3.30.48 \(\int \frac {16 x^3+24 x^4+12 x^5+2 x^6+e^{\frac {2-x}{2 x+x^2}} (-4-4 x+5 x^2+4 x^3+x^4)}{8 x^4+8 x^5+2 x^6+e^{\frac {2-x}{2 x+x^2}} (4 x^2+4 x^3+x^4)} \, dx\)

Optimal. Leaf size=26 \[ -3+x+\log \left (e^{\frac {2-x}{x (2+x)}}+2 x^2\right ) \]

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Rubi [F]  time = 3.22, 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 {16 x^3+24 x^4+12 x^5+2 x^6+e^{\frac {2-x}{2 x+x^2}} \left (-4-4 x+5 x^2+4 x^3+x^4\right )}{8 x^4+8 x^5+2 x^6+e^{\frac {2-x}{2 x+x^2}} \left (4 x^2+4 x^3+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16*x^3 + 24*x^4 + 12*x^5 + 2*x^6 + E^((2 - x)/(2*x + x^2))*(-4 - 4*x + 5*x^2 + 4*x^3 + x^4))/(8*x^4 + 8*x
^5 + 2*x^6 + E^((2 - x)/(2*x + x^2))*(4*x^2 + 4*x^3 + x^4)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2+x}{x}+\frac {e^{\frac {2}{x (2+x)}} \left (-4-12 x-7 x^2-2 x^3\right )}{x^2 (2+x)^2 \left (e^{\frac {2}{2 x+x^2}}+2 e^{\frac {1}{2+x}} x^2\right )}\right ) \, dx\\ &=\int \frac {2+x}{x} \, dx+\int \frac {e^{\frac {2}{x (2+x)}} \left (-4-12 x-7 x^2-2 x^3\right )}{x^2 (2+x)^2 \left (e^{\frac {2}{2 x+x^2}}+2 e^{\frac {1}{2+x}} x^2\right )} \, dx\\ &=\int \left (1+\frac {2}{x}\right ) \, dx+\int \left (-\frac {e^{\frac {2}{x (2+x)}}}{x^2 \left (e^{\frac {2}{2 x+x^2}}+2 e^{\frac {1}{2+x}} x^2\right )}-\frac {2 e^{\frac {2}{x (2+x)}}}{x \left (e^{\frac {2}{2 x+x^2}}+2 e^{\frac {1}{2+x}} x^2\right )}+\frac {2 e^{\frac {2}{x (2+x)}}}{(2+x)^2 \left (e^{\frac {2}{2 x+x^2}}+2 e^{\frac {1}{2+x}} x^2\right )}\right ) \, dx\\ &=x+2 \log (x)-2 \int \frac {e^{\frac {2}{x (2+x)}}}{x \left (e^{\frac {2}{2 x+x^2}}+2 e^{\frac {1}{2+x}} x^2\right )} \, dx+2 \int \frac {e^{\frac {2}{x (2+x)}}}{(2+x)^2 \left (e^{\frac {2}{2 x+x^2}}+2 e^{\frac {1}{2+x}} x^2\right )} \, dx-\int \frac {e^{\frac {2}{x (2+x)}}}{x^2 \left (e^{\frac {2}{2 x+x^2}}+2 e^{\frac {1}{2+x}} x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 36, normalized size = 1.38 \begin {gather*} x-\frac {1}{2+x}+\log \left (e^{\frac {1}{x}-\frac {1}{2+x}}+2 e^{\frac {1}{2+x}} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x^3 + 24*x^4 + 12*x^5 + 2*x^6 + E^((2 - x)/(2*x + x^2))*(-4 - 4*x + 5*x^2 + 4*x^3 + x^4))/(8*x^4
+ 8*x^5 + 2*x^6 + E^((2 - x)/(2*x + x^2))*(4*x^2 + 4*x^3 + x^4)),x]

[Out]

x - (2 + x)^(-1) + Log[E^(x^(-1) - (2 + x)^(-1)) + 2*E^(2 + x)^(-1)*x^2]

