\(\int e^{x^2} x^3 \, dx\) [641]

Optimal result

Integrand size = 9, antiderivative size = 22 \[ \int e^{x^2} x^3 \, dx=-\frac {e^{x^2}}{2}+\frac {1}{2} e^{x^2} x^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2243, 2240} \[ \int e^{x^2} x^3 \, dx=\frac {1}{2} e^{x^2} x^2-\frac {e^{x^2}}{2} \]

[In]

Int[E^x^2*x^3,x]

[Out]

-1/2*E^x^2 + (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])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} e^{x^2} x^2-\int e^{x^2} x \, dx \\ & = -\frac {e^{x^2}}{2}+\frac {1}{2} e^{x^2} x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int e^{x^2} x^3 \, dx=\frac {1}{2} e^{x^2} \left (-1+x^2\right ) \]

[In]

Integrate[E^x^2*x^3,x]

[Out]

(E^x^2*(-1 + x^2))/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.55

method	result	size
gosper	\(\frac {\left (x^{2}-1\right ) {\mathrm e}^{x^{2}}}{2}\)	\(12\)
risch	\(\left (\frac {x^{2}}{2}-\frac {1}{2}\right ) {\mathrm e}^{x^{2}}\)	\(13\)
meijerg	\(\frac {1}{2}-\frac {\left (-2 x^{2}+2\right ) {\mathrm e}^{x^{2}}}{4}\)	\(16\)
derivativedivides	\(-\frac {{\mathrm e}^{x^{2}}}{2}+\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(17\)
default	\(-\frac {{\mathrm e}^{x^{2}}}{2}+\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(17\)
norman	\(-\frac {{\mathrm e}^{x^{2}}}{2}+\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(17\)
parallelrisch	\(-\frac {{\mathrm e}^{x^{2}}}{2}+\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\)	\(17\)
parts	\(\frac {\operatorname {erfi}\left (x \right ) \sqrt {\pi }\, x^{3}}{2}-\frac {3 \sqrt {\pi }\, \left (\frac {x^{3} \operatorname {erfi}\left (x \right )}{3}-\frac {2 \left (-\frac {{\mathrm e}^{x^{2}}}{2}+\frac {{\mathrm e}^{x^{2}} x^{2}}{2}\right )}{3 \sqrt {\pi }}\right )}{2}\)	\(46\)

[In]

int(exp(x^2)*x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50 \[ \int e^{x^2} x^3 \, dx=\frac {1}{2} \, {\left (x^{2} - 1\right )} e^{\left (x^{2}\right )} \]

[In]

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

[Out]

1/2*(x^2 - 1)*e^(x^2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45 \[ \int e^{x^2} x^3 \, dx=\frac {\left (x^{2} - 1\right ) e^{x^{2}}}{2} \]

[In]

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

[Out]

(x**2 - 1)*exp(x**2)/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50 \[ \int e^{x^2} x^3 \, dx=\frac {1}{2} \, {\left (x^{2} - 1\right )} e^{\left (x^{2}\right )} \]

[In]

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

[Out]

1/2*(x^2 - 1)*e^(x^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50 \[ \int e^{x^2} x^3 \, dx=\frac {1}{2} \, {\left (x^{2} - 1\right )} e^{\left (x^{2}\right )} \]

[In]

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

[Out]

1/2*(x^2 - 1)*e^(x^2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50 \[ \int e^{x^2} x^3 \, dx=\frac {{\mathrm {e}}^{x^2}\,\left (x^2-1\right )}{2} \]

[In]

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

[Out]

(exp(x^2)*(x^2 - 1))/2