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

Optimal. Leaf size=27 \[ \frac {-14-e^2+e^{-2 (1-x)+\frac {1}{x}}-x}{e^2} \]

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Rubi [A]  time = 0.14, antiderivative size = 17, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 14, 6706} \begin {gather*} e^{2 x+\frac {1}{x}-4}-\frac {x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-4 + x^(-1) + 2*x) - x/E^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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

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Mathematica [A]  time = 0.06, size = 17, normalized size = 0.63 \begin {gather*} e^{-4+\frac {1}{x}+2 x}-\frac {x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-4 + x^(-1) + 2*x) - x/E^2

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fricas [A]  time = 0.94, size = 23, normalized size = 0.85 \begin {gather*} -{\left (x - e^{\left (\frac {2 \, x^{2} - 2 \, x + 1}{x}\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x - e^((2*x^2 - 2*x + 1)/x))*e^(-2)

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giac [A]  time = 0.24, size = 23, normalized size = 0.85 \begin {gather*} -{\left (x - e^{\left (\frac {2 \, x^{2} - 2 \, x + 1}{x}\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x - e^((2*x^2 - 2*x + 1)/x))*e^(-2)

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maple [A]  time = 0.18, size = 22, normalized size = 0.81

method	result	size
risch	\(-x \,{\mathrm e}^{-2}+{\mathrm e}^{\frac {2 x^{2}-4 x +1}{x}}\)	\(22\)
norman	\(\frac {x \,{\mathrm e}^{-2} {\mathrm e}^{\frac {2 x^{2}-2 x +1}{x}}-x^{2} {\mathrm e}^{-2}}{x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-1)*exp((2*x^2-2*x+1)/x)-x^2)/x^2/exp(2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.82, size = 17, normalized size = 0.63 \begin {gather*} -{\left (x - e^{\left (2 \, x + \frac {1}{x} - 2\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x - e^(2*x + 1/x - 2))*e^(-2)

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mupad [B]  time = 2.12, size = 17, normalized size = 0.63 \begin {gather*} {\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{1/x}\,{\mathrm {e}}^{-4}-x\,{\mathrm {e}}^{-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*x)*exp(1/x)*exp(-4) - x*exp(-2)

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sympy [A]  time = 0.14, size = 20, normalized size = 0.74 \begin {gather*} - \frac {x}{e^{2}} + \frac {e^{\frac {2 x^{2} - 2 x + 1}{x}}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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