3.19.67 \(\int \frac {e^{-2+x} (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} (256-256 x-32 e^{2-x} x+64 x^2)+e^{\frac {e^{-2+x} (-8+2 x)}{x}} (256 x^2-256 x^3+64 x^4))}{x^4} \, dx\)

Optimal. Leaf size=26 \[ \frac {16 \left (e^{\frac {2 e^{-2+x} (-4+x)}{x}}+x^2\right )^2}{x^2} \]

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Rubi [F]  time = 2.69, 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^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-2 + x)*(32*E^(2 - x)*x^5 + E^((2*E^(-2 + x)*(-8 + 2*x))/x)*(256 - 256*x - 32*E^(2 - x)*x + 64*x^2) +
E^((E^(-2 + x)*(-8 + 2*x))/x)*(256*x^2 - 256*x^3 + 64*x^4)))/x^4,x]

[Out]

16*x^2 + (16*E^(-2 - (4*E^(-2 + x)*(4 - x))/x)*(4*E^x - 4*E^x*x + E^x*x^2))/(((E^(-2 + x)*(4 - x))/x^2 + E^(-2
+ x)/x - (E^(-2 + x)*(4 - x))/x)*x^4) + 64*Defer[Int][E^(-2 + (2*E^(-2 + x)*(-4 + x))/x + x), x] + 256*Defer[
Int][E^(-2 + (2*E^(-2 + x)*(-4 + x))/x + x)/x^2, x] - 256*Defer[Int][E^(-2 + (2*E^(-2 + x)*(-4 + x))/x + x)/x,
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {64 e^{-2+\frac {2 e^{-2+x} (-4+x)}{x}+x} (-2+x)^2}{x^2}+32 x+\frac {32 e^{-2+\frac {4 e^{-2+x} (-4+x)}{x}} \left (8 e^x-e^2 x-8 e^x x+2 e^x x^2\right )}{x^4}\right ) \, dx\\ &=16 x^2+32 \int \frac {e^{-2+\frac {4 e^{-2+x} (-4+x)}{x}} \left (8 e^x-e^2 x-8 e^x x+2 e^x x^2\right )}{x^4} \, dx+64 \int \frac {e^{-2+\frac {2 e^{-2+x} (-4+x)}{x}+x} (-2+x)^2}{x^2} \, dx\\ &=16 x^2+\frac {16 e^{-2-\frac {4 e^{-2+x} (4-x)}{x}} \left (4 e^x-4 e^x x+e^x x^2\right )}{\left (\frac {e^{-2+x} (4-x)}{x^2}+\frac {e^{-2+x}}{x}-\frac {e^{-2+x} (4-x)}{x}\right ) x^4}+64 \int \left (e^{-2+\frac {2 e^{-2+x} (-4+x)}{x}+x}+\frac {4 e^{-2+\frac {2 e^{-2+x} (-4+x)}{x}+x}}{x^2}-\frac {4 e^{-2+\frac {2 e^{-2+x} (-4+x)}{x}+x}}{x}\right ) \, dx\\ &=16 x^2+\frac {16 e^{-2-\frac {4 e^{-2+x} (4-x)}{x}} \left (4 e^x-4 e^x x+e^x x^2\right )}{\left (\frac {e^{-2+x} (4-x)}{x^2}+\frac {e^{-2+x}}{x}-\frac {e^{-2+x} (4-x)}{x}\right ) x^4}+64 \int e^{-2+\frac {2 e^{-2+x} (-4+x)}{x}+x} \, dx+256 \int \frac {e^{-2+\frac {2 e^{-2+x} (-4+x)}{x}+x}}{x^2} \, dx-256 \int \frac {e^{-2+\frac {2 e^{-2+x} (-4+x)}{x}+x}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 26, normalized size = 1.00 \begin {gather*} \frac {16 \left (e^{\frac {2 e^{-2+x} (-4+x)}{x}}+x^2\right )^2}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 + x)*(32*E^(2 - x)*x^5 + E^((2*E^(-2 + x)*(-8 + 2*x))/x)*(256 - 256*x - 32*E^(2 - x)*x + 64*x
^2) + E^((E^(-2 + x)*(-8 + 2*x))/x)*(256*x^2 - 256*x^3 + 64*x^4)))/x^4,x]

[Out]

(16*(E^((2*E^(-2 + x)*(-4 + x))/x) + x^2)^2)/x^2

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fricas [A]  time = 0.81, size = 44, normalized size = 1.69 \begin {gather*} \frac {16 \, {\left (x^{4} + 2 \, x^{2} e^{\left (\frac {2 \, {\left (x - 4\right )} e^{\left (x + \log \relax (x) - 2\right )}}{x^{2}}\right )} + e^{\left (\frac {4 \, {\left (x - 4\right )} e^{\left (x + \log \relax (x) - 2\right )}}{x^{2}}\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^2*exp(-log(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp(-log(x)+2-x))^2+(64*x^4-256*x^3+256
*x^2)*exp((2*x-8)/x^2/exp(-log(x)+2-x))+32*x^6*exp(-log(x)+2-x))/x^5/exp(-log(x)+2-x),x, algorithm="fricas")

[Out]

