\(\int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} (-1+512 e^{4 x} x^3+e^{2 x} (-16 x+32 x^2))}{4 x^3} \, dx\) [6173]

Optimal result

Integrand size = 63, antiderivative size = 23 \[ \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{4 x^3} \, dx=2+e^{2 \left (4 e^{2 x}+\frac {1}{4 x}\right )^2} \]

[Out]

exp((4*exp(2*x)+1/4/x)^2)^2+2

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {12, 6838} \[ \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{4 x^3} \, dx=e^{\frac {256 e^{4 x} x^2+32 e^{2 x} x+1}{8 x^2}} \]

[In]

Int[(E^((1 + 32*E^(2*x)*x + 256*E^(4*x)*x^2)/(8*x^2))*(-1 + 512*E^(4*x)*x^3 + E^(2*x)*(-16*x + 32*x^2)))/(4*x^
3),x]

[Out]

E^((1 + 32*E^(2*x)*x + 256*E^(4*x)*x^2)/(8*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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{x^3} \, dx \\ & = e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{4 x^3} \, dx=e^{\frac {\left (1+16 e^{2 x} x\right )^2}{8 x^2}} \]

[In]

Integrate[(E^((1 + 32*E^(2*x)*x + 256*E^(4*x)*x^2)/(8*x^2))*(-1 + 512*E^(4*x)*x^3 + E^(2*x)*(-16*x + 32*x^2)))
/(4*x^3),x]

[Out]

E^((1 + 16*E^(2*x)*x)^2/(8*x^2))

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
risch	\({\mathrm e}^{\frac {256 x^{2} {\mathrm e}^{4 x}+32 x \,{\mathrm e}^{2 x}+1}{8 x^{2}}}\)	\(25\)
parallelrisch	\({\mathrm e}^{\frac {256 x^{2} {\mathrm e}^{4 x}+32 x \,{\mathrm e}^{2 x}+1}{8 x^{2}}}\)	\(29\)

[In]

int(1/4*(512*x^3*exp(2*x)^2+(32*x^2-16*x)*exp(2*x)-1)*exp(1/16*(256*x^2*exp(2*x)^2+32*x*exp(2*x)+1)/x^2)^2/x^3
,x,method=_RETURNVERBOSE)

[Out]

exp(1/8*(256*x^2*exp(4*x)+32*x*exp(2*x)+1)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{4 x^3} \, dx=e^{\left (\frac {256 \, x^{2} e^{\left (4 \, x\right )} + 32 \, x e^{\left (2 \, x\right )} + 1}{8 \, x^{2}}\right )} \]

[In]

integrate(1/4*(512*x^3*exp(2*x)^2+(32*x^2-16*x)*exp(2*x)-1)*exp(1/16*(256*x^2*exp(2*x)^2+32*x*exp(2*x)+1)/x^2)
^2/x^3,x, algorithm="fricas")

[Out]

e^(1/8*(256*x^2*e^(4*x) + 32*x*e^(2*x) + 1)/x^2)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{4 x^3} \, dx=e^{\frac {2 \cdot \left (16 x^{2} e^{4 x} + 2 x e^{2 x} + \frac {1}{16}\right )}{x^{2}}} \]

[In]

integrate(1/4*(512*x**3*exp(2*x)**2+(32*x**2-16*x)*exp(2*x)-1)*exp(1/16*(256*x**2*exp(2*x)**2+32*x*exp(2*x)+1)
/x**2)**2/x**3,x)

[Out]

exp(2*(16*x**2*exp(4*x) + 2*x*exp(2*x) + 1/16)/x**2)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{4 x^3} \, dx=e^{\left (\frac {4 \, e^{\left (2 \, x\right )}}{x} + \frac {1}{8 \, x^{2}} + 32 \, e^{\left (4 \, x\right )}\right )} \]

[In]

integrate(1/4*(512*x^3*exp(2*x)^2+(32*x^2-16*x)*exp(2*x)-1)*exp(1/16*(256*x^2*exp(2*x)^2+32*x*exp(2*x)+1)/x^2)
^2/x^3,x, algorithm="maxima")

[Out]

e^(4*e^(2*x)/x + 1/8/x^2 + 32*e^(4*x))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{4 x^3} \, dx=e^{\left (\frac {4 \, e^{\left (2 \, x\right )}}{x} + \frac {1}{8 \, x^{2}} + 32 \, e^{\left (4 \, x\right )}\right )} \]

[In]

integrate(1/4*(512*x^3*exp(2*x)^2+(32*x^2-16*x)*exp(2*x)-1)*exp(1/16*(256*x^2*exp(2*x)^2+32*x*exp(2*x)+1)/x^2)
^2/x^3,x, algorithm="giac")

[Out]

e^(4*e^(2*x)/x + 1/8/x^2 + 32*e^(4*x))

Mupad [B] (verification not implemented)

Time = 11.60 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {1+32 e^{2 x} x+256 e^{4 x} x^2}{8 x^2}} \left (-1+512 e^{4 x} x^3+e^{2 x} \left (-16 x+32 x^2\right )\right )}{4 x^3} \, dx={\mathrm {e}}^{32\,{\mathrm {e}}^{4\,x}}\,{\mathrm {e}}^{\frac {1}{8\,x^2}}\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{2\,x}}{x}} \]

[In]

int(-(exp((2*(2*x*exp(2*x) + 16*x^2*exp(4*x) + 1/16))/x^2)*(exp(2*x)*(16*x - 32*x^2) - 512*x^3*exp(4*x) + 1))/
(4*x^3),x)

[Out]

exp(32*exp(4*x))*exp(1/(8*x^2))*exp((4*exp(2*x))/x)