3.79.56 \(\int \frac {e^{4-\frac {2 e^4}{x}} (32 e^4+16 e^7+2 e^{10})}{9 x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{9} e^{4-\frac {2 e^4}{x}} \left (4+e^3\right )^2 \]

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 2209} \begin {gather*} \frac {1}{9} \left (4+e^3\right )^2 e^{4-\frac {2 e^4}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4 - (2*E^4)/x)*(32*E^4 + 16*E^7 + 2*E^10))/(9*x^2),x]

[Out]

(E^(4 - (2*E^4)/x)*(4 + E^3)^2)/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \begin {gather*} \frac {1}{9} e^{4-\frac {2 e^4}{x}} \left (4+e^3\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4 - (2*E^4)/x)*(32*E^4 + 16*E^7 + 2*E^10))/(9*x^2),x]

[Out]

(E^(4 - (2*E^4)/x)*(4 + E^3)^2)/9

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fricas [A]  time = 0.60, size = 24, normalized size = 1.04 \begin {gather*} \frac {1}{9} \, {\left (e^{6} + 8 \, e^{3} + 16\right )} e^{\left (\frac {2 \, {\left (2 \, x - e^{4}\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(2*exp(2)^2*exp(3)^2+16*exp(2)^2*exp(3)+32*exp(2)^2)*exp(4)/x^2/exp(exp(4)/x)^2,x, algorithm="fr
icas")

[Out]

1/9*(e^6 + 8*e^3 + 16)*e^(2*(2*x - e^4)/x)

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giac [A]  time = 0.12, size = 21, normalized size = 0.91 \begin {gather*} \frac {1}{9} \, {\left (e^{10} + 8 \, e^{7} + 16 \, e^{4}\right )} e^{\left (-\frac {2 \, e^{4}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(2*exp(2)^2*exp(3)^2+16*exp(2)^2*exp(3)+32*exp(2)^2)*exp(4)/x^2/exp(exp(4)/x)^2,x, algorithm="gi
ac")

[Out]

1/9*(e^10 + 8*e^7 + 16*e^4)*e^(-2*e^4/x)

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maple [A]  time = 0.04, size = 26, normalized size = 1.13

method	result	size
gosper	\(\frac {{\mathrm e}^{4} \left ({\mathrm e}^{6}+8 \,{\mathrm e}^{3}+16\right ) {\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}}{9}\)	\(26\)
norman	\(\frac {{\mathrm e}^{4} \left ({\mathrm e}^{6}+8 \,{\mathrm e}^{3}+16\right ) {\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}}{9}\)	\(26\)
default	\(\frac {\left ({\mathrm e}^{4} {\mathrm e}^{6}+8 \,{\mathrm e}^{3} {\mathrm e}^{4}+16 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}}{9}\)	\(36\)
derivativedivides	\(-\frac {\left (-\frac {2 \,{\mathrm e}^{4} {\mathrm e}^{6}}{9}-\frac {16 \,{\mathrm e}^{3} {\mathrm e}^{4}}{9}-\frac {32 \,{\mathrm e}^{4}}{9}\right ) {\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}}{2}\)	\(37\)
risch	\(\frac {{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}} {\mathrm e}^{10}}{9}+\frac {8 \,{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}} {\mathrm e}^{7}}{9}+\frac {16 \,{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}} {\mathrm e}^{4}}{9}\)	\(38\)
meijerg	\(-\frac {{\mathrm e}^{10} \left (1-{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}\right )}{9}-\frac {8 \,{\mathrm e}^{7} \left (1-{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}\right )}{9}-\frac {16 \,{\mathrm e}^{4} \left (1-{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}\right )}{9}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(2*exp(2)^2*exp(3)^2+16*exp(2)^2*exp(3)+32*exp(2)^2)*exp(4)/x^2/exp(exp(4)/x)^2,x,method=_RETURNVERBOS
E)

[Out]

1/9*exp(2)^2*(exp(3)^2+8*exp(3)+16)/exp(exp(4)/x)^2

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maxima [A]  time = 0.37, size = 21, normalized size = 0.91 \begin {gather*} \frac {1}{9} \, {\left (e^{10} + 8 \, e^{7} + 16 \, e^{4}\right )} e^{\left (-\frac {2 \, e^{4}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(2*exp(2)^2*exp(3)^2+16*exp(2)^2*exp(3)+32*exp(2)^2)*exp(4)/x^2/exp(exp(4)/x)^2,x, algorithm="ma
xima")

[Out]

1/9*(e^10 + 8*e^7 + 16*e^4)*e^(-2*e^4/x)

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mupad [B]  time = 5.61, size = 18, normalized size = 0.78 \begin {gather*} \frac {{\mathrm {e}}^{4-\frac {2\,{\mathrm {e}}^4}{x}}\,{\left ({\mathrm {e}}^3+4\right )}^2}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(2*exp(4))/x)*exp(4)*(32*exp(4) + 16*exp(7) + 2*exp(10)))/(9*x^2),x)

[Out]

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

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sympy [A]  time = 0.20, size = 22, normalized size = 0.96 \begin {gather*} \frac {\left (16 e^{4} + 8 e^{7} + e^{10}\right ) e^{- \frac {2 e^{4}}{x}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(2*exp(2)**2*exp(3)**2+16*exp(2)**2*exp(3)+32*exp(2)**2)*exp(4)/x**2/exp(exp(4)/x)**2,x)

[Out]

(16*exp(4) + 8*exp(7) + exp(10))*exp(-2*exp(4)/x)/9

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