3.26.100 \(\int \frac {-8+x+4 x^2-2 x^3}{27 e^{\frac {-8+2 x-4 x^2+x^3}{x}} x+3 x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac {1}{3} \log \left (3+\frac {1}{3} e^{-\left ((-4+x) \left (\frac {2}{x}+x\right )\right )} x\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 1.53, 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+x+4 x^2-2 x^3}{27 e^{\frac {-8+2 x-4 x^2+x^3}{x}} x+3 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 + x + 4*x^2 - 2*x^3)/(27*E^((-8 + 2*x - 4*x^2 + x^3)/x)*x + 3*x^2),x]

[Out]

(8*Defer[Int][E^((4*(2 + x^2))/x)/(x*(-9*E^(2 + x^2) - E^(8/x + 4*x)*x)), x])/3 + (2*Defer[Int][(E^((4*(2 + x^
2))/x)*x^2)/(-9*E^(2 + x^2) - E^(8/x + 4*x)*x), x])/3 + Defer[Int][E^((4*(2 + x^2))/x)/(9*E^(2 + x^2) + E^(8/x
+ 4*x)*x), x]/3 + (4*Defer[Int][(E^((4*(2 + x^2))/x)*x)/(9*E^(2 + x^2) + E^(8/x + 4*x)*x), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {8 e^{\frac {4 \left (2+x^2\right )}{x}}}{3 x \left (-9 e^{2+x^2}-e^{\frac {8}{x}+4 x} x\right )}+\frac {2 e^{\frac {4 \left (2+x^2\right )}{x}} x^2}{3 \left (-9 e^{2+x^2}-e^{\frac {8}{x}+4 x} x\right )}+\frac {e^{\frac {4 \left (2+x^2\right )}{x}}}{3 \left (9 e^{2+x^2}+e^{\frac {8}{x}+4 x} x\right )}+\frac {4 e^{\frac {4 \left (2+x^2\right )}{x}} x}{3 \left (9 e^{2+x^2}+e^{\frac {8}{x}+4 x} x\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {4 \left (2+x^2\right )}{x}}}{9 e^{2+x^2}+e^{\frac {8}{x}+4 x} x} \, dx+\frac {2}{3} \int \frac {e^{\frac {4 \left (2+x^2\right )}{x}} x^2}{-9 e^{2+x^2}-e^{\frac {8}{x}+4 x} x} \, dx+\frac {4}{3} \int \frac {e^{\frac {4 \left (2+x^2\right )}{x}} x}{9 e^{2+x^2}+e^{\frac {8}{x}+4 x} x} \, dx+\frac {8}{3} \int \frac {e^{\frac {4 \left (2+x^2\right )}{x}}}{x \left (-9 e^{2+x^2}-e^{\frac {8}{x}+4 x} x\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 1.06, size = 34, normalized size = 1.31 \begin {gather*} \frac {1}{3} \left (-x^2+\log \left (9 e^{2+x^2}+e^{\frac {8}{x}+4 x} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + x + 4*x^2 - 2*x^3)/(27*E^((-8 + 2*x - 4*x^2 + x^3)/x)*x + 3*x^2),x]

[Out]

(-x^2 + Log[9*E^(2 + x^2) + E^(8/x + 4*x)*x])/3

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 41, normalized size = 1.58 \begin {gather*} -\frac {x^{3} - 4 \, x^{2} - x \log \left (x + 9 \, e^{\left (\frac {x^{3} - 4 \, x^{2} + 2 \, x - 8}{x}\right )}\right ) - 8}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3+4*x^2+x-8)/(27*x*exp((x^3-4*x^2+2*x-8)/x)+3*x^2),x, algorithm="fricas")

[Out]

-1/3*(x^3 - 4*x^2 - x*log(x + 9*e^((x^3 - 4*x^2 + 2*x - 8)/x)) - 8)/x

________________________________________________________________________________________

giac [A]  time = 0.19, size = 41, normalized size = 1.58 \begin {gather*} -\frac {x^{3} - 4 \, x^{2} - x \log \left (x + 9 \, e^{\left (\frac {x^{3} - 4 \, x^{2} + 2 \, x - 8}{x}\right )}\right ) - 8}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3+4*x^2+x-8)/(27*x*exp((x^3-4*x^2+2*x-8)/x)+3*x^2),x, algorithm="giac")

[Out]

-1/3*(x^3 - 4*x^2 - x*log(x + 9*e^((x^3 - 4*x^2 + 2*x - 8)/x)) - 8)/x

________________________________________________________________________________________

maple [A]  time = 0.14, size = 40, normalized size = 1.54

method	result	size
risch	\(-\frac {x^{3}-4 x^{2}+2 x -8}{3 x}+\frac {\ln \left (\frac {x}{9}+{\mathrm e}^{\frac {\left (x -4\right ) \left (x^{2}+2\right )}{x}}\right )}{3}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^3+4*x^2+x-8)/(27*x*exp((x^3-4*x^2+2*x-8)/x)+3*x^2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(x^3-4*x^2+2*x-8)/x+1/3*ln(1/9*x+exp((x-4)*(x^2+2)/x))

________________________________________________________________________________________

maxima [A]  time = 0.55, size = 34, normalized size = 1.31 \begin {gather*} -\frac {1}{3} \, x^{2} + \frac {1}{3} \, \log \left (\frac {1}{9} \, {\left (x e^{\left (4 \, x + \frac {8}{x}\right )} + 9 \, e^{\left (x^{2} + 2\right )}\right )} e^{\left (-2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3+4*x^2+x-8)/(27*x*exp((x^3-4*x^2+2*x-8)/x)+3*x^2),x, algorithm="maxima")

[Out]

-1/3*x^2 + 1/3*log(1/9*(x*e^(4*x + 8/x) + 9*e^(x^2 + 2))*e^(-2))

________________________________________________________________________________________

mupad [B]  time = 1.62, size = 37, normalized size = 1.42 \begin {gather*} \frac {4\,x}{3}+\frac {\ln \left (x+9\,{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^2\,{\mathrm {e}}^{-\frac {8}{x}}\right )}{3}+\frac {8}{3\,x}-\frac {x^2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 4*x^2 - 2*x^3 - 8)/(27*x*exp((2*x - 4*x^2 + x^3 - 8)/x) + 3*x^2),x)

[Out]

(4*x)/3 + log(x + 9*exp(-4*x)*exp(x^2)*exp(2)*exp(-8/x))/3 + 8/(3*x) - x^2/3

________________________________________________________________________________________

sympy [B]  time = 0.23, size = 37, normalized size = 1.42 \begin {gather*} - \frac {x^{2}}{3} + \frac {4 x}{3} + \frac {\log {\left (\frac {x}{9} + e^{\frac {x^{3} - 4 x^{2} + 2 x - 8}{x}} \right )}}{3} + \frac {8}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**3+4*x**2+x-8)/(27*x*exp((x**3-4*x**2+2*x-8)/x)+3*x**2),x)

[Out]

-x**2/3 + 4*x/3 + log(x/9 + exp((x**3 - 4*x**2 + 2*x - 8)/x))/3 + 8/(3*x)

________________________________________________________________________________________