3.40.94 \(\int \frac {1}{4} e^{-3+x^2} (e^{3-x^2} (8 e+8 x)+e^{30 e^{-3+x^2}} (e^{3-x^2} x^3+15 x^5)+e^{15 e^{-3+x^2}} (60 e x^3+60 x^4+e^{3-x^2} (4 e x+6 x^2))) \, dx\)

Optimal. Leaf size=23 \[ \left (e+x+\frac {1}{4} e^{15 e^{-3+x^2}} x^2\right )^2 \]

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Rubi [A]  time = 0.28, antiderivative size = 28, normalized size of antiderivative = 1.22, number of steps used = 4, number of rules used = 3, integrand size = 104, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {12, 6688, 6686} \begin {gather*} \frac {1}{16} \left (e^{15 e^{x^2-3}} x^2+4 x+4 e\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-3 + x^2)*(E^(3 - x^2)*(8*E + 8*x) + E^(30*E^(-3 + x^2))*(E^(3 - x^2)*x^3 + 15*x^5) + E^(15*E^(-3 + x^
2))*(60*E*x^3 + 60*x^4 + E^(3 - x^2)*(4*E*x + 6*x^2))))/4,x]

[Out]

(4*E + 4*x + E^(15*E^(-3 + x^2))*x^2)^2/16

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{-3+x^2} \left (e^{3-x^2} (8 e+8 x)+e^{30 e^{-3+x^2}} \left (e^{3-x^2} x^3+15 x^5\right )+e^{15 e^{-3+x^2}} \left (60 e x^3+60 x^4+e^{3-x^2} \left (4 e x+6 x^2\right )\right )\right ) \, dx\\ &=\frac {1}{4} \int \frac {\left (4 e+4 x+e^{15 e^{-3+x^2}} x^2\right ) \left (2 e^3+e^{3+15 e^{-3+x^2}} x+15 e^{15 e^{-3+x^2}+x^2} x^3\right )}{e^3} \, dx\\ &=\frac {\int \left (4 e+4 x+e^{15 e^{-3+x^2}} x^2\right ) \left (2 e^3+e^{3+15 e^{-3+x^2}} x+15 e^{15 e^{-3+x^2}+x^2} x^3\right ) \, dx}{4 e^3}\\ &=\frac {1}{16} \left (4 e+4 x+e^{15 e^{-3+x^2}} x^2\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 28, normalized size = 1.22 \begin {gather*} \frac {1}{16} \left (4 e+4 x+e^{15 e^{-3+x^2}} x^2\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 + x^2)*(E^(3 - x^2)*(8*E + 8*x) + E^(30*E^(-3 + x^2))*(E^(3 - x^2)*x^3 + 15*x^5) + E^(15*E^(-
3 + x^2))*(60*E*x^3 + 60*x^4 + E^(3 - x^2)*(4*E*x + 6*x^2))))/4,x]

[Out]

(4*E + 4*x + E^(15*E^(-3 + x^2))*x^2)^2/16

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fricas [A]  time = 0.92, size = 44, normalized size = 1.91 \begin {gather*} \frac {1}{16} \, x^{4} e^{\left (30 \, e^{\left (x^{2} - 3\right )}\right )} + x^{2} + 2 \, x e + \frac {1}{2} \, {\left (x^{3} + x^{2} e\right )} e^{\left (15 \, e^{\left (x^{2} - 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^3*exp(-x^2+3)+15*x^5)*exp(15/exp(-x^2+3))^2+((4*x*exp(1)+6*x^2)*exp(-x^2+3)+60*x^3*exp(1)+60
*x^4)*exp(15/exp(-x^2+3))+(8*exp(1)+8*x)*exp(-x^2+3))/exp(-x^2+3),x, algorithm="fricas")

[Out]

1/16*x^4*e^(30*e^(x^2 - 3)) + x^2 + 2*x*e + 1/2*(x^3 + x^2*e)*e^(15*e^(x^2 - 3))