16*(x^4 + 2*x^2*e^(2*(x - 4)*e^(x + log(x) - 2)/x^2) + e^(4*(x - 4)*e^(x + log(x) - 2)/x^2))/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {32 \, {\left (x^{6} e^{\left (-x - \log \relax (x) + 2\right )} - {\left (x^{2} e^{\left (-x - \log \relax (x) + 2\right )} - 2 \, x^{2} + 8 \, x - 8\right )} e^{\left (\frac {4 \, {\left (x - 4\right )} e^{\left (x + \log \relax (x) - 2\right )}}{x^{2}}\right )} + 2 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )} e^{\left (\frac {2 \, {\left (x - 4\right )} e^{\left (x + \log \relax (x) - 2\right )}}{x^{2}}\right )}\right )} e^{\left (x + \log \relax (x) - 2\right )}}{x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^2*exp(-log(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp(-log(x)+2-x))^2+(64*x^4-256*x^3+256
*x^2)*exp((2*x-8)/x^2/exp(-log(x)+2-x))+32*x^6*exp(-log(x)+2-x))/x^5/exp(-log(x)+2-x),x, algorithm="giac")

[Out]

integrate(32*(x^6*e^(-x - log(x) + 2) - (x^2*e^(-x - log(x) + 2) - 2*x^2 + 8*x - 8)*e^(4*(x - 4)*e^(x + log(x)
- 2)/x^2) + 2*(x^4 - 4*x^3 + 4*x^2)*e^(2*(x - 4)*e^(x + log(x) - 2)/x^2))*e^(x + log(x) - 2)/x^5, x)

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maple [A]  time = 0.10, size = 40, normalized size = 1.54

method	result	size
risch	\(16 x^{2}+\frac {16 \,{\mathrm e}^{\frac {4 \left (x -4\right ) {\mathrm e}^{x -2}}{x}}}{x^{2}}+32 \,{\mathrm e}^{\frac {2 \left (x -4\right ) {\mathrm e}^{x -2}}{x}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x^2*exp(-ln(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp(-ln(x)+2-x))^2+(64*x^4-256*x^3+256*x^2)*ex
p((2*x-8)/x^2/exp(-ln(x)+2-x))+32*x^6*exp(-ln(x)+2-x))/x^5/exp(-ln(x)+2-x),x,method=_RETURNVERBOSE)

[Out]

16*x^2+16/x^2*exp(4*(x-4)*exp(x-2)/x)+32*exp(2*(x-4)*exp(x-2)/x)

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maxima [A]  time = 0.69, size = 51, normalized size = 1.96 \begin {gather*} 16 \, x^{2} + \frac {16 \, {\left (2 \, x^{2} e^{\left (-\frac {8 \, e^{\left (x - 2\right )}}{x} + 2 \, e^{\left (x - 2\right )}\right )} + e^{\left (-\frac {16 \, e^{\left (x - 2\right )}}{x} + 4 \, e^{\left (x - 2\right )}\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^2*exp(-log(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp(-log(x)+2-x))^2+(64*x^4-256*x^3+256
*x^2)*exp((2*x-8)/x^2/exp(-log(x)+2-x))+32*x^6*exp(-log(x)+2-x))/x^5/exp(-log(x)+2-x),x, algorithm="maxima")

[Out]

16*x^2 + 16*(2*x^2*e^(-8*e^(x - 2)/x + 2*e^(x - 2)) + e^(-16*e^(x - 2)/x + 4*e^(x - 2)))/x^2

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mupad [B]  time = 1.26, size = 47, normalized size = 1.81 \begin {gather*} 32\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {8\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x}}+16\,x^2+\frac {16\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {16\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x}}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(x) - 2)*(32*x^6*exp(2 - log(x) - x) + exp((exp(x + log(x) - 2)*(2*x - 8))/x^2)*(256*x^2 - 256
*x^3 + 64*x^4) - exp((2*exp(x + log(x) - 2)*(2*x - 8))/x^2)*(256*x + 32*x^2*exp(2 - log(x) - x) - 64*x^2 - 256
)))/x^5,x)

[Out]

32*exp(2*exp(-2)*exp(x))*exp(-(8*exp(-2)*exp(x))/x) + 16*x^2 + (16*exp(4*exp(-2)*exp(x))*exp(-(16*exp(-2)*exp(
x))/x))/x^2

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sympy [A]  time = 0.27, size = 42, normalized size = 1.62 \begin {gather*} 16 x^{2} + \frac {32 x^{2} e^{\frac {\left (2 x - 8\right ) e^{x - 2}}{x}} + 16 e^{\frac {2 \left (2 x - 8\right ) e^{x - 2}}{x}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x**2*exp(-ln(x)+2-x)+64*x**2-256*x+256)*exp((2*x-8)/x**2/exp(-ln(x)+2-x))**2+(64*x**4-256*x**3
+256*x**2)*exp((2*x-8)/x**2/exp(-ln(x)+2-x))+32*x**6*exp(-ln(x)+2-x))/x**5/exp(-ln(x)+2-x),x)

[Out]

16*x**2 + (32*x**2*exp((2*x - 8)*exp(x - 2)/x) + 16*exp(2*(2*x - 8)*exp(x - 2)/x))/x**2

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