\(\int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} (8 x+8 x^2+e^2 (-8+8 x+16 x^2+4 x^3))}{4 x^3+4 x^4+x^5} \, dx\) [7571]

Optimal result

Integrand size = 75, antiderivative size = 30 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {\left (-e^2+x\right ) \left (4 e^{\frac {1}{(-2-x) x}}+3 x\right )}{x} \]

[Out]

(x-exp(1)^2)/x*(3*x+4*exp(1/x/(-2-x)))

Rubi [F]

\[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx \]

[In]

Int[(12*x^3 + 12*x^4 + 3*x^5 + (8*x + 8*x^2 + E^2*(-8 + 8*x + 16*x^2 + 4*x^3))/E^(2*x + x^2)^(-1))/(4*x^3 + 4*
x^4 + x^5),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{x^3 \left (4+4 x+x^2\right )} \, dx \\ & = \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{x^3 (2+x)^2} \, dx \\ & = \int \left (3+\frac {4 e^{-\frac {1}{x (2+x)}} \left (-2 e^2+2 \left (1+e^2\right ) x+2 \left (1+2 e^2\right ) x^2+e^2 x^3\right )}{x^3 (2+x)^2}\right ) \, dx \\ & = 3 x+4 \int \frac {e^{-\frac {1}{x (2+x)}} \left (-2 e^2+2 \left (1+e^2\right ) x+2 \left (1+2 e^2\right ) x^2+e^2 x^3\right )}{x^3 (2+x)^2} \, dx \\ & = 3 x+4 \int \left (-\frac {e^{2-\frac {1}{x (2+x)}}}{2 x^3}+\frac {e^{-\frac {1}{x (2+x)}} \left (1+2 e^2\right )}{2 x^2}+\frac {e^{2-\frac {1}{x (2+x)}}}{8 x}+\frac {e^{-\frac {1}{x (2+x)}} \left (-2-e^2\right )}{4 (2+x)^2}-\frac {e^{2-\frac {1}{x (2+x)}}}{8 (2+x)}\right ) \, dx \\ & = 3 x+\frac {1}{2} \int \frac {e^{2-\frac {1}{x (2+x)}}}{x} \, dx-\frac {1}{2} \int \frac {e^{2-\frac {1}{x (2+x)}}}{2+x} \, dx-2 \int \frac {e^{2-\frac {1}{x (2+x)}}}{x^3} \, dx+\left (-2-e^2\right ) \int \frac {e^{-\frac {1}{x (2+x)}}}{(2+x)^2} \, dx+\left (2 \left (1+2 e^2\right )\right ) \int \frac {e^{-\frac {1}{x (2+x)}}}{x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=3 x+\frac {4 e^{-\frac {1}{2 x+x^2}} \left (-e^2+x\right )}{x} \]

[In]

Integrate[(12*x^3 + 12*x^4 + 3*x^5 + (8*x + 8*x^2 + E^2*(-8 + 8*x + 16*x^2 + 4*x^3))/E^(2*x + x^2)^(-1))/(4*x^
3 + 4*x^4 + x^5),x]

[Out]

3*x + (4*(-E^2 + x))/(E^(2*x + x^2)^(-1)*x)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

method	result	size
risch	\(3 x -\frac {4 \left ({\mathrm e}^{2}-x \right ) {\mathrm e}^{-\frac {1}{x \left (2+x \right )}}}{x}\)	\(27\)
parallelrisch	\(-\frac {4 \,{\mathrm e}^{2} {\mathrm e}^{-\frac {1}{x \left (2+x \right )}}-3 x^{2}-4 \,{\mathrm e}^{-\frac {1}{x \left (2+x \right )}} x +24 x}{x}\)	\(46\)
parts	\(3 x +\frac {\left (-4 \,{\mathrm e}^{2}+8\right ) x^{2} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}+4 x^{3} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}-8 x \,{\mathrm e}^{2} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}}{x^{2} \left (2+x \right )}\)	\(75\)
norman	\(\frac {-12 x^{2}+\left (-4 \,{\mathrm e}^{2}+8\right ) x^{2} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}+3 x^{4}+4 x^{3} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}-8 x \,{\mathrm e}^{2} {\mathrm e}^{-\frac {1}{x^{2}+2 x}}}{x^{2} \left (2+x \right )}\)	\(81\)

[In]

int((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x^5+12*x^4+12*x^3)/(x^5+4*x^4+4*x^3),x,meth
od=_RETURNVERBOSE)

[Out]

3*x-4*(exp(2)-x)/x*exp(-1/x/(2+x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {3 \, x^{2} + 4 \, {\left (x - e^{2}\right )} e^{\left (-\frac {1}{x^{2} + 2 \, x}\right )}}{x} \]

[In]

integrate((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x^5+12*x^4+12*x^3)/(x^5+4*x^4+4*x^3),
x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=3 x + \frac {\left (4 x - 4 e^{2}\right ) e^{- \frac {1}{x^{2} + 2 x}}}{x} \]

[In]

integrate((((4*x**3+16*x**2+8*x-8)*exp(1)**2+8*x**2+8*x)*exp(-1/(x**2+2*x))+3*x**5+12*x**4+12*x**3)/(x**5+4*x*
*4+4*x**3),x)

[Out]

3*x + (4*x - 4*exp(2))*exp(-1/(x**2 + 2*x))/x

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=3 \, x + \frac {4 \, {\left (x - e^{2}\right )} e^{\left (\frac {1}{2 \, {\left (x + 2\right )}} - \frac {1}{2 \, x}\right )}}{x} \]

[In]

integrate((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x^5+12*x^4+12*x^3)/(x^5+4*x^4+4*x^3),
x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {3 \, x^{2} - 4 \, e^{\left (\frac {2 \, x^{2} + 4 \, x - 1}{x^{2} + 2 \, x}\right )}}{x} + 4 \, e^{\left (-\frac {1}{x^{2} + 2 \, x}\right )} \]

[In]

integrate((((4*x^3+16*x^2+8*x-8)*exp(1)^2+8*x^2+8*x)*exp(-1/(x^2+2*x))+3*x^5+12*x^4+12*x^3)/(x^5+4*x^4+4*x^3),
x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.62 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {12 x^3+12 x^4+3 x^5+e^{-\frac {1}{2 x+x^2}} \left (8 x+8 x^2+e^2 \left (-8+8 x+16 x^2+4 x^3\right )\right )}{4 x^3+4 x^4+x^5} \, dx=3\,x+4\,{\mathrm {e}}^{-\frac {1}{x^2+2\,x}}-\frac {4\,{\mathrm {e}}^{-\frac {1}{x^2+2\,x}}\,{\mathrm {e}}^2}{x} \]

[In]

int((exp(-1/(2*x + x^2))*(8*x + exp(2)*(8*x + 16*x^2 + 4*x^3 - 8) + 8*x^2) + 12*x^3 + 12*x^4 + 3*x^5)/(4*x^3 +
4*x^4 + x^5),x)

[Out]

3*x + 4*exp(-1/(2*x + x^2)) - (4*exp(-1/(2*x + x^2))*exp(2))/x