\(\int \frac {e^{9 e^{\frac {5 x}{\log (e^{e^x}-\log (x))}}+\frac {5 x}{\log (e^{e^x}-\log (x))}} (45-45 e^{e^x+x} x+(45 e^{e^x}-45 \log (x)) \log (e^{e^x}-\log (x)))}{(e^{e^x}-\log (x)) \log ^2(e^{e^x}-\log (x))} \, dx\) [4062]

Optimal result

Integrand size = 101, antiderivative size = 22 \[ \int \frac {e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}} \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx=e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}} \]

[Out]

exp(9*exp(5*x/ln(exp(exp(x))-ln(x))))

Rubi [F]

\[ \int \frac {e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}} \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx=\int \frac {\exp \left (9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx \]

[In]

Int[(E^(9*E^((5*x)/Log[E^E^x - Log[x]]) + (5*x)/Log[E^E^x - Log[x]])*(45 - 45*E^(E^x + x)*x + (45*E^E^x - 45*L
og[x])*Log[E^E^x - Log[x]]))/((E^E^x - Log[x])*Log[E^E^x - Log[x]]^2),x]

[Out]

45*Defer[Int][E^(9*E^((5*x)/Log[E^E^x - Log[x]]) + (5*x)/Log[E^E^x - Log[x]])/((E^E^x - Log[x])*Log[E^E^x - Lo
g[x]]^2), x] - 45*Defer[Int][(E^(E^x + 9*E^((5*x)/Log[E^E^x - Log[x]]) + x + (5*x)/Log[E^E^x - Log[x]])*x)/((E
^E^x - Log[x])*Log[E^E^x - Log[x]]^2), x] + 45*Defer[Int][E^(9*E^((5*x)/Log[E^E^x - Log[x]]) + (5*x)/Log[E^E^x
- Log[x]])/Log[E^E^x - Log[x]], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {45 \exp \left (e^x+9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+x+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) x}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )}+\frac {45 \exp \left (9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) \left (1+e^{e^x} \log \left (e^{e^x}-\log (x)\right )-\log (x) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )}\right ) \, dx \\ & = -\left (45 \int \frac {\exp \left (e^x+9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+x+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) x}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx\right )+45 \int \frac {\exp \left (9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) \left (1+e^{e^x} \log \left (e^{e^x}-\log (x)\right )-\log (x) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx \\ & = -\left (45 \int \frac {\exp \left (e^x+9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+x+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) x}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx\right )+45 \int \frac {\exp \left (9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) \left (1+\left (e^{e^x}-\log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx \\ & = 45 \int \left (\frac {\exp \left (9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )}+\frac {\exp \left (9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right )}{\log \left (e^{e^x}-\log (x)\right )}\right ) \, dx-45 \int \frac {\exp \left (e^x+9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+x+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) x}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx \\ & = 45 \int \frac {\exp \left (9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx-45 \int \frac {\exp \left (e^x+9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+x+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right ) x}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx+45 \int \frac {\exp \left (9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}\right )}{\log \left (e^{e^x}-\log (x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}} \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx=e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}} \]

[In]

Integrate[(E^(9*E^((5*x)/Log[E^E^x - Log[x]]) + (5*x)/Log[E^E^x - Log[x]])*(45 - 45*E^(E^x + x)*x + (45*E^E^x
- 45*Log[x])*Log[E^E^x - Log[x]]))/((E^E^x - Log[x])*Log[E^E^x - Log[x]]^2),x]

[Out]

E^(9*E^((5*x)/Log[E^E^x - Log[x]]))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

\[{\mathrm e}^{9 \,{\mathrm e}^{\frac {5 x}{\ln \left ({\mathrm e}^{{\mathrm e}^{x}}-\ln \left (x \right )\right )}}}\]

[In]

int(((45*exp(exp(x))-45*ln(x))*ln(exp(exp(x))-ln(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/ln(exp(exp(x))-ln(x))
)*exp(9*exp(5*x/ln(exp(exp(x))-ln(x))))/(exp(exp(x))-ln(x))/ln(exp(exp(x))-ln(x))^2,x)

[Out]

exp(9*exp(5*x/ln(exp(exp(x))-ln(x))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (18) = 36\).

