3.31.58 \(\int \frac {e^{\frac {-2 e^{e^x}+2 x-2 x^2}{3 x+3 x^2}} (-4 x^2+e^{e^x} (2+4 x+e^x (-2 x-2 x^2)))}{3 x^2+6 x^3+3 x^4} \, dx\)

Optimal. Leaf size=29 \[ 3+e^{\frac {2 \left (1-\frac {e^{e^x}+x^2}{x}\right )}{3 (1+x)}} \]

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Rubi [F]  time = 12.87, 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 {e^{\frac {-2 e^{e^x}+2 x-2 x^2}{3 x+3 x^2}} \left (-4 x^2+e^{e^x} \left (2+4 x+e^x \left (-2 x-2 x^2\right )\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-2*E^E^x + 2*x - 2*x^2)/(3*x + 3*x^2))*(-4*x^2 + E^E^x*(2 + 4*x + E^x*(-2*x - 2*x^2))))/(3*x^2 + 6*x^
3 + 3*x^4),x]

[Out]

(-2*Defer[Int][E^(E^x + x - (2*(E^E^x - x + x^2))/(x*(3 + 3*x)))/(-1 - x), x])/3 + (2*Defer[Int][E^(E^x - (2*(
E^E^x - x + x^2))/(x*(3 + 3*x)))/x^2, x])/3 - (2*Defer[Int][E^(E^x + x - (2*(E^E^x - x + x^2))/(x*(3 + 3*x)))/
x, x])/3 - (4*Defer[Int][1/(E^((2*(E^E^x - x + x^2))/(x*(3 + 3*x)))*(1 + x)^2), x])/3 - (2*Defer[Int][E^(E^x -
(2*(E^E^x - x + x^2))/(x*(3 + 3*x)))/(1 + x)^2, x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {-2 e^{e^x}+2 x-2 x^2}{3 x+3 x^2}} \left (-4 x^2+e^{e^x} \left (2+4 x+e^x \left (-2 x-2 x^2\right )\right )\right )}{x^2 \left (3+6 x+3 x^2\right )} \, dx\\ &=\int \frac {e^{\frac {-2 e^{e^x}+2 x-2 x^2}{3 x+3 x^2}} \left (-4 x^2+e^{e^x} \left (2+4 x+e^x \left (-2 x-2 x^2\right )\right )\right )}{3 x^2 (1+x)^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {-2 e^{e^x}+2 x-2 x^2}{3 x+3 x^2}} \left (-4 x^2+e^{e^x} \left (2+4 x+e^x \left (-2 x-2 x^2\right )\right )\right )}{x^2 (1+x)^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}} \left (-4 x^2+e^{e^x} \left (2+4 x+e^x \left (-2 x-2 x^2\right )\right )\right )}{x^2 (1+x)^2} \, dx\\ &=\frac {1}{3} \int \left (-\frac {2 e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{x (1+x)}-\frac {2 e^{-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}} \left (-e^{e^x}-2 e^{e^x} x+2 x^2\right )}{x^2 (1+x)^2}\right ) \, dx\\ &=-\left (\frac {2}{3} \int \frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{x (1+x)} \, dx\right )-\frac {2}{3} \int \frac {e^{-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}} \left (-e^{e^x}-2 e^{e^x} x+2 x^2\right )}{x^2 (1+x)^2} \, dx\\ &=-\left (\frac {2}{3} \int \left (\frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{-1-x}+\frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{x}\right ) \, dx\right )-\frac {2}{3} \int \left (\frac {2 e^{-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{(1+x)^2}-\frac {e^{e^x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}} (1+2 x)}{x^2 (1+x)^2}\right ) \, dx\\ &=-\left (\frac {2}{3} \int \frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{-1-x} \, dx\right )-\frac {2}{3} \int \frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{x} \, dx+\frac {2}{3} \int \frac {e^{e^x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}} (1+2 x)}{x^2 (1+x)^2} \, dx-\frac {4}{3} \int \frac {e^{-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{(1+x)^2} \, dx\\ &=-\left (\frac {2}{3} \int \frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{-1-x} \, dx\right )-\frac {2}{3} \int \frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{x} \, dx+\frac {2}{3} \int \left (\frac {e^{e^x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{x^2}-\frac {e^{e^x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{(1+x)^2}\right ) \, dx-\frac {4}{3} \int \frac {e^{-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{(1+x)^2} \, dx\\ &=-\left (\frac {2}{3} \int \frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{-1-x} \, dx\right )+\frac {2}{3} \int \frac {e^{e^x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{x^2} \, dx-\frac {2}{3} \int \frac {e^{e^x+x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{x} \, dx-\frac {2}{3} \int \frac {e^{e^x-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{(1+x)^2} \, dx-\frac {4}{3} \int \frac {e^{-\frac {2 \left (e^{e^x}-x+x^2\right )}{x (3+3 x)}}}{(1+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.75, size = 25, normalized size = 0.86 \begin {gather*} e^{-\frac {2 \left (e^{e^x}+(-1+x) x\right )}{3 x (1+x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-2*E^E^x + 2*x - 2*x^2)/(3*x + 3*x^2))*(-4*x^2 + E^E^x*(2 + 4*x + E^x*(-2*x - 2*x^2))))/(3*x^2
+ 6*x^3 + 3*x^4),x]

