \, dx\right )-2 \int \left (\frac {\exp \left (32+e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right )}{(-5+\log (x))^2}+\frac {\exp \left (32+e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right )}{x (-5+\log (x))^2}\right ) \, dx \\ & = -\left (2 \int \exp \left (e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right ) \, dx\right )-2 \int \frac {\exp \left (32+e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right )}{(-5+\log (x))^2} \, dx-2 \int \frac {\exp \left (32+e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}\right )}{x (-5+\log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{-2 e^{-\frac {e^{32}}{-5+\log (x)}} (1+x)} \]

[In]

Integrate[(E^((-2 - 2*x)/E^(E^32/(-5 + Log[x])) - E^32/(-5 + Log[x]))*(E^32*(-2 - 2*x) - 50*x + 20*x*Log[x] -
2*x*Log[x]^2))/(25*x - 10*x*Log[x] + x*Log[x]^2),x]

[Out]

E^((-2*(1 + x))/E^(E^32/(-5 + Log[x])))

Maple [A] (verified)

Time = 12.39 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
risch	\({\mathrm e}^{-2 \left (1+x \right ) {\mathrm e}^{-\frac {{\mathrm e}^{32}}{\ln \left (x \right )-5}}}\)	\(18\)
parallelrisch	\({\mathrm e}^{\left (-2-2 x \right ) {\mathrm e}^{-\frac {{\mathrm e}^{32}}{\ln \left (x \right )-5}}}\)	\(21\)

[In]

int((-2*x*ln(x)^2+20*x*ln(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16)^2/(ln(x)-5))*exp((-2-2*x)*exp(-exp(16)^2/(l
n(x)-5)))/(x*ln(x)^2-10*x*ln(x)+25*x),x,method=_RETURNVERBOSE)

[Out]

exp(-2*(1+x)*exp(-exp(32)/(ln(x)-5)))

Fricas [F(-1)]

Timed out. \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=\text {Timed out} \]

[In]

integrate((-2*x*log(x)^2+20*x*log(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16)^2/(log(x)-5))*exp((-2-2*x)*exp(-exp
(16)^2/(log(x)-5)))/(x*log(x)^2-10*x*log(x)+25*x),x, algorithm="fricas")

[Out]

Timed out

Sympy [A] (verification not implemented)

Time = 1.72 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{\left (- 2 x - 2\right ) e^{- \frac {e^{32}}{\log {\left (x \right )} - 5}}} \]

[In]

integrate((-2*x*ln(x)**2+20*x*ln(x)+(-2-2*x)*exp(16)**2-50*x)*exp(-exp(16)**2/(ln(x)-5))*exp((-2-2*x)*exp(-exp
(16)**2/(ln(x)-5)))/(x*ln(x)**2-10*x*ln(x)+25*x),x)

[Out]

exp((-2*x - 2)*exp(-exp(32)/(log(x) - 5)))

Maxima [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=e^{\left (-2 \, x e^{\left (-\frac {e^{32}}{\log \left (x\right ) - 5}\right )} - 2 \, e^{\left (-\frac {e^{32}}{\log \left (x\right ) - 5}\right )}\right )} \]

[In]

integrate((-2*x*log(x)^2+20*x*log(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16)^2/(log(x)-5))*exp((-2-2*x)*exp(-exp
(16)^2/(log(x)-5)))/(x*log(x)^2-10*x*log(x)+25*x),x, algorithm="maxima")

[Out]

e^(-2*x*e^(-e^32/(log(x) - 5)) - 2*e^(-e^32/(log(x) - 5)))

Giac [F]

\[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx=\int { -\frac {2 \, {\left (x \log \left (x\right )^{2} + {\left (x + 1\right )} e^{32} - 10 \, x \log \left (x\right ) + 25 \, x\right )} e^{\left (-2 \, {\left (x + 1\right )} e^{\left (-\frac {e^{32}}{\log \left (x\right ) - 5}\right )} - \frac {e^{32}}{\log \left (x\right ) - 5}\right )}}{x \log \left (x\right )^{2} - 10 \, x \log \left (x\right ) + 25 \, x} \,d x } \]

[In]

integrate((-2*x*log(x)^2+20*x*log(x)+(-2-2*x)*exp(16)^2-50*x)*exp(-exp(16)^2/(log(x)-5))*exp((-2-2*x)*exp(-exp
(16)^2/(log(x)-5)))/(x*log(x)^2-10*x*log(x)+25*x),x, algorithm="giac")

[Out]

integrate(-2*(x*log(x)^2 + (x + 1)*e^32 - 10*x*log(x) + 25*x)*e^(-2*(x + 1)*e^(-e^32/(log(x) - 5)) - e^32/(log
(x) - 5))/(x*log(x)^2 - 10*x*log(x) + 25*x), x)

Mupad [B] (verification not implemented)

Time = 13.86 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e^{e^{-\frac {e^{32}}{-5+\log (x)}} (-2-2 x)-\frac {e^{32}}{-5+\log (x)}} \left (e^{32} (-2-2 x)-50 x+20 x \log (x)-2 x \log ^2(x)\right )}{25 x-10 x \log (x)+x \log ^2(x)} \, dx={\mathrm {e}}^{-2\,x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{32}}{\ln \left (x\right )-5}}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{32}}{\ln \left (x\right )-5}}} \]

[In]

int(-(exp(-exp(-exp(32)/(log(x) - 5))*(2*x + 2))*exp(-exp(32)/(log(x) - 5))*(50*x + 2*x*log(x)^2 - 20*x*log(x)
+ exp(32)*(2*x + 2)))/(25*x + x*log(x)^2 - 10*x*log(x)),x)

[Out]

exp(-2*x*exp(-exp(32)/(log(x) - 5)))*exp(-2*exp(-exp(32)/(log(x) - 5)))