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

Optimal. Leaf size=29 \[ \frac {-25+e^{x+\frac {3+2 x-x^2}{2 x}}}{4+x} \]

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Rubi [F]  time = 0.96, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 22, normalized size = 0.76 \begin {gather*} \frac {-25+e^{\frac {1}{2} \left (2+\frac {3}{x}+x\right )}}{4+x} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.53, size = 22, normalized size = 0.76 \begin {gather*} \frac {e^{\left (\frac {x^{2} + 2 \, x + 3}{2 \, x}\right )} - 25}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.24, size = 22, normalized size = 0.76 \begin {gather*} \frac {e^{\left (\frac {x^{2} + 2 \, x + 3}{2 \, x}\right )} - 25}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.10, size = 29, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maxima [A]  time = 0.44, size = 25, normalized size = 0.86 \begin {gather*} \frac {e^{\left (\frac {1}{2} \, x + \frac {3}{2 \, x} + 1\right )}}{x + 4} - \frac {25}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.52, size = 19, normalized size = 0.66 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x}{2}+\frac {3}{2\,x}+1}-25}{x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.15, size = 20, normalized size = 0.69 \begin {gather*} \frac {e^{\frac {\frac {x^{2}}{2} + x + \frac {3}{2}}{x}}}{x + 4} - \frac {25}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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