3.12.97 \(\int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log (x^2)}{2+x}} (-8 x+24 x^2+30 x^3+8 x^4+e^8 (64+32 x)-16 e^8 x \log (x^2))}{4 x+4 x^2+x^3} \, dx\)

Optimal. Leaf size=27 \[ 2 e^{-x+2 x^2+\frac {8 e^8 \log \left (x^2\right )}{2+x}} \]

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Rubi [F]  time = 5.31, 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 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \left (-8 x+24 x^2+30 x^3+8 x^4+e^8 (64+32 x)-16 e^8 x \log \left (x^2\right )\right )}{4 x+4 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-2*x + 3*x^2 + 2*x^3 + 8*E^8*Log[x^2])/(2 + x))*(-8*x + 24*x^2 + 30*x^3 + 8*x^4 + E^8*(64 + 32*x) - 1
6*E^8*x*Log[x^2]))/(4*x + 4*x^2 + x^3),x]

[Out]

-2*Defer[Int][E^((-2*x + 3*x^2 + 2*x^3 + 8*E^8*Log[x^2])/(2 + x)), x] + 16*Defer[Int][E^((16 + 6*x + 3*x^2 + 2
*x^3 + 8*E^8*Log[x^2])/(2 + x))/x, x] + 8*Defer[Int][E^((-2*x + 3*x^2 + 2*x^3 + 8*E^8*Log[x^2])/(2 + x))*x, x]
- 16*Defer[Int][E^((16 + 6*x + 3*x^2 + 2*x^3 + 8*E^8*Log[x^2])/(2 + x))/(2 + x), x] - 16*Defer[Int][(E^((16 +
6*x + 3*x^2 + 2*x^3 + 8*E^8*Log[x^2])/(2 + x))*Log[x^2])/(2 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \left (-8 x+24 x^2+30 x^3+8 x^4+e^8 (64+32 x)-16 e^8 x \log \left (x^2\right )\right )}{x \left (4+4 x+x^2\right )} \, dx\\ &=\int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \left (-8 x+24 x^2+30 x^3+8 x^4+e^8 (64+32 x)-16 e^8 x \log \left (x^2\right )\right )}{x (2+x)^2} \, dx\\ &=\int \left (-\frac {8 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2}+\frac {24 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x}{(2+x)^2}+\frac {30 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x^2}{(2+x)^2}+\frac {8 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x^3}{(2+x)^2}+\frac {32 \exp \left (8+\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}\right )}{x (2+x)}-\frac {16 \exp \left (8+\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}\right ) \log \left (x^2\right )}{(2+x)^2}\right ) \, dx\\ &=-\left (8 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx\right )+8 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x^3}{(2+x)^2} \, dx-16 \int \frac {\exp \left (8+\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}\right ) \log \left (x^2\right )}{(2+x)^2} \, dx+24 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x}{(2+x)^2} \, dx+30 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x^2}{(2+x)^2} \, dx+32 \int \frac {\exp \left (8+\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}\right )}{x (2+x)} \, dx\\ &=-\left (8 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx\right )+8 \int \left (-4 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}+e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x-\frac {8 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2}+\frac {12 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x}\right ) \, dx-16 \int \frac {e^{\frac {16+6 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \log \left (x^2\right )}{(2+x)^2} \, dx+24 \int \left (-\frac {2 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2}+\frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x}\right ) \, dx+30 \int \left (e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}+\frac {4 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2}-\frac {4 e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x}\right ) \, dx+32 \int \frac {e^{\frac {16+6 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{x (2+x)} \, dx\\ &=8 \int e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x \, dx-8 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx-16 \int \frac {e^{\frac {16+6 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \log \left (x^2\right )}{(2+x)^2} \, dx+24 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x} \, dx+30 \int e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \, dx-32 \int e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \, dx+32 \int \left (\frac {e^{\frac {16+6 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2 x}-\frac {e^{\frac {16+6 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2 (2+x)}\right ) \, dx-48 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx-64 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx+96 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x} \, dx+120 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx-120 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x} \, dx\\ &=8 \int e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} x \, dx-8 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx+16 \int \frac {e^{\frac {16+6 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{x} \, dx-16 \int \frac {e^{\frac {16+6 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x} \, dx-16 \int \frac {e^{\frac {16+6 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \log \left (x^2\right )}{(2+x)^2} \, dx+24 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x} \, dx+30 \int e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \, dx-32 \int e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}} \, dx-48 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx-64 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx+96 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x} \, dx+120 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{(2+x)^2} \, dx-120 \int \frac {e^{\frac {-2 x+3 x^2+2 x^3+8 e^8 \log \left (x^2\right )}{2+x}}}{2+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 27, normalized size = 1.00 \begin {gather*} 2 e^{-x+2 x^2} \left (x^2\right )^{\frac {8 e^8}{2+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-2*x + 3*x^2 + 2*x^3 + 8*E^8*Log[x^2])/(2 + x))*(-8*x + 24*x^2 + 30*x^3 + 8*x^4 + E^8*(64 + 32*
x) - 16*E^8*x*Log[x^2]))/(4*x + 4*x^2 + x^3),x]

