3.71.47 \(\int \frac {e^{\frac {-1-x^2+e (x+x^3)+e^4 (-x^2-x^4)}{2 x}} (1-x^2+2 e x^3+e^4 (-x^2-3 x^4))}{2 x^2} \, dx\)

Optimal. Leaf size=29 \[ e^{\frac {1}{2} \left (-e^4-\frac {-e+\frac {1}{x}}{x}\right ) \left (x+x^3\right )} \]

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Rubi [A]  time = 0.71, antiderivative size = 33, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {12, 6706} \begin {gather*} \exp \left (-\frac {-e \left (x^3+x\right )+x^2+e^4 \left (x^4+x^2\right )+1}{2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-1 - x^2 + E*(x + x^3) + E^4*(-x^2 - x^4))/(2*x))*(1 - x^2 + 2*E*x^3 + E^4*(-x^2 - 3*x^4)))/(2*x^2),x
]

[Out]

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

Rule 12

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

Rule 6706

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

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Mathematica [A]  time = 0.19, size = 27, normalized size = 0.93 \begin {gather*} e^{-\frac {\left (1+x^2\right ) \left (1-e x+e^4 x^2\right )}{2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-1 - x^2 + E*(x + x^3) + E^4*(-x^2 - x^4))/(2*x))*(1 - x^2 + 2*E*x^3 + E^4*(-x^2 - 3*x^4)))/(2*
x^2),x]

[Out]

E^(-1/2*((1 + x^2)*(1 - E*x + E^4*x^2))/x)

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fricas [A]  time = 0.56, size = 30, normalized size = 1.03 \begin {gather*} e^{\left (-\frac {x^{2} + {\left (x^{4} + x^{2}\right )} e^{4} - {\left (x^{3} + x\right )} e + 1}{2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 33, normalized size = 1.14 \begin {gather*} e^{\left (-\frac {1}{2} \, x^{3} e^{4} + \frac {1}{2} \, x^{2} e - \frac {1}{2} \, x e^{4} - \frac {1}{2} \, x - \frac {1}{2 \, x} + \frac {1}{2} \, e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 25, normalized size = 0.86

method	result	size
risch	\({\mathrm e}^{\frac {\left (x^{2}+1\right ) \left (-x^{2} {\mathrm e}^{4}+x \,{\mathrm e}-1\right )}{2 x}}\)	\(25\)
norman	\({\mathrm e}^{\frac {\left (-x^{4}-x^{2}\right ) {\mathrm e}^{4}+\left (x^{3}+x \right ) {\mathrm e}-x^{2}-1}{2 x}}\)	\(36\)
gosper	\({\mathrm e}^{\frac {-x^{4} {\mathrm e}^{4}+x^{3} {\mathrm e}-x^{2} {\mathrm e}^{4}+x \,{\mathrm e}-x^{2}-1}{2 x}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-3*x^4-x^2)*exp(4)+2*x^3*exp(1)-x^2+1)*exp(1/2*((-x^4-x^2)*exp(4)+(x^3+x)*exp(1)-x^2-1)/x)/x^2,x,met
hod=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.60, size = 33, normalized size = 1.14 \begin {gather*} e^{\left (-\frac {1}{2} \, x^{3} e^{4} + \frac {1}{2} \, x^{2} e - \frac {1}{2} \, x e^{4} - \frac {1}{2} \, x - \frac {1}{2 \, x} + \frac {1}{2} \, e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.31, size = 38, normalized size = 1.31 \begin {gather*} {\mathrm {e}}^{\frac {x^2\,\mathrm {e}}{2}}\,{\mathrm {e}}^{-\frac {x^3\,{\mathrm {e}}^4}{2}}\,{\mathrm {e}}^{\frac {\mathrm {e}}{2}}\,{\mathrm {e}}^{-\frac {x}{2}}\,{\mathrm {e}}^{-\frac {1}{2\,x}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^4}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.23, size = 34, normalized size = 1.17 \begin {gather*} e^{\frac {- \frac {x^{2}}{2} + \frac {e \left (x^{3} + x\right )}{2} + \frac {\left (- x^{4} - x^{2}\right ) e^{4}}{2} - \frac {1}{2}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-3*x**4-x**2)*exp(4)+2*x**3*exp(1)-x**2+1)*exp(1/2*((-x**4-x**2)*exp(4)+(x**3+x)*exp(1)-x**2-1
)/x)/x**2,x)

[Out]

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

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