\(\int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} (-4-29 x+19 x^2+5 x^3))}{4 x^2+4 x^3+x^4} \, dx\) [519]

Optimal result

Integrand size = 103, antiderivative size = 31 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\frac {5+e^{\frac {5 \left (3-x+x^2\right )}{2+x}}+x-x^2}{x}} \]

[Out]

exp(1/x*(x+exp(5*(x^2-x+3)/(2+x))-x^2+5))

Rubi [F]

\[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=\int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx \]

[In]

Int[(E^((5 + E^((15 - 5*x + 5*x^2)/(2 + x)) + x - x^2)/x)*(-20 - 20*x - 9*x^2 - 4*x^3 - x^4 + E^((15 - 5*x + 5
*x^2)/(2 + x))*(-4 - 29*x + 19*x^2 + 5*x^3)))/(4*x^2 + 4*x^3 + x^4),x]

[Out]

-Defer[Int][E^(1 + 5/x + E^((15 - 5*x + 5*x^2)/(2 + x))/x - x), x] - 5*Defer[Int][E^(1 + 5/x + E^((15 - 5*x +
5*x^2)/(2 + x))/x - x)/x^2, x] - Defer[Int][E^(1 + 5/x + E^((15 - 5*x + 5*x^2)/(2 + x))/x - x + (5*(3 - x + x^
2))/(2 + x))/x^2, x] - (25*Defer[Int][E^(1 + 5/x + E^((15 - 5*x + 5*x^2)/(2 + x))/x - x + (5*(3 - x + x^2))/(2
+ x))/x, x])/4 + (45*Defer[Int][E^(1 + 5/x + E^((15 - 5*x + 5*x^2)/(2 + x))/x - x + (5*(3 - x + x^2))/(2 + x)
)/(2 + x)^2, x])/2 + (45*Defer[Int][E^(1 + 5/x + E^((15 - 5*x + 5*x^2)/(2 + x))/x - x + (5*(3 - x + x^2))/(2 +
x))/(2 + x), x])/4

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{x^2 \left (4+4 x+x^2\right )} \, dx \\ & = \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{x^2 (2+x)^2} \, dx \\ & = \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{x^2 (2+x)^2} \, dx \\ & = \int \left (-\frac {9 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2}-\frac {20 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x^2 (2+x)^2}-\frac {20 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x (2+x)^2}-\frac {4 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} x}{(2+x)^2}-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} x^2}{(2+x)^2}+\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right ) \left (-4-29 x+19 x^2+5 x^3\right )}{x^2 (2+x)^2}\right ) \, dx \\ & = -\left (4 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} x}{(2+x)^2} \, dx\right )-9 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx-20 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x^2 (2+x)^2} \, dx-20 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x (2+x)^2} \, dx-\int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} x^2}{(2+x)^2} \, dx+\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right ) \left (-4-29 x+19 x^2+5 x^3\right )}{x^2 (2+x)^2} \, dx \\ & = -\left (4 \int \left (-\frac {2 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{2+x}\right ) \, dx\right )-9 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx-20 \int \left (\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 x}-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{2 (2+x)^2}-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 (2+x)}\right ) \, dx-20 \int \left (\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 x^2}-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 x}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 (2+x)^2}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 (2+x)}\right ) \, dx-\int \left (e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}+\frac {4 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2}-\frac {4 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{2+x}\right ) \, dx+\int \left (-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}}}{x^2}-\frac {25 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}}}{4 x}+\frac {45 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}}}{2 (2+x)^2}+\frac {45 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}}}{4 (2+x)}\right ) \, dx \\ & = -\left (4 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx\right )-5 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x^2} \, dx-5 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx-\frac {25}{4} \int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right )}{x} \, dx+8 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx-9 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx+10 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx+\frac {45}{4} \int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right )}{2+x} \, dx+\frac {45}{2} \int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right )}{(2+x)^2} \, dx-\int e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} \, dx-\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right )}{x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{1+\frac {5}{x}+\frac {e^{-15+5 x+\frac {45}{2+x}}}{x}-x} \]

[In]

Integrate[(E^((5 + E^((15 - 5*x + 5*x^2)/(2 + x)) + x - x^2)/x)*(-20 - 20*x - 9*x^2 - 4*x^3 - x^4 + E^((15 - 5
*x + 5*x^2)/(2 + x))*(-4 - 29*x + 19*x^2 + 5*x^3)))/(4*x^2 + 4*x^3 + x^4),x]

[Out]

