\(\int \frac {e^{\frac {1}{16} (e^9-8 e^5 x+16 e x^2)} (-2-e^5 x+4 e x^2)}{5 x^2} \, dx\) [9629]

Optimal result

Integrand size = 43, antiderivative size = 24 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 e^{e \left (\frac {e^4}{4}-x\right )^2}}{5 x} \]

[Out]

2/5*exp((1/4*exp(2)^2-x)^2*exp(1))/x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(24)=48\).

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 2326} \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 e^{\frac {1}{16} \left (16 e x^2-8 e^5 x+e^9\right )-1} \left (e^5 x-4 e x^2\right )}{5 \left (e^4-4 x\right ) x^2} \]

[In]

Int[(E^((E^9 - 8*E^5*x + 16*E*x^2)/16)*(-2 - E^5*x + 4*E*x^2))/(5*x^2),x]

[Out]

(2*E^(-1 + (E^9 - 8*E^5*x + 16*E*x^2)/16)*(E^5*x - 4*E*x^2))/(5*(E^4 - 4*x)*x^2)

Rule 12

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

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{x^2} \, dx \\ & = \frac {2 e^{-1+\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (e^5 x-4 e x^2\right )}{5 \left (e^4-4 x\right ) x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 e^{\frac {1}{16} e \left (e^4-4 x\right )^2}}{5 x} \]

[In]

Integrate[(E^((E^9 - 8*E^5*x + 16*E*x^2)/16)*(-2 - E^5*x + 4*E*x^2))/(5*x^2),x]

[Out]

(2*E^((E*(E^4 - 4*x)^2)/16))/(5*x)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{\frac {{\mathrm e}^{9}}{16}-\frac {x \,{\mathrm e}^{5}}{2}+x^{2} {\mathrm e}}}{5 x}\)	\(23\)
parallelrisch	\(\frac {2 \,{\mathrm e}^{\frac {{\mathrm e} \left ({\mathrm e}^{8}-8 x \,{\mathrm e}^{4}+16 x^{2}\right )}{16}}}{5 x}\)	\(28\)
gosper	\(\frac {2 \,{\mathrm e}^{\frac {{\mathrm e} \,{\mathrm e}^{8}}{16}-\frac {x \,{\mathrm e} \,{\mathrm e}^{4}}{2}+x^{2} {\mathrm e}}}{5 x}\)	\(31\)
norman	\(\frac {2 \,{\mathrm e}^{\frac {{\mathrm e} \,{\mathrm e}^{8}}{16}-\frac {x \,{\mathrm e} \,{\mathrm e}^{4}}{2}+x^{2} {\mathrm e}}}{5 x}\)	\(31\)

[In]

int(1/5*(-x*exp(1)*exp(2)^2+4*x^2*exp(1)-2)*exp(1/16*exp(1)*exp(2)^4-1/2*x*exp(1)*exp(2)^2+x^2*exp(1))/x^2,x,m
ethod=_RETURNVERBOSE)

[Out]

2/5/x*exp(1/16*exp(9)-1/2*x*exp(5)+x^2*exp(1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 \, e^{\left (x^{2} e - \frac {1}{2} \, x e^{5} + \frac {1}{16} \, e^{9}\right )}}{5 \, x} \]

[In]

integrate(1/5*(-x*exp(1)*exp(2)^2+4*x^2*exp(1)-2)*exp(1/16*exp(1)*exp(2)^4-1/2*x*exp(1)*exp(2)^2+x^2*exp(1))/x
^2,x, algorithm="fricas")

[Out]

2/5*e^(x^2*e - 1/2*x*e^5 + 1/16*e^9)/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 e^{e x^{2} - \frac {x e^{5}}{2} + \frac {e^{9}}{16}}}{5 x} \]

[In]

integrate(1/5*(-x*exp(1)*exp(2)**2+4*x**2*exp(1)-2)*exp(1/16*exp(1)*exp(2)**4-1/2*x*exp(1)*exp(2)**2+x**2*exp(
1))/x**2,x)

[Out]

2*exp(E*x**2 - x*exp(5)/2 + exp(9)/16)/(5*x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 \, e^{\left (x^{2} e - \frac {1}{2} \, x e^{5} + \frac {1}{16} \, e^{9}\right )}}{5 \, x} \]

[In]

integrate(1/5*(-x*exp(1)*exp(2)^2+4*x^2*exp(1)-2)*exp(1/16*exp(1)*exp(2)^4-1/2*x*exp(1)*exp(2)^2+x^2*exp(1))/x
^2,x, algorithm="maxima")

[Out]

2/5*e^(x^2*e - 1/2*x*e^5 + 1/16*e^9)/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 \, e^{\left (x^{2} e - \frac {1}{2} \, x e^{5} + \frac {1}{16} \, e^{9}\right )}}{5 \, x} \]

[In]

integrate(1/5*(-x*exp(1)*exp(2)^2+4*x^2*exp(1)-2)*exp(1/16*exp(1)*exp(2)^4-1/2*x*exp(1)*exp(2)^2+x^2*exp(1))/x
^2,x, algorithm="giac")

[Out]

2/5*e^(x^2*e - 1/2*x*e^5 + 1/16*e^9)/x

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2\,{\mathrm {e}}^{x^2\,\mathrm {e}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^9}{16}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{2}}}{5\,x} \]

[In]

int(-(exp(exp(9)/16 - (x*exp(5))/2 + x^2*exp(1))*(x*exp(5) - 4*x^2*exp(1) + 2))/(5*x^2),x)

[Out]

(2*exp(x^2*exp(1))*exp(exp(9)/16)*exp(-(x*exp(5))/2))/(5*x)