3.55.13 \(\int \frac {e^{\frac {1}{2} (24 x^2-e^3 x^2+x^3)} (48 x-2 e^3 x+3 x^2)}{2+2 e^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^{\frac {1}{2} x^2 \left (24-e^3+x\right )}}{1+e^2} \]

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Rubi [A]  time = 0.23, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6, 12, 1593, 6688, 6706} \begin {gather*} \frac {e^{\frac {1}{2} x^2 \left (x-e^3+24\right )}}{1+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((24*x^2 - E^3*x^2 + x^3)/2)*(48*x - 2*E^3*x + 3*x^2))/(2 + 2*E^2),x]

[Out]

E^((x^2*(24 - E^3 + x))/2)/(1 + E^2)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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 {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (\left (48-2 e^3\right ) x+3 x^2\right )}{2+2 e^2} \, dx\\ &=\frac {\int e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (\left (48-2 e^3\right ) x+3 x^2\right ) \, dx}{2 \left (1+e^2\right )}\\ &=\frac {\int e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} x \left (48-2 e^3+3 x\right ) \, dx}{2 \left (1+e^2\right )}\\ &=\frac {\int e^{\frac {1}{2} x^2 \left (24-e^3+x\right )} x \left (48-2 e^3+3 x\right ) \, dx}{2 \left (1+e^2\right )}\\ &=\frac {e^{\frac {1}{2} x^2 \left (24-e^3+x\right )}}{1+e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 25, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {1}{2} x^2 \left (24-e^3+x\right )}}{1+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((24*x^2 - E^3*x^2 + x^3)/2)*(48*x - 2*E^3*x + 3*x^2))/(2 + 2*E^2),x]

[Out]

E^((x^2*(24 - E^3 + x))/2)/(1 + E^2)

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fricas [A]  time = 0.53, size = 26, normalized size = 1.04 \begin {gather*} \frac {e^{\left (\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} e^{3} + 12 \, x^{2}\right )}}{e^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(3)+3*x^2+48*x)*exp(-1/2*x^2*exp(3)+1/2*x^3+12*x^2)/(2*exp(2)+2),x, algorithm="fricas")

[Out]

e^(1/2*x^3 - 1/2*x^2*e^3 + 12*x^2)/(e^2 + 1)

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giac [A]  time = 0.13, size = 26, normalized size = 1.04 \begin {gather*} \frac {e^{\left (\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} e^{3} + 12 \, x^{2}\right )}}{e^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(3)+3*x^2+48*x)*exp(-1/2*x^2*exp(3)+1/2*x^3+12*x^2)/(2*exp(2)+2),x, algorithm="giac")

[Out]

e^(1/2*x^3 - 1/2*x^2*e^3 + 12*x^2)/(e^2 + 1)

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maple [A]  time = 0.06, size = 24, normalized size = 0.96

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{-\frac {x^{2} \left ({\mathrm e}^{3}-x -24\right )}{2}}}{2 \,{\mathrm e}^{2}+2}\)	\(24\)
gosper	\(\frac {{\mathrm e}^{-\frac {x^{2} {\mathrm e}^{3}}{2}+\frac {x^{3}}{2}+12 x^{2}}}{{\mathrm e}^{2}+1}\)	\(27\)
norman	\(\frac {{\mathrm e}^{-\frac {x^{2} {\mathrm e}^{3}}{2}+\frac {x^{3}}{2}+12 x^{2}}}{{\mathrm e}^{2}+1}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*exp(3)+3*x^2+48*x)*exp(-1/2*x^2*exp(3)+1/2*x^3+12*x^2)/(2*exp(2)+2),x,method=_RETURNVERBOSE)

[Out]

2/(2*exp(2)+2)*exp(-1/2*x^2*(exp(3)-x-24))

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maxima [A]  time = 0.36, size = 26, normalized size = 1.04 \begin {gather*} \frac {e^{\left (\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} e^{3} + 12 \, x^{2}\right )}}{e^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(3)+3*x^2+48*x)*exp(-1/2*x^2*exp(3)+1/2*x^3+12*x^2)/(2*exp(2)+2),x, algorithm="maxima")

[Out]

e^(1/2*x^3 - 1/2*x^2*e^3 + 12*x^2)/(e^2 + 1)

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mupad [B]  time = 0.14, size = 27, normalized size = 1.08 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^3}{2}}\,{\mathrm {e}}^{\frac {x^3}{2}}\,{\mathrm {e}}^{12\,x^2}}{{\mathrm {e}}^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(12*x^2 - (x^2*exp(3))/2 + x^3/2)*(48*x - 2*x*exp(3) + 3*x^2))/(2*exp(2) + 2),x)

[Out]

(exp(-(x^2*exp(3))/2)*exp(x^3/2)*exp(12*x^2))/(exp(2) + 1)

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sympy [A]  time = 0.14, size = 24, normalized size = 0.96 \begin {gather*} \frac {e^{\frac {x^{3}}{2} - \frac {x^{2} e^{3}}{2} + 12 x^{2}}}{1 + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(3)+3*x**2+48*x)*exp(-1/2*x**2*exp(3)+1/2*x**3+12*x**2)/(2*exp(2)+2),x)

[Out]

exp(x**3/2 - x**2*exp(3)/2 + 12*x**2)/(1 + exp(2))

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