3.96.22 \(\int \frac {x^2+e^{\frac {-2+2 x+x^2}{x}} (-2-x^2)}{10 x^2} \, dx\)

Optimal. Leaf size=18 \[ \frac {1}{10} \left (-e^{2-\frac {2}{x}+x}+x\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 14, 6706} \begin {gather*} \frac {x}{10}-\frac {1}{10} e^{x-\frac {2}{x}+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + E^((-2 + 2*x + x^2)/x)*(-2 - x^2))/(10*x^2),x]

[Out]

-1/10*E^(2 - 2/x + x) + x/10

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 {1}{10} \int \frac {x^2+e^{\frac {-2+2 x+x^2}{x}} \left (-2-x^2\right )}{x^2} \, dx\\ &=\frac {1}{10} \int \left (1-\frac {e^{2-\frac {2}{x}+x} \left (2+x^2\right )}{x^2}\right ) \, dx\\ &=\frac {x}{10}-\frac {1}{10} \int \frac {e^{2-\frac {2}{x}+x} \left (2+x^2\right )}{x^2} \, dx\\ &=-\frac {1}{10} e^{2-\frac {2}{x}+x}+\frac {x}{10}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{10} \left (-e^{2-\frac {2}{x}+x}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + E^((-2 + 2*x + x^2)/x)*(-2 - x^2))/(10*x^2),x]

[Out]

(-E^(2 - 2/x + x) + x)/10

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fricas [A]  time = 0.66, size = 19, normalized size = 1.06 \begin {gather*} \frac {1}{10} \, x - \frac {1}{10} \, e^{\left (\frac {x^{2} + 2 \, x - 2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x^2-2)*exp((x^2+2*x-2)/x)+x^2)/x^2,x, algorithm="fricas")

[Out]

1/10*x - 1/10*e^((x^2 + 2*x - 2)/x)

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giac [A]  time = 0.17, size = 19, normalized size = 1.06 \begin {gather*} \frac {1}{10} \, x - \frac {1}{10} \, e^{\left (\frac {x^{2} + 2 \, x - 2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x^2-2)*exp((x^2+2*x-2)/x)+x^2)/x^2,x, algorithm="giac")

[Out]

1/10*x - 1/10*e^((x^2 + 2*x - 2)/x)

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maple [A]  time = 0.14, size = 20, normalized size = 1.11

method	result	size
risch	\(\frac {x}{10}-\frac {{\mathrm e}^{\frac {x^{2}+2 x -2}{x}}}{10}\)	\(20\)
norman	\(\frac {\frac {x^{2}}{10}-\frac {x \,{\mathrm e}^{\frac {x^{2}+2 x -2}{x}}}{10}}{x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((-x^2-2)*exp((x^2+2*x-2)/x)+x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/10*x-1/10*exp((x^2+2*x-2)/x)

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maxima [A]  time = 0.41, size = 15, normalized size = 0.83 \begin {gather*} \frac {1}{10} \, x - \frac {1}{10} \, e^{\left (x - \frac {2}{x} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x^2-2)*exp((x^2+2*x-2)/x)+x^2)/x^2,x, algorithm="maxima")

[Out]

1/10*x - 1/10*e^(x - 2/x + 2)

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mupad [B]  time = 5.84, size = 16, normalized size = 0.89 \begin {gather*} \frac {x}{10}-\frac {{\mathrm {e}}^2\,{\mathrm {e}}^{-\frac {2}{x}}\,{\mathrm {e}}^x}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp((2*x + x^2 - 2)/x)*(x^2 + 2))/10 - x^2/10)/x^2,x)

[Out]

x/10 - (exp(2)*exp(-2/x)*exp(x))/10

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sympy [A]  time = 0.12, size = 15, normalized size = 0.83 \begin {gather*} \frac {x}{10} - \frac {e^{\frac {x^{2} + 2 x - 2}{x}}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x**2-2)*exp((x**2+2*x-2)/x)+x**2)/x**2,x)

[Out]

x/10 - exp((x**2 + 2*x - 2)/x)/10

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