\(\int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4)}{x^3} \, dx\) [4100]

Optimal result

Integrand size = 70, antiderivative size = 24 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{8-\left (-\frac {e^x}{x}+x\right )^2-\log ^2(2)} \]

[Out]

exp(8-ln(2)^2-(x-exp(x)/x)^2)

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=\exp \left (-\frac {x^4-2 e^x x^2-8 x^2+x^2 \log ^2(2)+e^{2 x}}{x^2}\right ) \]

[In]

Int[(E^((-E^(2*x) + 8*x^2 + 2*E^x*x^2 - x^4 - x^2*Log[2]^2)/x^2)*(E^(2*x)*(2 - 2*x) + 2*E^x*x^3 - 2*x^4))/x^3,
x]

[Out]

E^(-((E^(2*x) - 8*x^2 - 2*E^x*x^2 + x^4 + x^2*Log[2]^2)/x^2))

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}& = \exp \left (-\frac {e^{2 x}-8 x^2-2 e^x x^2+x^4+x^2 \log ^2(2)}{x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{8+2 e^x-\frac {e^{2 x}}{x^2}-x^2-\log ^2(2)} \]

[In]

Integrate[(E^((-E^(2*x) + 8*x^2 + 2*E^x*x^2 - x^4 - x^2*Log[2]^2)/x^2)*(E^(2*x)*(2 - 2*x) + 2*E^x*x^3 - 2*x^4)
)/x^3,x]

[Out]

E^(8 + 2*E^x - E^(2*x)/x^2 - x^2 - Log[2]^2)

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46

method	result	size
parallelrisch	\({\mathrm e}^{-\frac {x^{2} \ln \left (2\right )^{2}+x^{4}-2 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{2 x}-8 x^{2}}{x^{2}}}\)	\(35\)
norman	\({\mathrm e}^{\frac {-{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x^{2}-x^{2} \ln \left (2\right )^{2}-x^{4}+8 x^{2}}{x^{2}}}\)	\(39\)
risch	\({\mathrm e}^{\frac {-{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x^{2}-x^{2} \ln \left (2\right )^{2}-x^{4}+8 x^{2}}{x^{2}}}\)	\(39\)

[In]

int(((2-2*x)*exp(x)^2+2*exp(x)*x^3-2*x^4)*exp((-exp(x)^2+2*exp(x)*x^2-x^2*ln(2)^2-x^4+8*x^2)/x^2)/x^3,x,method
=_RETURNVERBOSE)

[Out]

exp(-(x^2*ln(2)^2+x^4-2*exp(x)*x^2+exp(x)^2-8*x^2)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{\left (-\frac {x^{4} + x^{2} \log \left (2\right )^{2} - 2 \, x^{2} e^{x} - 8 \, x^{2} + e^{\left (2 \, x\right )}}{x^{2}}\right )} \]

[In]

integrate(((2-2*x)*exp(x)^2+2*exp(x)*x^3-2*x^4)*exp((-exp(x)^2+2*exp(x)*x^2-x^2*log(2)^2-x^4+8*x^2)/x^2)/x^3,x
, algorithm="fricas")

[Out]

e^(-(x^4 + x^2*log(2)^2 - 2*x^2*e^x - 8*x^2 + e^(2*x))/x^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{\frac {- x^{4} + 2 x^{2} e^{x} - x^{2} \log {\left (2 \right )}^{2} + 8 x^{2} - e^{2 x}}{x^{2}}} \]

[In]

integrate(((2-2*x)*exp(x)**2+2*exp(x)*x**3-2*x**4)*exp((-exp(x)**2+2*exp(x)*x**2-x**2*ln(2)**2-x**4+8*x**2)/x*
*2)/x**3,x)

[Out]

exp((-x**4 + 2*x**2*exp(x) - x**2*log(2)**2 + 8*x**2 - exp(2*x))/x**2)

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{\left (-x^{2} - \log \left (2\right )^{2} - \frac {e^{\left (2 \, x\right )}}{x^{2}} + 2 \, e^{x} + 8\right )} \]

[In]

integrate(((2-2*x)*exp(x)^2+2*exp(x)*x^3-2*x^4)*exp((-exp(x)^2+2*exp(x)*x^2-x^2*log(2)^2-x^4+8*x^2)/x^2)/x^3,x
, algorithm="maxima")

[Out]

e^(-x^2 - log(2)^2 - e^(2*x)/x^2 + 2*e^x + 8)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{\left (-x^{2} - \log \left (2\right )^{2} - \frac {e^{\left (2 \, x\right )}}{x^{2}} + 2 \, e^{x} + 8\right )} \]

[In]

integrate(((2-2*x)*exp(x)^2+2*exp(x)*x^3-2*x^4)*exp((-exp(x)^2+2*exp(x)*x^2-x^2*log(2)^2-x^4+8*x^2)/x^2)/x^3,x
, algorithm="giac")

[Out]

e^(-x^2 - log(2)^2 - e^(2*x)/x^2 + 2*e^x + 8)

Mupad [B] (verification not implemented)

Time = 10.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx={\mathrm {e}}^8\,{\mathrm {e}}^{-{\ln \left (2\right )}^2}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{2\,x}}{x^2}} \]

[In]

int(-(exp(-(exp(2*x) + x^2*log(2)^2 - 2*x^2*exp(x) - 8*x^2 + x^4)/x^2)*(exp(2*x)*(2*x - 2) - 2*x^3*exp(x) + 2*
x^4))/x^3,x)

[Out]

exp(8)*exp(-log(2)^2)*exp(-x^2)*exp(2*exp(x))*exp(-exp(2*x)/x^2)