\(\int \frac {e^{e^{\frac {e^x (-1+e^4+20 x)}{5 x}}+x+\frac {e^x (-1+e^4+20 x)}{5 x}} (1+e^4 (-1+x)-x+20 x^2)}{5 x^2} \, dx\) [10285]

Optimal result

Integrand size = 66, antiderivative size = 22 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{e^{e^x \left (4+\frac {-1+e^4}{5 x}\right )}} \]

[Out]

exp(exp(exp(x)*(1/5*(exp(4)-1)/x+4)))

Rubi [F]

\[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=\int \frac {\exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right ) \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx \]

[In]

Int[(E^(E^((E^x*(-1 + E^4 + 20*x))/(5*x)) + x + (E^x*(-1 + E^4 + 20*x))/(5*x))*(1 + E^4*(-1 + x) - x + 20*x^2)
)/(5*x^2),x]

[Out]

4*Defer[Int][E^(E^((E^x*(-1 + E^4 + 20*x))/(5*x)) + x + (E^x*(-1 + E^4 + 20*x))/(5*x)), x] + ((1 - E^4)*Defer[
Int][E^(E^((E^x*(-1 + E^4 + 20*x))/(5*x)) + x + (E^x*(-1 + E^4 + 20*x))/(5*x))/x^2, x])/5 - ((1 - E^4)*Defer[I
nt][E^(E^((E^x*(-1 + E^4 + 20*x))/(5*x)) + x + (E^x*(-1 + E^4 + 20*x))/(5*x))/x, x])/5

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {\exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right ) \left (1+e^4 (-1+x)-x+20 x^2\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \frac {\exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right ) \left (1-e^4-\left (1-e^4\right ) x+20 x^2\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \left (20 \exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right )+\frac {\exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right ) \left (1-e^4\right )}{x^2}+\frac {\exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right ) \left (-1+e^4\right )}{x}\right ) \, dx \\ & = 4 \int \exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right ) \, dx+\frac {1}{5} \left (1-e^4\right ) \int \frac {\exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right )}{x^2} \, dx+\frac {1}{5} \left (-1+e^4\right ) \int \frac {\exp \left (e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}\right )}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}} \]

[In]

Integrate[(E^(E^((E^x*(-1 + E^4 + 20*x))/(5*x)) + x + (E^x*(-1 + E^4 + 20*x))/(5*x))*(1 + E^4*(-1 + x) - x + 2
0*x^2))/(5*x^2),x]

[Out]

E^E^((E^x*(-1 + E^4 + 20*x))/(5*x))

Maple [A] (verified)

Time = 5.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
norman	\({\mathrm e}^{{\mathrm e}^{\frac {\left ({\mathrm e}^{4}+20 x -1\right ) {\mathrm e}^{x}}{5 x}}}\)	\(17\)
risch	\({\mathrm e}^{{\mathrm e}^{\frac {\left ({\mathrm e}^{4}+20 x -1\right ) {\mathrm e}^{x}}{5 x}}}\)	\(17\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{\frac {\left ({\mathrm e}^{4}+20 x -1\right ) {\mathrm e}^{x}}{5 x}}}\)	\(17\)

[In]

int(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*exp(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)*exp(x)
/x))/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(exp(1/5*(exp(4)+20*x-1)*exp(x)/x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (16) = 32\).

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.64 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{\left (-x - \frac {{\left (20 \, x + e^{4} - 1\right )} e^{x}}{5 \, x} + \frac {5 \, x^{2} + {\left (20 \, x + e^{4} - 1\right )} e^{x} + 5 \, x e^{\left (\frac {{\left (20 \, x + e^{4} - 1\right )} e^{x}}{5 \, x}\right )}}{5 \, x}\right )} \]

[In]

integrate(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*exp(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)*
exp(x)/x))/x^2,x, algorithm="fricas")

[Out]

e^(-x - 1/5*(20*x + e^4 - 1)*e^x/x + 1/5*(5*x^2 + (20*x + e^4 - 1)*e^x + 5*x*e^(1/5*(20*x + e^4 - 1)*e^x/x))/x
)

Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{e^{\frac {\left (4 x - \frac {1}{5} + \frac {e^{4}}{5}\right ) e^{x}}{x}}} \]

[In]

integrate(1/5*((-1+x)*exp(4)+20*x**2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*exp(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)
*exp(x)/x))/x**2,x)

[Out]

exp(exp((4*x - 1/5 + exp(4)/5)*exp(x)/x))

Maxima [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{\left (e^{\left (\frac {e^{\left (x + 4\right )}}{5 \, x} - \frac {e^{x}}{5 \, x} + 4 \, e^{x}\right )}\right )} \]

[In]

integrate(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*exp(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)*
exp(x)/x))/x^2,x, algorithm="maxima")

[Out]

e^(e^(1/5*e^(x + 4)/x - 1/5*e^x/x + 4*e^x))

Giac [F]

\[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=\int { \frac {{\left (20 \, x^{2} + {\left (x - 1\right )} e^{4} - x + 1\right )} e^{\left (x + \frac {{\left (20 \, x + e^{4} - 1\right )} e^{x}}{5 \, x} + e^{\left (\frac {{\left (20 \, x + e^{4} - 1\right )} e^{x}}{5 \, x}\right )}\right )}}{5 \, x^{2}} \,d x } \]

[In]

integrate(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*exp(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)*
exp(x)/x))/x^2,x, algorithm="giac")

[Out]

integrate(1/5*(20*x^2 + (x - 1)*e^4 - x + 1)*e^(x + 1/5*(20*x + e^4 - 1)*e^x/x + e^(1/5*(20*x + e^4 - 1)*e^x/x
))/x^2, x)

Mupad [B] (verification not implemented)

Time = 19.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx={\mathrm {e}}^{{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{5\,x}}\,{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^4\,{\mathrm {e}}^x}{5\,x}}} \]

[In]

int((exp((exp(x)*(20*x + exp(4) - 1))/(5*x))*exp(exp((exp(x)*(20*x + exp(4) - 1))/(5*x)))*exp(x)*(exp(4)*(x -
1) - x + 20*x^2 + 1))/(5*x^2),x)

[Out]

exp(exp(-exp(x)/(5*x))*exp(4*exp(x))*exp((exp(4)*exp(x))/(5*x)))