3.65 \(\int e^{c+b^2 x^2} x^3 \text{Erf}(b x) \, dx\)

Optimal. Leaf size=79 \[ \frac{x^2 e^{b^2 x^2+c} \text{Erf}(b x)}{2 b^2}-\frac{e^{b^2 x^2+c} \text{Erf}(b x)}{2 b^4}+\frac{e^c x}{\sqrt{\pi } b^3}-\frac{e^c x^3}{3 \sqrt{\pi } b} \]

[Out]

(E^c*x)/(b^3*Sqrt[Pi]) - (E^c*x^3)/(3*b*Sqrt[Pi]) - (E^(c + b^2*x^2)*Erf[b*x])/(2*b^4) + (E^(c + b^2*x^2)*x^2*
Erf[b*x])/(2*b^2)

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Rubi [A]  time = 0.0799707, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {6385, 6382, 8, 12, 30} \[ \frac{x^2 e^{b^2 x^2+c} \text{Erf}(b x)}{2 b^2}-\frac{e^{b^2 x^2+c} \text{Erf}(b x)}{2 b^4}+\frac{e^c x}{\sqrt{\pi } b^3}-\frac{e^c x^3}{3 \sqrt{\pi } b} \]

Antiderivative was successfully verified.

[In]

Int[E^(c + b^2*x^2)*x^3*Erf[b*x],x]

[Out]

(E^c*x)/(b^3*Sqrt[Pi]) - (E^c*x^3)/(3*b*Sqrt[Pi]) - (E^(c + b^2*x^2)*Erf[b*x])/(2*b^4) + (E^(c + b^2*x^2)*x^2*
Erf[b*x])/(2*b^2)

Rule 6385

Int[E^((c_.) + (d_.)*(x_)^2)*Erf[(a_.) + (b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m - 1)*E^(c + d*x^2)*Erf
[a + b*x])/(2*d), x] + (-Dist[(m - 1)/(2*d), Int[x^(m - 2)*E^(c + d*x^2)*Erf[a + b*x], x], x] - Dist[b/(d*Sqrt
[Pi]), Int[x^(m - 1)*E^(-a^2 + c - 2*a*b*x - (b^2 - d)*x^2), x], x]) /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1]

Rule 6382

Int[E^((c_.) + (d_.)*(x_)^2)*Erf[(a_.) + (b_.)*(x_)]*(x_), x_Symbol] :> Simp[(E^(c + d*x^2)*Erf[a + b*x])/(2*d
), x] - Dist[b/(d*Sqrt[Pi]), Int[E^(-a^2 + c - 2*a*b*x - (b^2 - d)*x^2), x], x] /; FreeQ[{a, b, c, d}, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int e^{c+b^2 x^2} x^3 \text{erf}(b x) \, dx &=\frac{e^{c+b^2 x^2} x^2 \text{erf}(b x)}{2 b^2}-\frac{\int e^{c+b^2 x^2} x \text{erf}(b x) \, dx}{b^2}-\frac{\int e^c x^2 \, dx}{b \sqrt{\pi }}\\ &=-\frac{e^{c+b^2 x^2} \text{erf}(b x)}{2 b^4}+\frac{e^{c+b^2 x^2} x^2 \text{erf}(b x)}{2 b^2}+\frac{\int e^c \, dx}{b^3 \sqrt{\pi }}-\frac{e^c \int x^2 \, dx}{b \sqrt{\pi }}\\ &=\frac{e^c x}{b^3 \sqrt{\pi }}-\frac{e^c x^3}{3 b \sqrt{\pi }}-\frac{e^{c+b^2 x^2} \text{erf}(b x)}{2 b^4}+\frac{e^{c+b^2 x^2} x^2 \text{erf}(b x)}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0348345, size = 57, normalized size = 0.72 \[ \frac{e^c \left (3 \sqrt{\pi } e^{b^2 x^2} \left (b^2 x^2-1\right ) \text{Erf}(b x)-2 b^3 x^3+6 b x\right )}{6 \sqrt{\pi } b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c + b^2*x^2)*x^3*Erf[b*x],x]

[Out]

(E^c*(6*b*x - 2*b^3*x^3 + 3*E^(b^2*x^2)*Sqrt[Pi]*(-1 + b^2*x^2)*Erf[b*x]))/(6*b^4*Sqrt[Pi])

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Maple [A]  time = 0.218, size = 66, normalized size = 0.8 \begin{align*}{\frac{1}{b} \left ({\frac{{\it Erf} \left ( bx \right ){{\rm e}^{c}}}{{b}^{3}} \left ({\frac{{b}^{2}{x}^{2}{{\rm e}^{{b}^{2}{x}^{2}}}}{2}}-{\frac{{{\rm e}^{{b}^{2}{x}^{2}}}}{2}} \right ) }-{\frac{{{\rm e}^{c}}}{{b}^{3}\sqrt{\pi }} \left ({\frac{{x}^{3}{b}^{3}}{3}}-bx \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b^2*x^2+c)*x^3*erf(b*x),x)

[Out]

(erf(b*x)/b^3*exp(c)*(1/2*b^2*x^2*exp(b^2*x^2)-1/2*exp(b^2*x^2))-1/Pi^(1/2)/b^3*exp(c)*(1/3*x^3*b^3-b*x))/b

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Maxima [A]  time = 1.01245, size = 80, normalized size = 1.01 \begin{align*} -\frac{2 \, b^{3} x^{3} e^{c} - 3 \,{\left (\sqrt{\pi } b^{2} x^{2} e^{c} - \sqrt{\pi } e^{c}\right )} \operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2}\right )} - 6 \, b x e^{c}}{6 \, \sqrt{\pi } b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b^2*x^2+c)*x^3*erf(b*x),x, algorithm="maxima")

[Out]

-1/6*(2*b^3*x^3*e^c - 3*(sqrt(pi)*b^2*x^2*e^c - sqrt(pi)*e^c)*erf(b*x)*e^(b^2*x^2) - 6*b*x*e^c)/(sqrt(pi)*b^4)

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Fricas [A]  time = 2.71925, size = 131, normalized size = 1.66 \begin{align*} -\frac{3 \,{\left (\pi - \pi b^{2} x^{2}\right )} \operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )} + 2 \, \sqrt{\pi }{\left (b^{3} x^{3} - 3 \, b x\right )} e^{c}}{6 \, \pi b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b^2*x^2+c)*x^3*erf(b*x),x, algorithm="fricas")

[Out]

-1/6*(3*(pi - pi*b^2*x^2)*erf(b*x)*e^(b^2*x^2 + c) + 2*sqrt(pi)*(b^3*x^3 - 3*b*x)*e^c)/(pi*b^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b**2*x**2+c)*x**3*erf(b*x),x)

[Out]

Timed out

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Giac [A]  time = 1.27213, size = 78, normalized size = 0.99 \begin{align*} \frac{{\left (b^{2} x^{2} - 1\right )} \operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}}{2 \, b^{4}} - \frac{\sqrt{\pi } b^{2} x^{3} e^{c} - 3 \, \sqrt{\pi } x e^{c}}{3 \, \pi b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b^2*x^2+c)*x^3*erf(b*x),x, algorithm="giac")

[Out]

1/2*(b^2*x^2 - 1)*erf(b*x)*e^(b^2*x^2 + c)/b^4 - 1/3*(sqrt(pi)*b^2*x^3*e^c - 3*sqrt(pi)*x*e^c)/(pi*b^3)