[Out]

E^((-2*(E^E^x + (-1 + x)*x))/(3*x*(1 + x)))

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fricas [A]  time = 0.67, size = 20, normalized size = 0.69 \begin {gather*} e^{\left (-\frac {2 \, {\left (x^{2} - x + e^{\left (e^{x}\right )}\right )}}{3 \, {\left (x^{2} + x\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(x)+4*x+2)*exp(exp(x))-4*x^2)*exp((-2*exp(exp(x))-2*x^2+2*x)/(3*x^2+3*x))/(3*x^4+6
*x^3+3*x^2),x, algorithm="fricas")

[Out]

e^(-2/3*(x^2 - x + e^(e^x))/(x^2 + x))

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giac [A]  time = 0.36, size = 36, normalized size = 1.24 \begin {gather*} e^{\left (-\frac {2 \, x^{2}}{3 \, {\left (x^{2} + x\right )}} + \frac {2 \, x}{3 \, {\left (x^{2} + x\right )}} - \frac {2 \, e^{\left (e^{x}\right )}}{3 \, {\left (x^{2} + x\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(x)+4*x+2)*exp(exp(x))-4*x^2)*exp((-2*exp(exp(x))-2*x^2+2*x)/(3*x^2+3*x))/(3*x^4+6
*x^3+3*x^2),x, algorithm="giac")

[Out]

e^(-2/3*x^2/(x^2 + x) + 2/3*x/(x^2 + x) - 2/3*e^(e^x)/(x^2 + x))

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maple [A]  time = 0.09, size = 22, normalized size = 0.76

method	result	size
risch	\({\mathrm e}^{-\frac {2 \left (x^{2}+{\mathrm e}^{{\mathrm e}^{x}}-x \right )}{3 \left (x +1\right ) x}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2-2*x)*exp(x)+4*x+2)*exp(exp(x))-4*x^2)*exp((-2*exp(exp(x))-2*x^2+2*x)/(3*x^2+3*x))/(3*x^4+6*x^3+3
*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.91, size = 28, normalized size = 0.97 \begin {gather*} e^{\left (\frac {2 \, e^{\left (e^{x}\right )}}{3 \, {\left (x + 1\right )}} - \frac {2 \, e^{\left (e^{x}\right )}}{3 \, x} + \frac {4}{3 \, {\left (x + 1\right )}} - \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-2*x)*exp(x)+4*x+2)*exp(exp(x))-4*x^2)*exp((-2*exp(exp(x))-2*x^2+2*x)/(3*x^2+3*x))/(3*x^4+6
*x^3+3*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.26, size = 39, normalized size = 1.34 \begin {gather*} {\mathrm {e}}^{\frac {2}{3\,x+3}}\,{\mathrm {e}}^{-\frac {2\,x}{3\,x+3}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{{\mathrm {e}}^x}}{3\,x^2+3\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(2*exp(exp(x)) - 2*x + 2*x^2)/(3*x + 3*x^2))*(exp(exp(x))*(4*x - exp(x)*(2*x + 2*x^2) + 2) - 4*x^2))
/(3*x^2 + 6*x^3 + 3*x^4),x)

[Out]

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

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sympy [A]  time = 0.60, size = 24, normalized size = 0.83 \begin {gather*} e^{\frac {- 2 x^{2} + 2 x - 2 e^{e^{x}}}{3 x^{2} + 3 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2-2*x)*exp(x)+4*x+2)*exp(exp(x))-4*x**2)*exp((-2*exp(exp(x))-2*x**2+2*x)/(3*x**2+3*x))/(3*x
**4+6*x**3+3*x**2),x)

[Out]

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

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