\(\int \frac {e^{4-\frac {2 e^4}{x}} (32 e^4+16 e^7+2 e^{10})}{9 x^2} \, dx\) [7856]

Optimal result

Integrand size = 35, antiderivative size = 23 \[ \int \frac {e^{4-\frac {2 e^4}{x}} \left (32 e^4+16 e^7+2 e^{10}\right )}{9 x^2} \, dx=\frac {1}{9} e^{4-\frac {2 e^4}{x}} \left (4+e^3\right )^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 2240} \[ \int \frac {e^{4-\frac {2 e^4}{x}} \left (32 e^4+16 e^7+2 e^{10}\right )}{9 x^2} \, dx=\frac {1}{9} \left (4+e^3\right )^2 e^{4-\frac {2 e^4}{x}} \]

[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 2240

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4-\frac {2 e^4}{x}} \left (32 e^4+16 e^7+2 e^{10}\right )}{9 x^2} \, dx=\frac {1}{9} e^{4-\frac {2 e^4}{x}} \left (4+e^3\right )^2 \]

[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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 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	\(-\frac {\left (-\frac {2 \,{\mathrm e}^{4} {\mathrm e}^{6}}{9}-\frac {16 \,{\mathrm e}^{3} {\mathrm e}^{4}}{9}-\frac {32 \,{\mathrm e}^{4}}{9}\right ) {\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}}{2}\)	\(37\)
parallelrisch	\(\frac {\left (2 \,{\mathrm e}^{4} {\mathrm e}^{6}+16 \,{\mathrm e}^{3} {\mathrm e}^{4}+32 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}}{18}\)	\(37\)
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	\(-\frac {{\mathrm e}^{10} \left (1-{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}\right )}{9}-\frac {8 \,{\mathrm e}^{7} \left (1-{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}\right )}{9}-\frac {16 \,{\mathrm e}^{4} \left (1-{\mathrm e}^{-\frac {2 \,{\mathrm e}^{4}}{x}}\right )}{9}\)	\(50\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{4-\frac {2 e^4}{x}} \left (32 e^4+16 e^7+2 e^{10}\right )}{9 x^2} \, dx=\frac {1}{9} \, {\left (e^{6} + 8 \, e^{3} + 16\right )} e^{\left (\frac {2 \, {\left (2 \, x - e^{4}\right )}}{x}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4-\frac {2 e^4}{x}} \left (32 e^4+16 e^7+2 e^{10}\right )}{9 x^2} \, dx=\frac {\left (16 e^{4} + 8 e^{7} + e^{10}\right ) e^{- \frac {2 e^{4}}{x}}}{9} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{4-\frac {2 e^4}{x}} \left (32 e^4+16 e^7+2 e^{10}\right )}{9 x^2} \, dx=\frac {1}{9} \, {\left (e^{10} + 8 \, e^{7} + 16 \, e^{4}\right )} e^{\left (-\frac {2 \, e^{4}}{x}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{4-\frac {2 e^4}{x}} \left (32 e^4+16 e^7+2 e^{10}\right )}{9 x^2} \, dx=\frac {1}{9} \, {\left (e^{10} + 8 \, e^{7} + 16 \, e^{4}\right )} e^{\left (-\frac {2 \, e^{4}}{x}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 12.40 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{4-\frac {2 e^4}{x}} \left (32 e^4+16 e^7+2 e^{10}\right )}{9 x^2} \, dx=\frac {{\mathrm {e}}^{4-\frac {2\,{\mathrm {e}}^4}{x}}\,{\left ({\mathrm {e}}^3+4\right )}^2}{9} \]

[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