\(\int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} (16+4 x+x^2)}{16 x+8 x^2+x^3} \, dx\) [5965]

Optimal result

Integrand size = 60, antiderivative size = 15 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=e^{e^{\frac {x}{-4-x}} x} \]

[Out]

exp(exp(x/(-4-x)+ln(x)))

Rubi [F]

\[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=\int \frac {\exp \left (e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}\right ) \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx \]

[In]

Int[(E^(E^((-x + (4 + x)*Log[x])/(4 + x)) + (-x + (4 + x)*Log[x])/(4 + x))*(16 + 4*x + x^2))/(16*x + 8*x^2 + x
^3),x]

[Out]

Defer[Int][E^(x/E^(x/(4 + x)) - x/(4 + x)), x] + 16*Defer[Int][E^(x/E^(x/(4 + x)) - x/(4 + x))/(4 + x)^2, x] -
4*Defer[Int][E^(x/E^(x/(4 + x)) - x/(4 + x))/(4 + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}\right ) \left (16+4 x+x^2\right )}{x \left (16+8 x+x^2\right )} \, dx \\ & = \int \frac {\exp \left (e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}\right ) \left (16+4 x+x^2\right )}{x (4+x)^2} \, dx \\ & = \int \frac {e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}} \left (16+4 x+x^2\right )}{(4+x)^2} \, dx \\ & = \int \left (e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}+\frac {16 e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}}{(4+x)^2}-\frac {4 e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}}{4+x}\right ) \, dx \\ & = -\left (4 \int \frac {e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}}{4+x} \, dx\right )+16 \int \frac {e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}}}{(4+x)^2} \, dx+\int e^{e^{-\frac {x}{4+x}} x-\frac {x}{4+x}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=e^{e^{-1+\frac {4}{4+x}} x} \]

[In]

Integrate[(E^(E^((-x + (4 + x)*Log[x])/(4 + x)) + (-x + (4 + x)*Log[x])/(4 + x))*(16 + 4*x + x^2))/(16*x + 8*x
^2 + x^3),x]

[Out]

E^(E^(-1 + 4/(4 + x))*x)

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27

method	result	size
parallelrisch	\({\mathrm e}^{{\mathrm e}^{\frac {\left (4+x \right ) \ln \left (x \right )-x}{4+x}}}\)	\(19\)
risch	\({\mathrm e}^{{\mathrm e}^{\frac {x \ln \left (x \right )+4 \ln \left (x \right )-x}{4+x}}}\)	\(21\)

[In]

int((x^2+4*x+16)*exp(((4+x)*ln(x)-x)/(4+x))*exp(exp(((4+x)*ln(x)-x)/(4+x)))/(x^3+8*x^2+16*x),x,method=_RETURNV
ERBOSE)

[Out]

exp(exp(((4+x)*ln(x)-x)/(4+x)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (13) = 26\).

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 3.73 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=e^{\left (\frac {{\left (x + 4\right )} e^{\left (\frac {{\left (x + 4\right )} \log \left (x\right ) - x}{x + 4}\right )} + {\left (x + 4\right )} \log \left (x\right ) - x}{x + 4} - \frac {{\left (x + 4\right )} \log \left (x\right ) - x}{x + 4}\right )} \]

[In]

integrate((x^2+4*x+16)*exp(((4+x)*log(x)-x)/(4+x))*exp(exp(((4+x)*log(x)-x)/(4+x)))/(x^3+8*x^2+16*x),x, algori
thm="fricas")

[Out]

e^(((x + 4)*e^(((x + 4)*log(x) - x)/(x + 4)) + (x + 4)*log(x) - x)/(x + 4) - ((x + 4)*log(x) - x)/(x + 4))

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=\text {Timed out} \]

[In]

integrate((x**2+4*x+16)*exp(((4+x)*ln(x)-x)/(4+x))*exp(exp(((4+x)*ln(x)-x)/(4+x)))/(x**3+8*x**2+16*x),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=e^{\left (x e^{\left (\frac {4}{x + 4} - 1\right )}\right )} \]

[In]

integrate((x^2+4*x+16)*exp(((4+x)*log(x)-x)/(4+x))*exp(exp(((4+x)*log(x)-x)/(4+x)))/(x^3+8*x^2+16*x),x, algori
thm="maxima")

[Out]

e^(x*e^(4/(x + 4) - 1))

Giac [F]

\[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx=\int { \frac {{\left (x^{2} + 4 \, x + 16\right )} e^{\left (\frac {{\left (x + 4\right )} \log \left (x\right ) - x}{x + 4} + e^{\left (\frac {{\left (x + 4\right )} \log \left (x\right ) - x}{x + 4}\right )}\right )}}{x^{3} + 8 \, x^{2} + 16 \, x} \,d x } \]

[In]

integrate((x^2+4*x+16)*exp(((4+x)*log(x)-x)/(4+x))*exp(exp(((4+x)*log(x)-x)/(4+x)))/(x^3+8*x^2+16*x),x, algori
thm="giac")

[Out]

integrate((x^2 + 4*x + 16)*e^(((x + 4)*log(x) - x)/(x + 4) + e^(((x + 4)*log(x) - x)/(x + 4)))/(x^3 + 8*x^2 +
16*x), x)

Mupad [B] (verification not implemented)

Time = 11.89 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {e^{e^{\frac {-x+(4+x) \log (x)}{4+x}}+\frac {-x+(4+x) \log (x)}{4+x}} \left (16+4 x+x^2\right )}{16 x+8 x^2+x^3} \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^{-\frac {x}{x+4}}} \]

[In]

int((exp(exp(-(x - log(x)*(x + 4))/(x + 4)))*exp(-(x - log(x)*(x + 4))/(x + 4))*(4*x + x^2 + 16))/(16*x + 8*x^
2 + x^3),x)

[Out]

exp(x*exp(-x/(x + 4)))