3.54 \(\int e^{x^2} x^5 \, dx\)

Optimal. Leaf size=28 \[ \frac{1}{2} e^{x^2} x^4-e^{x^2} x^2+e^{x^2} \]

[Out]

E^x^2 - E^x^2*x^2 + (E^x^2*x^4)/2

________________________________________________________________________________________

Rubi [A]  time = 0.0337833, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2212, 2209} \[ \frac{1}{2} e^{x^2} x^4-e^{x^2} x^2+e^{x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^x^2*x^5,x]

[Out]

E^x^2 - E^x^2*x^2 + (E^x^2*x^4)/2

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^2} x^5 \, dx &=\frac{1}{2} e^{x^2} x^4-2 \int e^{x^2} x^3 \, dx\\ &=-e^{x^2} x^2+\frac{1}{2} e^{x^2} x^4+2 \int e^{x^2} x \, dx\\ &=e^{x^2}-e^{x^2} x^2+\frac{1}{2} e^{x^2} x^4\\ \end{align*}

Mathematica [A]  time = 0.0021054, size = 19, normalized size = 0.68 \[ \frac{1}{2} e^{x^2} \left (x^4-2 x^2+2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x^2*x^5,x]

[Out]

(E^x^2*(2 - 2*x^2 + x^4))/2

________________________________________________________________________________________

Maple [A]  time = 0.001, size = 17, normalized size = 0.6 \begin{align*}{\frac{ \left ({x}^{4}-2\,{x}^{2}+2 \right ){{\rm e}^{{x}^{2}}}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2)*x^5,x)

[Out]

1/2*(x^4-2*x^2+2)*exp(x^2)

________________________________________________________________________________________

Maxima [A]  time = 0.943267, size = 22, normalized size = 0.79 \begin{align*} \frac{1}{2} \,{\left (x^{4} - 2 \, x^{2} + 2\right )} e^{\left (x^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x^5,x, algorithm="maxima")

[Out]

1/2*(x^4 - 2*x^2 + 2)*e^(x^2)

________________________________________________________________________________________

Fricas [A]  time = 2.28847, size = 42, normalized size = 1.5 \begin{align*} \frac{1}{2} \,{\left (x^{4} - 2 \, x^{2} + 2\right )} e^{\left (x^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x^5,x, algorithm="fricas")

[Out]

1/2*(x^4 - 2*x^2 + 2)*e^(x^2)

________________________________________________________________________________________

Sympy [A]  time = 0.084731, size = 15, normalized size = 0.54 \begin{align*} \frac{\left (x^{4} - 2 x^{2} + 2\right ) e^{x^{2}}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x**2)*x**5,x)

[Out]

(x**4 - 2*x**2 + 2)*exp(x**2)/2

________________________________________________________________________________________

Giac [A]  time = 1.066, size = 22, normalized size = 0.79 \begin{align*} \frac{1}{2} \,{\left (x^{4} - 2 \, x^{2} + 2\right )} e^{\left (x^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x^5,x, algorithm="giac")

[Out]

1/2*(x^4 - 2*x^2 + 2)*e^(x^2)