3.62.69 \(\int \frac {8+8 x+2 x^2+e^{\frac {5 x}{2+x}} (18+28 x+2 x^2)}{4+4 x+x^2} \, dx\)

Optimal. Leaf size=20 \[ \left (2+\frac {1}{x}\right ) \left (x+e^{\frac {5 x}{2+x}} x\right ) \]

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Rubi [F]  time = 0.27, 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 {8+8 x+2 x^2+e^{\frac {5 x}{2+x}} \left (18+28 x+2 x^2\right )}{4+4 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8 + 8*x + 2*x^2 + E^((5*x)/(2 + x))*(18 + 28*x + 2*x^2))/(4 + 4*x + x^2),x]

[Out]

-3*E^(5 - 10/(2 + x)) + 2*x - 20*E^5*ExpIntegralEi[-10/(2 + x)] + 2*Defer[Int][E^((5*x)/(2 + x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8+8 x+2 x^2+e^{\frac {5 x}{2+x}} \left (18+28 x+2 x^2\right )}{(2+x)^2} \, dx\\ &=\int \left (2+\frac {2 e^{\frac {5 x}{2+x}} \left (9+14 x+x^2\right )}{(2+x)^2}\right ) \, dx\\ &=2 x+2 \int \frac {e^{\frac {5 x}{2+x}} \left (9+14 x+x^2\right )}{(2+x)^2} \, dx\\ &=2 x+2 \int \left (e^{\frac {5 x}{2+x}}-\frac {15 e^{\frac {5 x}{2+x}}}{(2+x)^2}+\frac {10 e^{\frac {5 x}{2+x}}}{2+x}\right ) \, dx\\ &=2 x+2 \int e^{\frac {5 x}{2+x}} \, dx+20 \int \frac {e^{\frac {5 x}{2+x}}}{2+x} \, dx-30 \int \frac {e^{\frac {5 x}{2+x}}}{(2+x)^2} \, dx\\ &=2 x+2 \int e^{\frac {5 x}{2+x}} \, dx+20 \int \frac {e^{5-\frac {10}{2+x}}}{2+x} \, dx-30 \int \frac {e^{5-\frac {10}{2+x}}}{(2+x)^2} \, dx\\ &=-3 e^{5-\frac {10}{2+x}}+2 x-20 e^5 \text {Ei}\left (-\frac {10}{2+x}\right )+2 \int e^{\frac {5 x}{2+x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 28, normalized size = 1.40 \begin {gather*} 2 x+\frac {1}{5} e^{-5 \left (-1+\frac {2}{2+x}\right )} (-15+10 (2+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + 8*x + 2*x^2 + E^((5*x)/(2 + x))*(18 + 28*x + 2*x^2))/(4 + 4*x + x^2),x]

[Out]

2*x + (-15 + 10*(2 + x))/(5*E^(5*(-1 + 2/(2 + x))))

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fricas [A]  time = 1.06, size = 19, normalized size = 0.95 \begin {gather*} {\left (2 \, x + 1\right )} e^{\left (\frac {5 \, x}{x + 2}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+28*x+18)*exp(5*x/(2+x))+2*x^2+8*x+8)/(x^2+4*x+4),x, algorithm="fricas")

[Out]

(2*x + 1)*e^(5*x/(x + 2)) + 2*x

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giac [A]  time = 0.14, size = 25, normalized size = 1.25 \begin {gather*} 2 \, x e^{\left (\frac {5 \, x}{x + 2}\right )} + 2 \, x + e^{\left (\frac {5 \, x}{x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+28*x+18)*exp(5*x/(2+x))+2*x^2+8*x+8)/(x^2+4*x+4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.54, size = 20, normalized size = 1.00

method	result	size
risch	\(2 x +\left (2 x +1\right ) {\mathrm e}^{\frac {5 x}{2+x}}\)	\(20\)
derivativedivides	\(2 \,{\mathrm e}^{5-\frac {10}{2+x}} \left (2+x \right )+4+2 x -3 \,{\mathrm e}^{5-\frac {10}{2+x}}\)	\(33\)
default	\(2 \,{\mathrm e}^{5-\frac {10}{2+x}} \left (2+x \right )+4+2 x -3 \,{\mathrm e}^{5-\frac {10}{2+x}}\)	\(33\)
norman	\(\frac {2 x^{2}+5 \,{\mathrm e}^{\frac {5 x}{2+x}} x +2 \,{\mathrm e}^{\frac {5 x}{2+x}} x^{2}+2 \,{\mathrm e}^{\frac {5 x}{2+x}}-8}{2+x}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+28*x+18)*exp(5*x/(2+x))+2*x^2+8*x+8)/(x^2+4*x+4),x,method=_RETURNVERBOSE)

[Out]

2*x+(2*x+1)*exp(5*x/(2+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, x e^{\left (-\frac {10}{x + 2} + 5\right )} + 2 \, x + \frac {9}{5} \, e^{\left (-\frac {10}{x + 2} + 5\right )} - 8 \, \int \frac {e^{\left (-\frac {10}{x + 2} + 5\right )}}{x^{2} + 4 \, x + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+28*x+18)*exp(5*x/(2+x))+2*x^2+8*x+8)/(x^2+4*x+4),x, algorithm="maxima")

[Out]

2*x*e^(-10/(x + 2) + 5) + 2*x + 9/5*e^(-10/(x + 2) + 5) - 8*integrate(e^(-10/(x + 2) + 5)/(x^2 + 4*x + 4), x)

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mupad [B]  time = 0.11, size = 25, normalized size = 1.25 \begin {gather*} {\mathrm {e}}^{\frac {5\,x}{x+2}}+x\,\left (2\,{\mathrm {e}}^{\frac {5\,x}{x+2}}+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + exp((5*x)/(x + 2))*(28*x + 2*x^2 + 18) + 2*x^2 + 8)/(4*x + x^2 + 4),x)

[Out]

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

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sympy [A]  time = 0.15, size = 15, normalized size = 0.75 \begin {gather*} 2 x + \left (2 x + 1\right ) e^{\frac {5 x}{x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+28*x+18)*exp(5*x/(2+x))+2*x**2+8*x+8)/(x**2+4*x+4),x)

[Out]

2*x + (2*x + 1)*exp(5*x/(x + 2))

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