3.86.63 \(\int (6 e^{\frac {2}{3} (-4+3 e^5+3 x^3)} x^2-12 e^{2+\frac {1}{3} (-4+3 e^5+3 x^3)} x^2) \, dx\)

Optimal. Leaf size=27 \[ \left (-2 e^2+e^{e^5+\frac {1}{3} \left (-1+3 \left (-1+x^3\right )\right )}\right )^2 \]

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Rubi [A]  time = 0.14, antiderivative size = 39, normalized size of antiderivative = 1.44, number of steps used = 5, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {2225, 2209} \begin {gather*} e^{2 x^3-\frac {2}{3} \left (4-3 e^5\right )}-4 e^{x^3+\frac {1}{3} \left (2+3 e^5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[6*E^((2*(-4 + 3*E^5 + 3*x^3))/3)*x^2 - 12*E^(2 + (-4 + 3*E^5 + 3*x^3)/3)*x^2,x]

[Out]

-4*E^((2 + 3*E^5)/3 + x^3) + E^((-2*(4 - 3*E^5))/3 + 2*x^3)

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]

Rule 2225

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && BinomialQ[v, x] &&  !(LinearMatchQ[u, x] && BinomialMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=6 \int e^{\frac {2}{3} \left (-4+3 e^5+3 x^3\right )} x^2 \, dx-12 \int e^{2+\frac {1}{3} \left (-4+3 e^5+3 x^3\right )} x^2 \, dx\\ &=6 \int e^{-\frac {2}{3} \left (4-3 e^5\right )+2 x^3} x^2 \, dx-12 \int e^{\frac {1}{3} \left (2+3 e^5\right )+x^3} x^2 \, dx\\ &=-4 e^{\frac {1}{3} \left (2+3 e^5\right )+x^3}+e^{-\frac {2}{3} \left (4-3 e^5\right )+2 x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 25, normalized size = 0.93 \begin {gather*} \frac {\left (-2 e^{10/3}+e^{e^5+x^3}\right )^2}{e^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[6*E^((2*(-4 + 3*E^5 + 3*x^3))/3)*x^2 - 12*E^(2 + (-4 + 3*E^5 + 3*x^3)/3)*x^2,x]

[Out]

(-2*E^(10/3) + E^(E^5 + x^3))^2/E^(8/3)

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fricas [A]  time = 0.68, size = 26, normalized size = 0.96 \begin {gather*} {\left (e^{\left (2 \, x^{3} + 2 \, e^{5} + \frac {4}{3}\right )} - 4 \, e^{\left (x^{3} + e^{5} + \frac {14}{3}\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6*x^2*exp(exp(5)+x^3-4/3)^2-12*x^2*exp(2)*exp(exp(5)+x^3-4/3),x, algorithm="fricas")

[Out]

(e^(2*x^3 + 2*e^5 + 4/3) - 4*e^(x^3 + e^5 + 14/3))*e^(-4)

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giac [A]  time = 0.26, size = 23, normalized size = 0.85 \begin {gather*} e^{\left (2 \, x^{3} + 2 \, e^{5} - \frac {8}{3}\right )} - 4 \, e^{\left (x^{3} + e^{5} + \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6*x^2*exp(exp(5)+x^3-4/3)^2-12*x^2*exp(2)*exp(exp(5)+x^3-4/3),x, algorithm="giac")

[Out]

e^(2*x^3 + 2*e^5 - 8/3) - 4*e^(x^3 + e^5 + 2/3)

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maple [A]  time = 0.03, size = 24, normalized size = 0.89

method	result	size
default	\({\mathrm e}^{2 \,{\mathrm e}^{5}+2 x^{3}-\frac {8}{3}}-4 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{5}+x^{3}-\frac {4}{3}}\)	\(24\)
norman	\({\mathrm e}^{2 \,{\mathrm e}^{5}+2 x^{3}-\frac {8}{3}}-4 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{5}+x^{3}-\frac {4}{3}}\)	\(24\)
risch	\({\mathrm e}^{2 \,{\mathrm e}^{5}+2 x^{3}-\frac {8}{3}}-4 \,{\mathrm e}^{\frac {2}{3}+{\mathrm e}^{5}+x^{3}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(6*x^2*exp(exp(5)+x^3-4/3)^2-12*x^2*exp(2)*exp(exp(5)+x^3-4/3),x,method=_RETURNVERBOSE)

[Out]

exp(exp(5)+x^3-4/3)^2-4*exp(2)*exp(exp(5)+x^3-4/3)

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maxima [A]  time = 0.35, size = 23, normalized size = 0.85 \begin {gather*} e^{\left (2 \, x^{3} + 2 \, e^{5} - \frac {8}{3}\right )} - 4 \, e^{\left (x^{3} + e^{5} + \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6*x^2*exp(exp(5)+x^3-4/3)^2-12*x^2*exp(2)*exp(exp(5)+x^3-4/3),x, algorithm="maxima")

[Out]

e^(2*x^3 + 2*e^5 - 8/3) - 4*e^(x^3 + e^5 + 2/3)

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mupad [B]  time = 0.09, size = 25, normalized size = 0.93 \begin {gather*} -{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-\frac {8}{3}}\,{\mathrm {e}}^{{\mathrm {e}}^5}\,\left (4\,{\mathrm {e}}^{10/3}-{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{{\mathrm {e}}^5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(6*x^2*exp(2*exp(5) + 2*x^3 - 8/3) - 12*x^2*exp(exp(5) + x^3 - 4/3)*exp(2),x)

[Out]

-exp(x^3)*exp(-8/3)*exp(exp(5))*(4*exp(10/3) - exp(x^3)*exp(exp(5)))

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sympy [A]  time = 0.12, size = 31, normalized size = 1.15 \begin {gather*} - 4 e^{2} e^{x^{3} - \frac {4}{3} + e^{5}} + e^{2 x^{3} - \frac {8}{3} + 2 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6*x**2*exp(exp(5)+x**3-4/3)**2-12*x**2*exp(2)*exp(exp(5)+x**3-4/3),x)

[Out]

-4*exp(2)*exp(x**3 - 4/3 + exp(5)) + exp(2*x**3 - 8/3 + 2*exp(5))

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