3.19.15 \(\int \frac {1}{2} (6+e^{\frac {1}{2} (-4+2 e^{x^2}+2 x+3 x^3)} (-2-4 e^{x^2} x-9 x^2)) \, dx\)

Optimal. Leaf size=23 \[ -e^{-2+e^{x^2}+x+\frac {3 x^3}{2}}+3 x \]

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Rubi [A]  time = 0.24, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 6688, 6706} \begin {gather*} 3 x-e^{\frac {3 x^3}{2}+e^{x^2}+x-2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + E^((-4 + 2*E^x^2 + 2*x + 3*x^3)/2)*(-2 - 4*E^x^2*x - 9*x^2))/2,x]

[Out]

-E^(-2 + E^x^2 + x + (3*x^3)/2) + 3*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 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} &=\frac {1}{2} \int \left (6+e^{\frac {1}{2} \left (-4+2 e^{x^2}+2 x+3 x^3\right )} \left (-2-4 e^{x^2} x-9 x^2\right )\right ) \, dx\\ &=3 x+\frac {1}{2} \int e^{\frac {1}{2} \left (-4+2 e^{x^2}+2 x+3 x^3\right )} \left (-2-4 e^{x^2} x-9 x^2\right ) \, dx\\ &=3 x+\frac {1}{2} \int e^{-2+e^{x^2}+x+\frac {3 x^3}{2}} \left (-2-4 e^{x^2} x-9 x^2\right ) \, dx\\ &=-e^{-2+e^{x^2}+x+\frac {3 x^3}{2}}+3 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 23, normalized size = 1.00 \begin {gather*} -e^{-2+e^{x^2}+x+\frac {3 x^3}{2}}+3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + E^((-4 + 2*E^x^2 + 2*x + 3*x^3)/2)*(-2 - 4*E^x^2*x - 9*x^2))/2,x]

[Out]

-E^(-2 + E^x^2 + x + (3*x^3)/2) + 3*x

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fricas [A]  time = 0.92, size = 19, normalized size = 0.83 \begin {gather*} 3 \, x - e^{\left (\frac {3}{2} \, x^{3} + x + e^{\left (x^{2}\right )} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-4*exp(x^2)*x-9*x^2-2)*exp(exp(x^2)+3/2*x^3+x-2)+3,x, algorithm="fricas")

[Out]

3*x - e^(3/2*x^3 + x + e^(x^2) - 2)

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giac [A]  time = 0.21, size = 19, normalized size = 0.83 \begin {gather*} 3 \, x - e^{\left (\frac {3}{2} \, x^{3} + x + e^{\left (x^{2}\right )} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-4*exp(x^2)*x-9*x^2-2)*exp(exp(x^2)+3/2*x^3+x-2)+3,x, algorithm="giac")

[Out]

3*x - e^(3/2*x^3 + x + e^(x^2) - 2)

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maple [A]  time = 0.09, size = 20, normalized size = 0.87

method	result	size
default	\(3 x -{\mathrm e}^{{\mathrm e}^{x^{2}}+\frac {3 x^{3}}{2}+x -2}\)	\(20\)
norman	\(3 x -{\mathrm e}^{{\mathrm e}^{x^{2}}+\frac {3 x^{3}}{2}+x -2}\)	\(20\)
risch	\(3 x -{\mathrm e}^{{\mathrm e}^{x^{2}}+\frac {3 x^{3}}{2}+x -2}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-4*exp(x^2)*x-9*x^2-2)*exp(exp(x^2)+3/2*x^3+x-2)+3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 19, normalized size = 0.83 \begin {gather*} 3 \, x - e^{\left (\frac {3}{2} \, x^{3} + x + e^{\left (x^{2}\right )} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-4*exp(x^2)*x-9*x^2-2)*exp(exp(x^2)+3/2*x^3+x-2)+3,x, algorithm="maxima")

[Out]

3*x - e^(3/2*x^3 + x + e^(x^2) - 2)

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mupad [B]  time = 0.10, size = 19, normalized size = 0.83 \begin {gather*} 3\,x-{\mathrm {e}}^{x+{\mathrm {e}}^{x^2}+\frac {3\,x^3}{2}-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3 - (exp(x + exp(x^2) + (3*x^3)/2 - 2)*(4*x*exp(x^2) + 9*x^2 + 2))/2,x)

[Out]

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

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sympy [A]  time = 0.30, size = 19, normalized size = 0.83 \begin {gather*} 3 x - e^{\frac {3 x^{3}}{2} + x + e^{x^{2}} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-4*exp(x**2)*x-9*x**2-2)*exp(exp(x**2)+3/2*x**3+x-2)+3,x)

[Out]

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

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