3.103.38 \(\int \frac {e^{\frac {x^2}{6+14 x+6 x^2+4 x^3}} (6 x+7 x^2-2 x^4)}{18+84 x+134 x^2+108 x^3+74 x^4+24 x^5+8 x^6} \, dx\)

Optimal. Leaf size=25 \[ e^{\frac {x}{2 (1+2 x) \left (x+\frac {3+x}{x}\right )}} \]

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Rubi [A]  time = 0.48, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {1594, 6688, 12, 6706} \begin {gather*} e^{\frac {x^2}{2 \left (2 x^3+3 x^2+7 x+3\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x^2/(6 + 14*x + 6*x^2 + 4*x^3))*(6*x + 7*x^2 - 2*x^4))/(18 + 84*x + 134*x^2 + 108*x^3 + 74*x^4 + 24*x^
5 + 8*x^6),x]

[Out]

E^(x^2/(2*(3 + 7*x + 3*x^2 + 2*x^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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\int \frac {e^{\frac {x^2}{6+14 x+6 x^2+4 x^3}} x \left (6+7 x-2 x^3\right )}{18+84 x+134 x^2+108 x^3+74 x^4+24 x^5+8 x^6} \, dx\\ &=\int \frac {e^{\frac {x^2}{6+14 x+6 x^2+4 x^3}} x \left (6+7 x-2 x^3\right )}{2 \left (3+7 x+3 x^2+2 x^3\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{\frac {x^2}{6+14 x+6 x^2+4 x^3}} x \left (6+7 x-2 x^3\right )}{\left (3+7 x+3 x^2+2 x^3\right )^2} \, dx\\ &=e^{\frac {x^2}{2 \left (3+7 x+3 x^2+2 x^3\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 23, normalized size = 0.92 \begin {gather*} e^{\frac {x^2}{6+14 x+6 x^2+4 x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x^2/(6 + 14*x + 6*x^2 + 4*x^3))*(6*x + 7*x^2 - 2*x^4))/(18 + 84*x + 134*x^2 + 108*x^3 + 74*x^4 +
24*x^5 + 8*x^6),x]

[Out]

E^(x^2/(6 + 14*x + 6*x^2 + 4*x^3))

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fricas [A]  time = 0.66, size = 23, normalized size = 0.92 \begin {gather*} e^{\left (\frac {x^{2}}{2 \, {\left (2 \, x^{3} + 3 \, x^{2} + 7 \, x + 3\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^4+7*x^2+6*x)*exp(x^2/(4*x^3+6*x^2+14*x+6))/(8*x^6+24*x^5+74*x^4+108*x^3+134*x^2+84*x+18),x, al
gorithm="fricas")

[Out]

e^(1/2*x^2/(2*x^3 + 3*x^2 + 7*x + 3))

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giac [A]  time = 0.23, size = 23, normalized size = 0.92 \begin {gather*} e^{\left (\frac {x^{2}}{2 \, {\left (2 \, x^{3} + 3 \, x^{2} + 7 \, x + 3\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^4+7*x^2+6*x)*exp(x^2/(4*x^3+6*x^2+14*x+6))/(8*x^6+24*x^5+74*x^4+108*x^3+134*x^2+84*x+18),x, al
gorithm="giac")

[Out]

e^(1/2*x^2/(2*x^3 + 3*x^2 + 7*x + 3))

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maple [A]  time = 0.14, size = 22, normalized size = 0.88

method	result	size
risch	\({\mathrm e}^{\frac {x^{2}}{2 \left (2 x +1\right ) \left (x^{2}+x +3\right )}}\)	\(22\)
gosper	\({\mathrm e}^{\frac {x^{2}}{4 x^{3}+6 x^{2}+14 x +6}}\)	\(24\)
norman	\(\frac {7 x \,{\mathrm e}^{\frac {x^{2}}{4 x^{3}+6 x^{2}+14 x +6}}+3 x^{2} {\mathrm e}^{\frac {x^{2}}{4 x^{3}+6 x^{2}+14 x +6}}+2 x^{3} {\mathrm e}^{\frac {x^{2}}{4 x^{3}+6 x^{2}+14 x +6}}+3 \,{\mathrm e}^{\frac {x^{2}}{4 x^{3}+6 x^{2}+14 x +6}}}{2 x^{3}+3 x^{2}+7 x +3}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^4+7*x^2+6*x)*exp(x^2/(4*x^3+6*x^2+14*x+6))/(8*x^6+24*x^5+74*x^4+108*x^3+134*x^2+84*x+18),x,method=_R
ETURNVERBOSE)

[Out]

exp(1/2*x^2/(2*x+1)/(x^2+x+3))

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maxima [A]  time = 0.46, size = 32, normalized size = 1.28 \begin {gather*} e^{\left (\frac {5 \, x}{22 \, {\left (x^{2} + x + 3\right )}} - \frac {3}{22 \, {\left (x^{2} + x + 3\right )}} + \frac {1}{22 \, {\left (2 \, x + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^4+7*x^2+6*x)*exp(x^2/(4*x^3+6*x^2+14*x+6))/(8*x^6+24*x^5+74*x^4+108*x^3+134*x^2+84*x+18),x, al
gorithm="maxima")

[Out]

e^(5/22*x/(x^2 + x + 3) - 3/22/(x^2 + x + 3) + 1/22/(2*x + 1))

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mupad [B]  time = 6.02, size = 23, normalized size = 0.92 \begin {gather*} {\mathrm {e}}^{\frac {1}{22\,\left (2\,x+1\right )}+\frac {\frac {5\,x}{22}-\frac {3}{22}}{x^2+x+3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2/(14*x + 6*x^2 + 4*x^3 + 6))*(6*x + 7*x^2 - 2*x^4))/(84*x + 134*x^2 + 108*x^3 + 74*x^4 + 24*x^5 +
8*x^6 + 18),x)

[Out]

exp(1/(22*(2*x + 1)) + ((5*x)/22 - 3/22)/(x + x^2 + 3))

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sympy [A]  time = 0.25, size = 19, normalized size = 0.76 \begin {gather*} e^{\frac {x^{2}}{4 x^{3} + 6 x^{2} + 14 x + 6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**4+7*x**2+6*x)*exp(x**2/(4*x**3+6*x**2+14*x+6))/(8*x**6+24*x**5+74*x**4+108*x**3+134*x**2+84*x
+18),x)

[Out]

exp(x**2/(4*x**3 + 6*x**2 + 14*x + 6))

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