3.6.38 \(\int e^{x^2} x (1+x^2) \, dx\) [538]

Optimal. Leaf size=12 \[ \frac {1}{2} e^{x^2} x^2 \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2258, 2240, 2243} \begin {gather*} \frac {1}{2} e^{x^2} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^x^2*x*(1 + x^2),x]

[Out]

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

Rule 2240

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 2243

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 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {align*} \int e^{x^2} x \left (1+x^2\right ) \, dx &=\int \left (e^{x^2} x+e^{x^2} x^3\right ) \, dx\\ &=\int e^{x^2} x \, dx+\int e^{x^2} x^3 \, dx\\ &=\frac {e^{x^2}}{2}+\frac {1}{2} e^{x^2} x^2-\int e^{x^2} x \, dx\\ &=\frac {1}{2} e^{x^2} x^2\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 12, normalized size = 1.00 \begin {gather*} \frac {1}{2} e^{x^2} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x^2*x*(1 + x^2),x]

[Out]

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

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Maple [A]
time = 0.02, size = 10, normalized size = 0.83

method	result	size
gosper	\(\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(10\)
derivativedivides	\(\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(10\)
default	\(\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(10\)
norman	\(\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(10\)
risch	\(\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(10\)
meijerg	\(-\frac {\left (-2 x^{2}+2\right ) {\mathrm e}^{x^{2}}}{4}+\frac {{\mathrm e}^{x^{2}}}{2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2)*x*(x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 2.33, size = 18, normalized size = 1.50 \begin {gather*} \frac {1}{2} \, {\left (x^{2} - 1\right )} e^{\left (x^{2}\right )} + \frac {1}{2} \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x*(x^2+1),x, algorithm="maxima")

[Out]

1/2*(x^2 - 1)*e^(x^2) + 1/2*e^(x^2)

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Fricas [A]
time = 0.99, size = 9, normalized size = 0.75 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x*(x^2+1),x, algorithm="fricas")

[Out]

1/2*x^2*e^(x^2)

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Sympy [A]
time = 0.02, size = 8, normalized size = 0.67 \begin {gather*} \frac {x^{2} e^{x^{2}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x**2)*x*(x**2+1),x)

[Out]

x**2*exp(x**2)/2

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Giac [A]
time = 0.73, size = 18, normalized size = 1.50 \begin {gather*} \frac {1}{2} \, {\left (x^{2} - 1\right )} e^{\left (x^{2}\right )} + \frac {1}{2} \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*x*(x^2+1),x, algorithm="giac")

[Out]

1/2*(x^2 - 1)*e^(x^2) + 1/2*e^(x^2)

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Mupad [B]
time = 0.05, size = 9, normalized size = 0.75 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{x^2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*exp(x^2)*(x^2 + 1),x)

[Out]

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

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