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fricas [A]  time = 0.86, size = 24, normalized size = 0.92 \begin {gather*} x + \log \left (2 \, x^{2} + e^{\left (-\frac {x - 2}{x^{2} + 2 \, x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+4*x^3+5*x^2-4*x-4)*exp((2-x)/(x^2+2*x))+2*x^6+12*x^5+24*x^4+16*x^3)/((x^4+4*x^3+4*x^2)*exp((2-
x)/(x^2+2*x))+2*x^6+8*x^5+8*x^4),x, algorithm="fricas")

[Out]

x + log(2*x^2 + e^(-(x - 2)/(x^2 + 2*x)))

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giac [A]  time = 0.26, size = 24, normalized size = 0.92 \begin {gather*} x + \log \left (2 \, x^{2} + e^{\left (-\frac {x - 2}{x^{2} + 2 \, x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+4*x^3+5*x^2-4*x-4)*exp((2-x)/(x^2+2*x))+2*x^6+12*x^5+24*x^4+16*x^3)/((x^4+4*x^3+4*x^2)*exp((2-
x)/(x^2+2*x))+2*x^6+8*x^5+8*x^4),x, algorithm="giac")

[Out]

x + log(2*x^2 + e^(-(x - 2)/(x^2 + 2*x)))

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maple [A]  time = 0.09, size = 41, normalized size = 1.58

method	result	size
norman	\(\frac {x^{3}-4 x}{x \left (2+x \right )}+\ln \left (2 x^{2}+{\mathrm e}^{\frac {2-x}{x^{2}+2 x}}\right )\)	\(41\)
risch	\(x +\frac {2-x}{\left (2+x \right ) x}-\frac {2-x}{x^{2}+2 x}+\ln \left (2 x^{2}+{\mathrm e}^{-\frac {x -2}{x \left (2+x \right )}}\right )\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4+4*x^3+5*x^2-4*x-4)*exp((2-x)/(x^2+2*x))+2*x^6+12*x^5+24*x^4+16*x^3)/((x^4+4*x^3+4*x^2)*exp((2-x)/(x^
2+2*x))+2*x^6+8*x^5+8*x^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.51, size = 51, normalized size = 1.96 \begin {gather*} \frac {x^{2} + 2 \, x - 1}{x + 2} + 2 \, \log \relax (x) + \log \left (\frac {{\left (2 \, x^{2} e^{\left (\frac {2}{x + 2}\right )} + e^{\frac {1}{x}}\right )} e^{\left (-\frac {1}{x + 2}\right )}}{2 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+4*x^3+5*x^2-4*x-4)*exp((2-x)/(x^2+2*x))+2*x^6+12*x^5+24*x^4+16*x^3)/((x^4+4*x^3+4*x^2)*exp((2-
x)/(x^2+2*x))+2*x^6+8*x^5+8*x^4),x, algorithm="maxima")

[Out]

(x^2 + 2*x - 1)/(x + 2) + 2*log(x) + log(1/2*(2*x^2*e^(2/(x + 2)) + e^(1/x))*e^(-1/(x + 2))/x^2)

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mupad [B]  time = 1.99, size = 23, normalized size = 0.88 \begin {gather*} x+\ln \left (\frac {{\mathrm {e}}^{-\frac {x-2}{x\,\left (x+2\right )}}}{2}+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^3 + 24*x^4 + 12*x^5 + 2*x^6 + exp(-(x - 2)/(2*x + x^2))*(5*x^2 - 4*x + 4*x^3 + x^4 - 4))/(exp(-(x -
2)/(2*x + x^2))*(4*x^2 + 4*x^3 + x^4) + 8*x^4 + 8*x^5 + 2*x^6),x)

[Out]

x + log(exp(-(x - 2)/(x*(x + 2)))/2 + x^2)

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sympy [A]  time = 0.36, size = 19, normalized size = 0.73 \begin {gather*} x + \log {\left (2 x^{2} + e^{\frac {2 - x}{x^{2} + 2 x}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**4+4*x**3+5*x**2-4*x-4)*exp((2-x)/(x**2+2*x))+2*x**6+12*x**5+24*x**4+16*x**3)/((x**4+4*x**3+4*x*
*2)*exp((2-x)/(x**2+2*x))+2*x**6+8*x**5+8*x**4),x)

[Out]

x + log(2*x**2 + exp((2 - x)/(x**2 + 2*x)))

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