3.59.91 \(\int \frac {e^{-\frac {2 (-5+x^2)}{x}} (5 e^{\frac {2 (-5+x^2)}{x}} x^2+e^{32 e^{-\frac {2 (-5+x^2)}{x}}} (320+64 x^2)+e^{16 e^{-\frac {2 (-5+x^2)}{x}}} (960+192 x^2))}{x^2} \, dx\) [5891]

Optimal. Leaf size=27 \[ -2-\left (3+e^{16 e^{-\frac {2 \left (-5+x^2\right )}{x}}}\right )^2+5 x \]

[Out]

-2+5*x-(exp(16/exp((x^2-5)/x)^2)+3)^2

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Rubi [A]
time = 0.76, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 2, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6874, 6818} \begin {gather*} 5 x-\left (e^{16 e^{\frac {10}{x}-2 x}}+3\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^((2*(-5 + x^2))/x)*x^2 + E^(32/E^((2*(-5 + x^2))/x))*(320 + 64*x^2) + E^(16/E^((2*(-5 + x^2))/x))*(96
0 + 192*x^2))/(E^((2*(-5 + x^2))/x)*x^2),x]

[Out]

-(3 + E^(16*E^(10/x - 2*x)))^2 + 5*x

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (5+\frac {64 \exp \left (16 e^{-\frac {2 \left (-5+x^2\right )}{x}}-\frac {2 \left (-5+x^2\right )}{x}\right ) \left (3+e^{16 e^{\frac {10}{x}-2 x}}\right ) \left (5+x^2\right )}{x^2}\right ) \, dx\\ &=5 x+64 \int \frac {\exp \left (16 e^{-\frac {2 \left (-5+x^2\right )}{x}}-\frac {2 \left (-5+x^2\right )}{x}\right ) \left (3+e^{16 e^{\frac {10}{x}-2 x}}\right ) \left (5+x^2\right )}{x^2} \, dx\\ &=-\left (3+e^{16 e^{\frac {10}{x}-2 x}}\right )^2+5 x\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.19, size = 38, normalized size = 1.41 \begin {gather*} -6 e^{16 e^{\frac {10}{x}-2 x}}-e^{32 e^{\frac {10}{x}-2 x}}+5 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^((2*(-5 + x^2))/x)*x^2 + E^(32/E^((2*(-5 + x^2))/x))*(320 + 64*x^2) + E^(16/E^((2*(-5 + x^2))/x
))*(960 + 192*x^2))/(E^((2*(-5 + x^2))/x)*x^2),x]

[Out]

-6*E^(16*E^(10/x - 2*x)) - E^(32*E^(10/x - 2*x)) + 5*x

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Maple [A]
time = 0.67, size = 37, normalized size = 1.37

method	result	size
risch	\(-{\mathrm e}^{32 \,{\mathrm e}^{-\frac {2 \left (x^{2}-5\right )}{x}}}+5 x -6 \,{\mathrm e}^{16 \,{\mathrm e}^{-\frac {2 \left (x^{2}-5\right )}{x}}}\)	\(37\)
norman	\(\frac {\left (5 x^{2} {\mathrm e}^{\frac {2 x^{2}-10}{x}}-x \,{\mathrm e}^{\frac {2 x^{2}-10}{x}} {\mathrm e}^{32 \,{\mathrm e}^{-\frac {2 \left (x^{2}-5\right )}{x}}}-6 \,{\mathrm e}^{\frac {2 x^{2}-10}{x}} {\mathrm e}^{16 \,{\mathrm e}^{-\frac {2 \left (x^{2}-5\right )}{x}}} x \right ) {\mathrm e}^{-\frac {2 \left (x^{2}-5\right )}{x}}}{x}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((64*x^2+320)*exp(16/exp((x^2-5)/x)^2)^2+(192*x^2+960)*exp(16/exp((x^2-5)/x)^2)+5*x^2*exp((x^2-5)/x)^2)/x^
2/exp((x^2-5)/x)^2,x,method=_RETURNVERBOSE)

[Out]

-exp(32*exp(-2*(x^2-5)/x))+5*x-6*exp(16*exp(-2*(x^2-5)/x))

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Maxima [A]
time = 0.37, size = 34, normalized size = 1.26 \begin {gather*} 5 \, x - e^{\left (32 \, e^{\left (-2 \, x + \frac {10}{x}\right )}\right )} - 6 \, e^{\left (16 \, e^{\left (-2 \, x + \frac {10}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+320)*exp(16/exp((x^2-5)/x)^2)^2+(192*x^2+960)*exp(16/exp((x^2-5)/x)^2)+5*x^2*exp((x^2-5)/x)
^2)/x^2/exp((x^2-5)/x)^2,x, algorithm="maxima")

[Out]

5*x - e^(32*e^(-2*x + 10/x)) - 6*e^(16*e^(-2*x + 10/x))

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Fricas [A]
time = 0.47, size = 36, normalized size = 1.33 \begin {gather*} 5 \, x - e^{\left (32 \, e^{\left (-\frac {2 \, {\left (x^{2} - 5\right )}}{x}\right )}\right )} - 6 \, e^{\left (16 \, e^{\left (-\frac {2 \, {\left (x^{2} - 5\right )}}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+320)*exp(16/exp((x^2-5)/x)^2)^2+(192*x^2+960)*exp(16/exp((x^2-5)/x)^2)+5*x^2*exp((x^2-5)/x)
^2)/x^2/exp((x^2-5)/x)^2,x, algorithm="fricas")

[Out]

5*x - e^(32*e^(-2*(x^2 - 5)/x)) - 6*e^(16*e^(-2*(x^2 - 5)/x))

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Sympy [A]
time = 0.14, size = 31, normalized size = 1.15 \begin {gather*} 5 x - e^{32 e^{- \frac {2 \left (x^{2} - 5\right )}{x}}} - 6 e^{16 e^{- \frac {2 \left (x^{2} - 5\right )}{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x**2+320)*exp(16/exp((x**2-5)/x)**2)**2+(192*x**2+960)*exp(16/exp((x**2-5)/x)**2)+5*x**2*exp((x
**2-5)/x)**2)/x**2/exp((x**2-5)/x)**2,x)

[Out]

5*x - exp(32*exp(-2*(x**2 - 5)/x)) - 6*exp(16*exp(-2*(x**2 - 5)/x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+320)*exp(16/exp((x^2-5)/x)^2)^2+(192*x^2+960)*exp(16/exp((x^2-5)/x)^2)+5*x^2*exp((x^2-5)/x)
^2)/x^2/exp((x^2-5)/x)^2,x, algorithm="giac")

[Out]

integrate((5*x^2*e^(2*(x^2 - 5)/x) + 64*(x^2 + 5)*e^(32*e^(-2*(x^2 - 5)/x)) + 192*(x^2 + 5)*e^(16*e^(-2*(x^2 -
5)/x)))*e^(-2*(x^2 - 5)/x)/x^2, x)

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Mupad [B]
time = 4.28, size = 34, normalized size = 1.26 \begin {gather*} 5\,x-6\,{\mathrm {e}}^{16\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{10/x}}-{\mathrm {e}}^{32\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{10/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(2*(x^2 - 5))/x)*(5*x^2*exp((2*(x^2 - 5))/x) + exp(32*exp(-(2*(x^2 - 5))/x))*(64*x^2 + 320) + exp(16
*exp(-(2*(x^2 - 5))/x))*(192*x^2 + 960)))/x^2,x)

[Out]

5*x - 6*exp(16*exp(-2*x)*exp(10/x)) - exp(32*exp(-2*x)*exp(10/x))

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