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

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Rubi [A]  time = 0.0423121, antiderivative size = 28, normalized size of antiderivative = 1., 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 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]

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 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{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*}

Mathematica [C]  time = 0.0013322, size = 13, normalized size = 0.46 \[ -\frac{2}{3} \text{Gamma}\left (2,-x^{3/2}\right ) \]

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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Maple [A]  time = 0.016, size = 17, normalized size = 0.6 \begin{align*} -{\frac{2}{3}{{\rm e}^{{x}^{{\frac{3}{2}}}}}}+{\frac{2}{3}{{\rm e}^{{x}^{{\frac{3}{2}}}}}{x}^{{\frac{3}{2}}}} \end{align*}

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 = 0.968502, size = 15, normalized size = 0.54 \begin{align*} \frac{2}{3} \,{\left (x^{\frac{3}{2}} - 1\right )} e^{\left (x^{\frac{3}{2}}\right )} \end{align*}

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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Fricas [A]  time = 0.695311, size = 42, normalized size = 1.5 \begin{align*} \frac{2}{3} \,{\left (x^{\frac{3}{2}} - 1\right )} e^{\left (x^{\frac{3}{2}}\right )} \end{align*}

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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Sympy [A]  time = 4.29522, size = 24, normalized size = 0.86 \begin{align*} \frac{2 x^{\frac{3}{2}} e^{x^{\frac{3}{2}}}}{3} - \frac{2 e^{x^{\frac{3}{2}}}}{3} \end{align*}

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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Giac [A]  time = 1.24377, size = 15, normalized size = 0.54 \begin{align*} \frac{2}{3} \,{\left (x^{\frac{3}{2}} - 1\right )} e^{\left (x^{\frac{3}{2}}\right )} \end{align*}

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))