E^(1 + 5/x + E^(-15 + 5*x + 45/(2 + x))/x - x)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97

method	result	size
parallelrisch	\({\mathrm e}^{\frac {{\mathrm e}^{\frac {5 x^{2}-5 x +15}{2+x}}-x^{2}+x +5}{x}}\)	\(30\)
risch	\({\mathrm e}^{-\frac {-x -{\mathrm e}^{\frac {5 x^{2}-5 x +15}{2+x}}+x^{2}-5}{x}}\)	\(33\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {5 x^{2}-5 x +15}{2+x}}-x^{2}+x +5}{x}}+2 x \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {5 x^{2}-5 x +15}{2+x}}-x^{2}+x +5}{x}}}{x \left (2+x \right )}\)	\(78\)

[In]

int(((5*x^3+19*x^2-29*x-4)*exp((5*x^2-5*x+15)/(2+x))-x^4-4*x^3-9*x^2-20*x-20)*exp((exp((5*x^2-5*x+15)/(2+x))-x
^2+x+5)/x)/(x^4+4*x^3+4*x^2),x,method=_RETURNVERBOSE)

[Out]

exp(1/x*(x+exp(5*(x^2-x+3)/(2+x))-x^2+5))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\left (-\frac {x^{2} - x - e^{\left (\frac {5 \, {\left (x^{2} - x + 3\right )}}{x + 2}\right )} - 5}{x}\right )} \]

[In]

integrate(((5*x^3+19*x^2-29*x-4)*exp((5*x^2-5*x+15)/(2+x))-x^4-4*x^3-9*x^2-20*x-20)*exp((exp((5*x^2-5*x+15)/(2
+x))-x^2+x+5)/x)/(x^4+4*x^3+4*x^2),x, algorithm="fricas")

[Out]

e^(-(x^2 - x - e^(5*(x^2 - x + 3)/(x + 2)) - 5)/x)

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\frac {- x^{2} + x + e^{\frac {5 x^{2} - 5 x + 15}{x + 2}} + 5}{x}} \]

[In]

integrate(((5*x**3+19*x**2-29*x-4)*exp((5*x**2-5*x+15)/(2+x))-x**4-4*x**3-9*x**2-20*x-20)*exp((exp((5*x**2-5*x
+15)/(2+x))-x**2+x+5)/x)/(x**4+4*x**3+4*x**2),x)

[Out]

exp((-x**2 + x + exp((5*x**2 - 5*x + 15)/(x + 2)) + 5)/x)

Maxima [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\left (-x + \frac {e^{\left (5 \, x + \frac {45}{x + 2} - 15\right )}}{x} + \frac {5}{x} + 1\right )} \]

[In]

integrate(((5*x^3+19*x^2-29*x-4)*exp((5*x^2-5*x+15)/(2+x))-x^4-4*x^3-9*x^2-20*x-20)*exp((exp((5*x^2-5*x+15)/(2
+x))-x^2+x+5)/x)/(x^4+4*x^3+4*x^2),x, algorithm="maxima")

[Out]

e^(-x + e^(5*x + 45/(x + 2) - 15)/x + 5/x + 1)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\left (-x + \frac {e^{\left (\frac {5 \, x^{2}}{x + 2} - \frac {5 \, x}{x + 2} + \frac {15}{x + 2}\right )}}{x} + \frac {5}{x} + 1\right )} \]

[In]

integrate(((5*x^3+19*x^2-29*x-4)*exp((5*x^2-5*x+15)/(2+x))-x^4-4*x^3-9*x^2-20*x-20)*exp((exp((5*x^2-5*x+15)/(2
+x))-x^2+x+5)/x)/(x^4+4*x^3+4*x^2),x, algorithm="giac")

[Out]

e^(-x + e^(5*x^2/(x + 2) - 5*x/(x + 2) + 15/(x + 2))/x + 5/x + 1)

Mupad [B] (verification not implemented)

Time = 7.83 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx={\mathrm {e}}^{-x}\,\mathrm {e}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-\frac {5\,x}{x+2}}\,{\mathrm {e}}^{\frac {5\,x^2}{x+2}}\,{\mathrm {e}}^{\frac {15}{x+2}}}{x}} \]

[In]

int(-(exp((x + exp((5*x^2 - 5*x + 15)/(x + 2)) - x^2 + 5)/x)*(20*x + exp((5*x^2 - 5*x + 15)/(x + 2))*(29*x - 1
9*x^2 - 5*x^3 + 4) + 9*x^2 + 4*x^3 + x^4 + 20))/(4*x^2 + 4*x^3 + x^4),x)

[Out]

exp(-x)*exp(1)*exp(5/x)*exp((exp(-(5*x)/(x + 2))*exp((5*x^2)/(x + 2))*exp(15/(x + 2)))/x)