\(\int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} (-12-3 x+2 x^2+x^3)}{32 x^2+16 x^3+2 x^4} \, dx\) [5639]

Optimal result

Integrand size = 56, antiderivative size = 29 \[ \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx=\frac {-25+e^{x+\frac {3+2 x-x^2}{2 x}}}{4+x} \]

[Out]

(exp(1/2*(-x^2+2*x+3)/x+x)-25)/(4+x)

Rubi [F]

\[ \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx=\int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx \]

[In]

Int[(50*x^2 + E^((3 + 2*x + x^2)/(2*x))*(-12 - 3*x + 2*x^2 + x^3))/(32*x^2 + 16*x^3 + 2*x^4),x]

[Out]

-25/(4 + x) - (3*Defer[Int][E^((3 + 2*x + x^2)/(2*x))/x^2, x])/8 + (3*Defer[Int][E^((3 + 2*x + x^2)/(2*x))/x,
x])/32 - Defer[Int][E^((3 + 2*x + x^2)/(2*x))/(4 + x)^2, x] + (13*Defer[Int][E^((3 + 2*x + x^2)/(2*x))/(4 + x)
, x])/32

Rubi steps \begin{align*} \text {integral}& = \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{x^2 \left (32+16 x+2 x^2\right )} \, dx \\ & = \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{2 x^2 (4+x)^2} \, dx \\ & = \frac {1}{2} \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{x^2 (4+x)^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {50}{(4+x)^2}+\frac {e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{x^2 (4+x)^2}\right ) \, dx \\ & = -\frac {25}{4+x}+\frac {1}{2} \int \frac {e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{x^2 (4+x)^2} \, dx \\ & = -\frac {25}{4+x}+\frac {1}{2} \int \left (-\frac {3 e^{\frac {3+2 x+x^2}{2 x}}}{4 x^2}+\frac {3 e^{\frac {3+2 x+x^2}{2 x}}}{16 x}-\frac {2 e^{\frac {3+2 x+x^2}{2 x}}}{(4+x)^2}+\frac {13 e^{\frac {3+2 x+x^2}{2 x}}}{16 (4+x)}\right ) \, dx \\ & = -\frac {25}{4+x}+\frac {3}{32} \int \frac {e^{\frac {3+2 x+x^2}{2 x}}}{x} \, dx-\frac {3}{8} \int \frac {e^{\frac {3+2 x+x^2}{2 x}}}{x^2} \, dx+\frac {13}{32} \int \frac {e^{\frac {3+2 x+x^2}{2 x}}}{4+x} \, dx-\int \frac {e^{\frac {3+2 x+x^2}{2 x}}}{(4+x)^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx=\frac {-25+e^{\frac {1}{2} \left (2+\frac {3}{x}+x\right )}}{4+x} \]

[In]

Integrate[(50*x^2 + E^((3 + 2*x + x^2)/(2*x))*(-12 - 3*x + 2*x^2 + x^3))/(32*x^2 + 16*x^3 + 2*x^4),x]

[Out]

(-25 + E^((2 + 3/x + x)/2))/(4 + x)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(\frac {-50+2 \,{\mathrm e}^{\frac {x^{2}+2 x +3}{2 x}}}{2 x +8}\)	\(26\)
risch	\(-\frac {25}{4+x}+\frac {{\mathrm e}^{\frac {x^{2}+2 x +3}{2 x}}}{4+x}\)	\(29\)
parts	\(-\frac {25}{4+x}+\frac {{\mathrm e}^{\frac {x^{2}+2 x +3}{2 x}}}{4+x}\)	\(29\)
norman	\(\frac {-25 x +{\mathrm e}^{\frac {x^{2}+2 x +3}{2 x}} x}{\left (4+x \right ) x}\)	\(30\)

[In]

int(((x^3+2*x^2-3*x-12)*exp(1/2*(x^2+2*x+3)/x)+50*x^2)/(2*x^4+16*x^3+32*x^2),x,method=_RETURNVERBOSE)

[Out]

1/2*(-50+2*exp(1/2*(x^2+2*x+3)/x))/(4+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx=\frac {e^{\left (\frac {x^{2} + 2 \, x + 3}{2 \, x}\right )} - 25}{x + 4} \]

[In]

integrate(((x^3+2*x^2-3*x-12)*exp(1/2*(x^2+2*x+3)/x)+50*x^2)/(2*x^4+16*x^3+32*x^2),x, algorithm="fricas")

[Out]

(e^(1/2*(x^2 + 2*x + 3)/x) - 25)/(x + 4)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx=\frac {e^{\frac {\frac {x^{2}}{2} + x + \frac {3}{2}}{x}}}{x + 4} - \frac {25}{x + 4} \]

[In]

integrate(((x**3+2*x**2-3*x-12)*exp(1/2*(x**2+2*x+3)/x)+50*x**2)/(2*x**4+16*x**3+32*x**2),x)

[Out]

exp((x**2/2 + x + 3/2)/x)/(x + 4) - 25/(x + 4)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx=\frac {e^{\left (\frac {1}{2} \, x + \frac {3}{2 \, x} + 1\right )}}{x + 4} - \frac {25}{x + 4} \]

[In]

integrate(((x^3+2*x^2-3*x-12)*exp(1/2*(x^2+2*x+3)/x)+50*x^2)/(2*x^4+16*x^3+32*x^2),x, algorithm="maxima")

[Out]

e^(1/2*x + 3/2/x + 1)/(x + 4) - 25/(x + 4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx=\frac {e^{\left (\frac {x^{2} + 2 \, x + 3}{2 \, x}\right )} - 25}{x + 4} \]

[In]

integrate(((x^3+2*x^2-3*x-12)*exp(1/2*(x^2+2*x+3)/x)+50*x^2)/(2*x^4+16*x^3+32*x^2),x, algorithm="giac")

[Out]

(e^(1/2*(x^2 + 2*x + 3)/x) - 25)/(x + 4)

Mupad [B] (verification not implemented)

Time = 11.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {50 x^2+e^{\frac {3+2 x+x^2}{2 x}} \left (-12-3 x+2 x^2+x^3\right )}{32 x^2+16 x^3+2 x^4} \, dx=\frac {{\mathrm {e}}^{\frac {x}{2}+\frac {3}{2\,x}+1}-25}{x+4} \]

[In]

int(-(exp((x + x^2/2 + 3/2)/x)*(3*x - 2*x^2 - x^3 + 12) - 50*x^2)/(32*x^2 + 16*x^3 + 2*x^4),x)

[Out]

(exp(x/2 + 3/(2*x) + 1) - 25)/(x + 4)