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 4.64 \[ \int \frac {e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}} \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx=e^{\left (\frac {9 \, e^{\left (\frac {5 \, x}{\log \left (-{\left (e^{x} \log \left (x\right ) - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right )}\right )} \log \left (-{\left (e^{x} \log \left (x\right ) - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right ) + 5 \, x}{\log \left (-{\left (e^{x} \log \left (x\right ) - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right )} - \frac {5 \, x}{\log \left (-{\left (e^{x} \log \left (x\right ) - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right )}\right )} \]

[In]

integrate(((45*exp(exp(x))-45*log(x))*log(exp(exp(x))-log(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/log(exp(exp(
x))-log(x)))*exp(9*exp(5*x/log(exp(exp(x))-log(x))))/(exp(exp(x))-log(x))/log(exp(exp(x))-log(x))^2,x, algorit
hm="fricas")

[Out]

e^((9*e^(5*x/log(-(e^x*log(x) - e^(x + e^x))*e^(-x)))*log(-(e^x*log(x) - e^(x + e^x))*e^(-x)) + 5*x)/log(-(e^x
*log(x) - e^(x + e^x))*e^(-x)) - 5*x/log(-(e^x*log(x) - e^(x + e^x))*e^(-x)))

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}} \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx=\text {Timed out} \]

[In]

integrate(((45*exp(exp(x))-45*ln(x))*ln(exp(exp(x))-ln(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/ln(exp(exp(x))-
ln(x)))*exp(9*exp(5*x/ln(exp(exp(x))-ln(x))))/(exp(exp(x))-ln(x))/ln(exp(exp(x))-ln(x))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}} \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx=e^{\left (9 \, e^{\left (\frac {5 \, x}{\log \left (e^{\left (e^{x}\right )} - \log \left (x\right )\right )}\right )}\right )} \]

[In]

integrate(((45*exp(exp(x))-45*log(x))*log(exp(exp(x))-log(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/log(exp(exp(
x))-log(x)))*exp(9*exp(5*x/log(exp(exp(x))-log(x))))/(exp(exp(x))-log(x))/log(exp(exp(x))-log(x))^2,x, algorit
hm="maxima")

[Out]

e^(9*e^(5*x/log(e^(e^x) - log(x))))

Giac [F]

\[ \int \frac {e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}} \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx=\int { -\frac {45 \, {\left (x e^{\left (x + e^{x}\right )} - {\left (e^{\left (e^{x}\right )} - \log \left (x\right )\right )} \log \left (e^{\left (e^{x}\right )} - \log \left (x\right )\right ) - 1\right )} e^{\left (\frac {5 \, x}{\log \left (e^{\left (e^{x}\right )} - \log \left (x\right )\right )} + 9 \, e^{\left (\frac {5 \, x}{\log \left (e^{\left (e^{x}\right )} - \log \left (x\right )\right )}\right )}\right )}}{{\left (e^{\left (e^{x}\right )} - \log \left (x\right )\right )} \log \left (e^{\left (e^{x}\right )} - \log \left (x\right )\right )^{2}} \,d x } \]

[In]

integrate(((45*exp(exp(x))-45*log(x))*log(exp(exp(x))-log(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/log(exp(exp(
x))-log(x)))*exp(9*exp(5*x/log(exp(exp(x))-log(x))))/(exp(exp(x))-log(x))/log(exp(exp(x))-log(x))^2,x, algorit
hm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 11.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {e^{9 e^{\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}}+\frac {5 x}{\log \left (e^{e^x}-\log (x)\right )}} \left (45-45 e^{e^x+x} x+\left (45 e^{e^x}-45 \log (x)\right ) \log \left (e^{e^x}-\log (x)\right )\right )}{\left (e^{e^x}-\log (x)\right ) \log ^2\left (e^{e^x}-\log (x)\right )} \, dx={\mathrm {e}}^{9\,{\mathrm {e}}^{\frac {5\,x}{\ln \left ({\mathrm {e}}^{{\mathrm {e}}^x}-\ln \left (x\right )\right )}}} \]

[In]

int((exp(9*exp((5*x)/log(exp(exp(x)) - log(x))))*exp((5*x)/log(exp(exp(x)) - log(x)))*(log(exp(exp(x)) - log(x
))*(45*exp(exp(x)) - 45*log(x)) - 45*x*exp(exp(x))*exp(x) + 45))/(log(exp(exp(x)) - log(x))^2*(exp(exp(x)) - l
og(x))),x)

[Out]

exp(9*exp((5*x)/log(exp(exp(x)) - log(x))))