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giac [B]  time = 0.14, size = 53, normalized size = 2.30 \begin {gather*} \frac {1}{16} \, x^{4} e^{\left (30 \, e^{\left (x^{2} - 3\right )}\right )} + \frac {1}{2} \, x^{3} e^{\left (15 \, e^{\left (x^{2} - 3\right )}\right )} + \frac {1}{2} \, x^{2} e^{\left (15 \, e^{\left (x^{2} - 3\right )} + 1\right )} + x^{2} + 2 \, x e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^3*exp(-x^2+3)+15*x^5)*exp(15/exp(-x^2+3))^2+((4*x*exp(1)+6*x^2)*exp(-x^2+3)+60*x^3*exp(1)+60
*x^4)*exp(15/exp(-x^2+3))+(8*exp(1)+8*x)*exp(-x^2+3))/exp(-x^2+3),x, algorithm="giac")

[Out]

1/16*x^4*e^(30*e^(x^2 - 3)) + 1/2*x^3*e^(15*e^(x^2 - 3)) + 1/2*x^2*e^(15*e^(x^2 - 3) + 1) + x^2 + 2*x*e

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maple [A]  time = 0.06, size = 42, normalized size = 1.83

method	result	size
risch	\(2 x \,{\mathrm e}+x^{2}+\frac {x^{4} {\mathrm e}^{30 \,{\mathrm e}^{x^{2}-3}}}{16}+\frac {x^{2} \left (x +{\mathrm e}\right ) {\mathrm e}^{15 \,{\mathrm e}^{x^{2}-3}}}{2}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((x^3*exp(-x^2+3)+15*x^5)*exp(15/exp(-x^2+3))^2+((4*x*exp(1)+6*x^2)*exp(-x^2+3)+60*x^3*exp(1)+60*x^4)*
exp(15/exp(-x^2+3))+(8*exp(1)+8*x)*exp(-x^2+3))/exp(-x^2+3),x,method=_RETURNVERBOSE)

[Out]

2*x*exp(1)+x^2+1/16*x^4*exp(30*exp(x^2-3))+1/2*x^2*(x+exp(1))*exp(15*exp(x^2-3))

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maxima [A]  time = 0.42, size = 44, normalized size = 1.91 \begin {gather*} \frac {1}{16} \, x^{4} e^{\left (30 \, e^{\left (x^{2} - 3\right )}\right )} + x^{2} + 2 \, x e + \frac {1}{2} \, {\left (x^{3} + x^{2} e\right )} e^{\left (15 \, e^{\left (x^{2} - 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^3*exp(-x^2+3)+15*x^5)*exp(15/exp(-x^2+3))^2+((4*x*exp(1)+6*x^2)*exp(-x^2+3)+60*x^3*exp(1)+60
*x^4)*exp(15/exp(-x^2+3))+(8*exp(1)+8*x)*exp(-x^2+3))/exp(-x^2+3),x, algorithm="maxima")

[Out]

1/16*x^4*e^(30*e^(x^2 - 3)) + x^2 + 2*x*e + 1/2*(x^3 + x^2*e)*e^(15*e^(x^2 - 3))

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mupad [B]  time = 2.59, size = 37, normalized size = 1.61 \begin {gather*} \frac {x\,\left (x\,{\mathrm {e}}^{15\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3}}+4\right )\,\left (4\,x+8\,\mathrm {e}+x^2\,{\mathrm {e}}^{15\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3}}\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2 - 3)*((exp(15*exp(x^2 - 3))*(exp(3 - x^2)*(4*x*exp(1) + 6*x^2) + 60*x^3*exp(1) + 60*x^4))/4 + (exp
(3 - x^2)*(8*x + 8*exp(1)))/4 + (exp(30*exp(x^2 - 3))*(x^3*exp(3 - x^2) + 15*x^5))/4),x)

[Out]

(x*(x*exp(15*exp(x^2)*exp(-3)) + 4)*(4*x + 8*exp(1) + x^2*exp(15*exp(x^2)*exp(-3))))/16

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x**3*exp(-x**2+3)+15*x**5)*exp(15/exp(-x**2+3))**2+((4*x*exp(1)+6*x**2)*exp(-x**2+3)+60*x**3*e
xp(1)+60*x**4)*exp(15/exp(-x**2+3))+(8*exp(1)+8*x)*exp(-x**2+3))/exp(-x**2+3),x)

[Out]

Timed out

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