[Out]

2*E^(-x + 2*x^2)*(x^2)^((8*E^8)/(2 + x))

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fricas [A]  time = 0.66, size = 31, normalized size = 1.15 \begin {gather*} 2 \, e^{\left (\frac {2 \, x^{3} + 3 \, x^{2} + 8 \, e^{8} \log \left (x^{2}\right ) - 2 \, x}{x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(8)*log(x^2)+(32*x+64)*exp(8)+8*x^4+30*x^3+24*x^2-8*x)*exp((8*exp(8)*log(x^2)+2*x^3+3*x^2-
2*x)/(2+x))/(x^3+4*x^2+4*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.47, size = 45, normalized size = 1.67 \begin {gather*} 2 \, e^{\left (\frac {2 \, x^{3}}{x + 2} + \frac {3 \, x^{2}}{x + 2} + \frac {8 \, e^{8} \log \left (x^{2}\right )}{x + 2} - \frac {2 \, x}{x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(8)*log(x^2)+(32*x+64)*exp(8)+8*x^4+30*x^3+24*x^2-8*x)*exp((8*exp(8)*log(x^2)+2*x^3+3*x^2-
2*x)/(2+x))/(x^3+4*x^2+4*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.28, size = 24, normalized size = 0.89

method	result	size
risch	\(2 \left (x^{2}\right )^{\frac {8 \,{\mathrm e}^{8}}{2+x}} {\mathrm e}^{x \left (2 x -1\right )}\)	\(24\)
default	\(\frac {2 x \,{\mathrm e}^{\frac {8 \,{\mathrm e}^{8} \ln \left (x^{2}\right )+2 x^{3}+3 x^{2}-2 x}{2+x}}+4 \,{\mathrm e}^{\frac {8 \,{\mathrm e}^{8} \ln \left (x^{2}\right )+2 x^{3}+3 x^{2}-2 x}{2+x}}}{2+x}\)	\(71\)
norman	\(\frac {2 x \,{\mathrm e}^{\frac {8 \,{\mathrm e}^{8} \ln \left (x^{2}\right )+2 x^{3}+3 x^{2}-2 x}{2+x}}+4 \,{\mathrm e}^{\frac {8 \,{\mathrm e}^{8} \ln \left (x^{2}\right )+2 x^{3}+3 x^{2}-2 x}{2+x}}}{2+x}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*x*exp(8)*ln(x^2)+(32*x+64)*exp(8)+8*x^4+30*x^3+24*x^2-8*x)*exp((8*exp(8)*ln(x^2)+2*x^3+3*x^2-2*x)/(2+
x))/(x^3+4*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

2*(x^2)^(8*exp(8)/(2+x))*exp(x*(2*x-1))

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maxima [A]  time = 0.56, size = 23, normalized size = 0.85 \begin {gather*} 2 \, e^{\left (2 \, x^{2} - x + \frac {16 \, e^{8} \log \relax (x)}{x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(8)*log(x^2)+(32*x+64)*exp(8)+8*x^4+30*x^3+24*x^2-8*x)*exp((8*exp(8)*log(x^2)+2*x^3+3*x^2-
2*x)/(2+x))/(x^3+4*x^2+4*x),x, algorithm="maxima")

[Out]

2*e^(2*x^2 - x + 16*e^8*log(x)/(x + 2))

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mupad [B]  time = 1.26, size = 46, normalized size = 1.70 \begin {gather*} 2\,{\mathrm {e}}^{-\frac {2\,x}{x+2}}\,{\mathrm {e}}^{\frac {3\,x^2}{x+2}}\,{\mathrm {e}}^{\frac {2\,x^3}{x+2}}\,{\left (x^2\right )}^{\frac {8\,{\mathrm {e}}^8}{x+2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((8*log(x^2)*exp(8) - 2*x + 3*x^2 + 2*x^3)/(x + 2))*(24*x^2 - 8*x + 30*x^3 + 8*x^4 + exp(8)*(32*x + 64
) - 16*x*log(x^2)*exp(8)))/(4*x + 4*x^2 + x^3),x)

[Out]

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

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sympy [A]  time = 0.48, size = 29, normalized size = 1.07 \begin {gather*} 2 e^{\frac {2 x^{3} + 3 x^{2} - 2 x + 8 e^{8} \log {\left (x^{2} \right )}}{x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(8)*ln(x**2)+(32*x+64)*exp(8)+8*x**4+30*x**3+24*x**2-8*x)*exp((8*exp(8)*ln(x**2)+2*x**3+3*
x**2-2*x)/(2+x))/(x**3+4*x**2+4*x),x)

[Out]

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

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