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

Optimal. Leaf size=22 \[ -3 e^2 \left (-2-e^{\frac {x^3}{2}}+x\right )+\log (4) \]

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2209} \begin {gather*} 3 e^{\frac {x^3}{2}+2}-3 e^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3*E^(2 + x^3/2) - 3*E^2*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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {3}{2} e^2 \left (2 e^{\frac {x^3}{2}}-2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*E^2*(2*E^(x^3/2) - 2*x))/2

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fricas [A]  time = 0.73, size = 16, normalized size = 0.73 \begin {gather*} -3 \, x e^{2} + 3 \, e^{\left (\frac {1}{2} \, x^{3} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x*e^2 + 3*e^(1/2*x^3 + 2)

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giac [A]  time = 0.25, size = 16, normalized size = 0.73 \begin {gather*} -3 \, x e^{2} + 3 \, e^{\left (\frac {1}{2} \, x^{3} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x*e^2 + 3*e^(1/2*x^3 + 2)

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maple [A]  time = 0.03, size = 17, normalized size = 0.77

method	result	size
default	\(3 \,{\mathrm e}^{2} {\mathrm e}^{\frac {x^{3}}{2}}-3 \,{\mathrm e}^{2} x\)	\(17\)
norman	\(3 \,{\mathrm e}^{2} {\mathrm e}^{\frac {x^{3}}{2}}-3 \,{\mathrm e}^{2} x\)	\(17\)
risch	\(3 \,{\mathrm e}^{2+\frac {x^{3}}{2}}-3 \,{\mathrm e}^{2} x\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 16, normalized size = 0.73 \begin {gather*} -3 \, x e^{2} + 3 \, e^{\left (\frac {1}{2} \, x^{3} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x*e^2 + 3*e^(1/2*x^3 + 2)

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mupad [B]  time = 1.12, size = 14, normalized size = 0.64 \begin {gather*} -3\,{\mathrm {e}}^2\,\left (x-{\mathrm {e}}^{\frac {x^3}{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*exp(2)*(x - exp(x^3/2))

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sympy [A]  time = 0.10, size = 17, normalized size = 0.77 \begin {gather*} - 3 x e^{2} + 3 e^{2} e^{\frac {x^{3}}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x*exp(2) + 3*exp(2)*exp(x**3/2)

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