3.663 \(\int e^{x^{3/2}} x^2 \, dx\)

Optimal. Leaf size=28 \[ \frac {2}{3} e^{x^{3/2}} x^{3/2}-\frac {2 e^{x^{3/2}}}{3} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2216, 2212, 2209} \[ \frac {2}{3} e^{x^{3/2}} x^{3/2}-\frac {2 e^{x^{3/2}}}{3} \]

Antiderivative was successfully verified.

[In]

Int[E^x^(3/2)*x^2,x]

[Out]

(-2*E^x^(3/2))/3 + (2*E^x^(3/2)*x^(3/2))/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 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2216

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> With[{k = Denomina
tor[n]}, Dist[k/d, Subst[Int[x^(k*(m + 1) - 1)*F^(a + b*x^(k*n)), x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{F, a,
b, c, d, m, n}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int e^{x^{3/2}} x^2 \, dx &=2 \operatorname {Subst}\left (\int e^{x^3} x^5 \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{3} e^{x^{3/2}} x^{3/2}-2 \operatorname {Subst}\left (\int e^{x^3} x^2 \, dx,x,\sqrt {x}\right )\\ &=-\frac {2}{3} e^{x^{3/2}}+\frac {2}{3} e^{x^{3/2}} x^{3/2}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 13, normalized size = 0.46 \[ -\frac {2}{3} \Gamma \left (2,-x^{3/2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x^(3/2)*x^2,x]

[Out]

(-2*Gamma[2, -x^(3/2)])/3

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fricas [A]  time = 0.40, size = 11, normalized size = 0.39 \[ \frac {2}{3} \, {\left (x^{\frac {3}{2}} - 1\right )} e^{\left (x^{\frac {3}{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(x^(3/2) - 1)*e^(x^(3/2))

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giac [A]  time = 0.25, size = 11, normalized size = 0.39 \[ \frac {2}{3} \, {\left (x^{\frac {3}{2}} - 1\right )} e^{\left (x^{\frac {3}{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(x^(3/2) - 1)*e^(x^(3/2))

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maple [A]  time = 0.03, size = 17, normalized size = 0.61 \[ \frac {2 x^{\frac {3}{2}} {\mathrm e}^{x^{\frac {3}{2}}}}{3}-\frac {2 \,{\mathrm e}^{x^{\frac {3}{2}}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.68, size = 11, normalized size = 0.39 \[ \frac {2}{3} \, {\left (x^{\frac {3}{2}} - 1\right )} e^{\left (x^{\frac {3}{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(x^(3/2) - 1)*e^(x^(3/2))

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mupad [B]  time = 3.57, size = 16, normalized size = 0.57 \[ \frac {2\,x^{3/2}\,{\mathrm {e}}^{x^{3/2}}}{3}-\frac {2\,{\mathrm {e}}^{x^{3/2}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.11, size = 24, normalized size = 0.86 \[ \frac {2 x^{\frac {3}{2}} e^{x^{\frac {3}{2}}}}{3} - \frac {2 e^{x^{\frac {3}{2}}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x**(3/2))*x**2,x)

[Out]

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

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