3.47.51 \(\int \frac {e^{-3+e-x} (12-2 x^2) (6+2 x-x^2)}{3 (-6+x^2)} \, dx\)

Optimal. Leaf size=18 \[ e^{-3+e-x} \left (4-\frac {2 x^2}{3}\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 26, normalized size of antiderivative = 1.44, number of steps used = 10, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {12, 21, 2196, 2194, 2176} \begin {gather*} 4 e^{-x+e-3}-\frac {2}{3} e^{-x+e-3} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-3 + E - x)*(12 - 2*x^2)*(6 + 2*x - x^2))/(3*(-6 + x^2)),x]

[Out]

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

Rule 12

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {e^{-3+e-x} \left (12-2 x^2\right ) \left (6+2 x-x^2\right )}{-6+x^2} \, dx\\ &=-\left (\frac {2}{3} \int e^{-3+e-x} \left (6+2 x-x^2\right ) \, dx\right )\\ &=-\left (\frac {2}{3} \int \left (6 e^{-3+e-x}+2 e^{-3+e-x} x-e^{-3+e-x} x^2\right ) \, dx\right )\\ &=\frac {2}{3} \int e^{-3+e-x} x^2 \, dx-\frac {4}{3} \int e^{-3+e-x} x \, dx-4 \int e^{-3+e-x} \, dx\\ &=4 e^{-3+e-x}+\frac {4}{3} e^{-3+e-x} x-\frac {2}{3} e^{-3+e-x} x^2-\frac {4}{3} \int e^{-3+e-x} \, dx+\frac {4}{3} \int e^{-3+e-x} x \, dx\\ &=\frac {16}{3} e^{-3+e-x}-\frac {2}{3} e^{-3+e-x} x^2+\frac {4}{3} \int e^{-3+e-x} \, dx\\ &=4 e^{-3+e-x}-\frac {2}{3} e^{-3+e-x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.06 \begin {gather*} \frac {2}{3} e^{-3+e-x} \left (6-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 + E - x)*(12 - 2*x^2)*(6 + 2*x - x^2))/(3*(-6 + x^2)),x]

[Out]

(2*E^(-3 + E - x)*(6 - x^2))/3

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fricas [A]  time = 0.63, size = 16, normalized size = 0.89 \begin {gather*} e^{\left (-x + e + \log \left (-\frac {2}{3} \, x^{2} + 4\right ) - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+2*x+6)*exp(log(-2/3*x^2+4)+exp(1)-3-x)/(x^2-6),x, algorithm="fricas")

[Out]

e^(-x + e + log(-2/3*x^2 + 4) - 3)

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giac [A]  time = 0.21, size = 24, normalized size = 1.33 \begin {gather*} -\frac {2}{3} \, x^{2} e^{\left (-x + e - 3\right )} + 4 \, e^{\left (-x + e - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+2*x+6)*exp(log(-2/3*x^2+4)+exp(1)-3-x)/(x^2-6),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.15, size = 17, normalized size = 0.94

method	result	size
gosper	\({\mathrm e}^{\ln \left (-\frac {2 x^{2}}{3}+4\right )+{\mathrm e}-3-x}\)	\(17\)
norman	\({\mathrm e}^{\ln \left (-\frac {2 x^{2}}{3}+4\right )+{\mathrm e}-3-x}\)	\(17\)
risch	\(\left (-\frac {2 x^{2}}{3}+4\right ) {\mathrm e}^{{\mathrm e}-3-x}\)	\(17\)
default	\(\text {Expression too large to display}\)	\(5846\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+2*x+6)*exp(ln(-2/3*x^2+4)+exp(1)-3-x)/(x^2-6),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 21, normalized size = 1.17 \begin {gather*} -\frac {2}{3} \, {\left (x^{2} e^{e} - 6 \, e^{e}\right )} e^{\left (-x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+2*x+6)*exp(log(-2/3*x^2+4)+exp(1)-3-x)/(x^2-6),x, algorithm="maxima")

[Out]

-2/3*(x^2*e^e - 6*e^e)*e^(-x - 3)

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mupad [B]  time = 0.14, size = 16, normalized size = 0.89 \begin {gather*} -\frac {2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{\mathrm {e}}\,\left (x^2-6\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(4 - (2*x^2)/3) - x + exp(1) - 3)*(2*x - x^2 + 6))/(x^2 - 6),x)

[Out]

-(2*exp(-x)*exp(-3)*exp(exp(1))*(x^2 - 6))/3

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sympy [A]  time = 0.12, size = 15, normalized size = 0.83 \begin {gather*} \frac {\left (12 - 2 x^{2}\right ) e^{- x - 3 + e}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2+2*x+6)*exp(ln(-2/3*x**2+4)+exp(1)-3-x)/(x**2-6),x)

[Out]

(12 - 2*x**2)*exp(-x - 3 + E)/3

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