\(\int \frac {e^{e^{x+e^{-x} (1+x)} x} (1+x+e^{x+e^{-x} (1+x)} (x+2 x^2+x^3-e^{-x} x^3 (1+x)))}{1+x} \, dx\) [719]

Optimal result

Integrand size = 63, antiderivative size = 19 \[ \int \frac {e^{e^{x+e^{-x} (1+x)} x} \left (1+x+e^{x+e^{-x} (1+x)} \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )\right )}{1+x} \, dx=e^{e^{x+e^{-x} (1+x)} x} x \]

[Out]

exp(x*exp(exp(ln(1+x)-x)+x))*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(105\) vs. \(2(19)=38\).

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 5.53, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {2326} \[ \int \frac {e^{e^{x+e^{-x} (1+x)} x} \left (1+x+e^{x+e^{-x} (1+x)} \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )\right )}{1+x} \, dx=\frac {\left (-e^{-x} (x+1) x^3+x^3+2 x^2+x\right ) \exp \left (e^{x+e^{-x} (x+1)} x+x+e^{-x} (x+1)\right )}{(x+1) \left (e^{x+e^{-x} (x+1)} x \left (-e^{-x} (x+1)+e^{-x}+1\right )+e^{x+e^{-x} (x+1)}\right )} \]

[In]

Int[(E^(E^(x + (1 + x)/E^x)*x)*(1 + x + E^(x + (1 + x)/E^x)*(x + 2*x^2 + x^3 - (x^3*(1 + x))/E^x)))/(1 + x),x]

[Out]

(E^(x + E^(x + (1 + x)/E^x)*x + (1 + x)/E^x)*(x + 2*x^2 + x^3 - (x^3*(1 + x))/E^x))/((1 + x)*(E^(x + (1 + x)/E
^x) + E^(x + (1 + x)/E^x)*x*(1 + E^(-x) - (1 + x)/E^x)))

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\exp \left (x+e^{x+e^{-x} (1+x)} x+e^{-x} (1+x)\right ) \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )}{(1+x) \left (e^{x+e^{-x} (1+x)}+e^{x+e^{-x} (1+x)} x \left (1+e^{-x}-e^{-x} (1+x)\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{e^{x+e^{-x} (1+x)} x} \left (1+x+e^{x+e^{-x} (1+x)} \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )\right )}{1+x} \, dx=e^{e^{e^{-x} \left (1+x+e^x x\right )} x} x \]

[In]

Integrate[(E^(E^(x + (1 + x)/E^x)*x)*(1 + x + E^(x + (1 + x)/E^x)*(x + 2*x^2 + x^3 - (x^3*(1 + x))/E^x)))/(1 +
x),x]

[Out]

E^(E^((1 + x + E^x*x)/E^x)*x)*x

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\({\mathrm e}^{x \,{\mathrm e}^{{\mathrm e}^{\ln \left (1+x \right )-x}+x}} x\)	\(18\)
risch	\(x \,{\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{-x}+{\mathrm e}^{-x}+x} x}\)	\(19\)

[In]

int(((-x^3*exp(ln(1+x)-x)+x^3+2*x^2+x)*exp(exp(ln(1+x)-x)+x)+x+1)*exp(x*exp(exp(ln(1+x)-x)+x))/(1+x),x,method=
_RETURNVERBOSE)

[Out]

exp(x*exp(exp(ln(1+x)-x)+x))*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^{e^{x+e^{-x} (1+x)} x} \left (1+x+e^{x+e^{-x} (1+x)} \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )\right )}{1+x} \, dx=x e^{\left (x e^{\left (x + e^{\left (-x + \log \left (x + 1\right )\right )}\right )}\right )} \]

[In]

integrate(((-x^3*exp(log(1+x)-x)+x^3+2*x^2+x)*exp(exp(log(1+x)-x)+x)+x+1)*exp(x*exp(exp(log(1+x)-x)+x))/(1+x),
x, algorithm="fricas")

[Out]

x*e^(x*e^(x + e^(-x + log(x + 1))))

Sympy [A] (verification not implemented)

Time = 2.40 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {e^{e^{x+e^{-x} (1+x)} x} \left (1+x+e^{x+e^{-x} (1+x)} \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )\right )}{1+x} \, dx=x e^{x e^{x + \left (x + 1\right ) e^{- x}}} \]

[In]

integrate(((-x**3*exp(ln(1+x)-x)+x**3+2*x**2+x)*exp(exp(ln(1+x)-x)+x)+x+1)*exp(x*exp(exp(ln(1+x)-x)+x))/(1+x),
x)

[Out]

x*exp(x*exp(x + (x + 1)*exp(-x)))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{e^{x+e^{-x} (1+x)} x} \left (1+x+e^{x+e^{-x} (1+x)} \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )\right )}{1+x} \, dx=x e^{\left (x e^{\left (x e^{\left (-x\right )} + x + e^{\left (-x\right )}\right )}\right )} \]

[In]

integrate(((-x^3*exp(log(1+x)-x)+x^3+2*x^2+x)*exp(exp(log(1+x)-x)+x)+x+1)*exp(x*exp(exp(log(1+x)-x)+x))/(1+x),
x, algorithm="maxima")

[Out]

x*e^(x*e^(x*e^(-x) + x + e^(-x)))

Giac [F]

\[ \int \frac {e^{e^{x+e^{-x} (1+x)} x} \left (1+x+e^{x+e^{-x} (1+x)} \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )\right )}{1+x} \, dx=\int { -\frac {{\left ({\left (x^{3} e^{\left (-x + \log \left (x + 1\right )\right )} - x^{3} - 2 \, x^{2} - x\right )} e^{\left (x + e^{\left (-x + \log \left (x + 1\right )\right )}\right )} - x - 1\right )} e^{\left (x e^{\left (x + e^{\left (-x + \log \left (x + 1\right )\right )}\right )}\right )}}{x + 1} \,d x } \]

[In]

integrate(((-x^3*exp(log(1+x)-x)+x^3+2*x^2+x)*exp(exp(log(1+x)-x)+x)+x+1)*exp(x*exp(exp(log(1+x)-x)+x))/(1+x),
x, algorithm="giac")

[Out]

integrate(-((x^3*e^(-x + log(x + 1)) - x^3 - 2*x^2 - x)*e^(x + e^(-x + log(x + 1))) - x - 1)*e^(x*e^(x + e^(-x
+ log(x + 1))))/(x + 1), x)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{x+e^{-x} (1+x)} x} \left (1+x+e^{x+e^{-x} (1+x)} \left (x+2 x^2+x^3-e^{-x} x^3 (1+x)\right )\right )}{1+x} \, dx=x\,{\mathrm {e}}^{x\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}\,{\mathrm {e}}^{{\mathrm {e}}^{-x}}\,{\mathrm {e}}^x} \]

[In]

int((exp(x*exp(x + exp(log(x + 1) - x)))*(x + exp(x + exp(log(x + 1) - x))*(x - x^3*exp(log(x + 1) - x) + 2*x^
2 + x^3) + 1))/(x + 1),x)

[Out]

x*exp(x*exp(x*exp(-x))*exp(exp(-x))*exp(x))