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

Optimal. Leaf size=76 \[ -\frac{e^c x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )}{2 \sqrt{\pi } b}+\frac{x e^{b^2 x^2+c} \text{Erf}(b x)}{2 b^2}-\frac{e^c x^2}{2 \sqrt{\pi } b} \]

[Out]

-(E^c*x^2)/(2*b*Sqrt[Pi]) + (E^(c + b^2*x^2)*x*Erf[b*x])/(2*b^2) - (E^c*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}
, b^2*x^2])/(2*b*Sqrt[Pi])

________________________________________________________________________________________

Rubi [A]  time = 0.0637064, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {6385, 6376, 12, 30} \[ -\frac{e^c x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi } b}+\frac{x e^{b^2 x^2+c} \text{Erf}(b x)}{2 b^2}-\frac{e^c x^2}{2 \sqrt{\pi } b} \]

Antiderivative was successfully verified.

[In]

Int[E^(c + b^2*x^2)*x^2*Erf[b*x],x]

[Out]

-(E^c*x^2)/(2*b*Sqrt[Pi]) + (E^(c + b^2*x^2)*x*Erf[b*x])/(2*b^2) - (E^c*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}
, b^2*x^2])/(2*b*Sqrt[Pi])

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 6376

Int[E^((c_.) + (d_.)*(x_)^2)*Erf[(b_.)*(x_)], x_Symbol] :> Simp[(b*E^c*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2},
b^2*x^2])/Sqrt[Pi], x] /; FreeQ[{b, c, d}, x] && EqQ[d, b^2]

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^2 \text{erf}(b x) \, dx &=\frac{e^{c+b^2 x^2} x \text{erf}(b x)}{2 b^2}-\frac{\int e^{c+b^2 x^2} \text{erf}(b x) \, dx}{2 b^2}-\frac{\int e^c x \, dx}{b \sqrt{\pi }}\\ &=\frac{e^{c+b^2 x^2} x \text{erf}(b x)}{2 b^2}-\frac{e^c x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 b \sqrt{\pi }}-\frac{e^c \int x \, dx}{b \sqrt{\pi }}\\ &=-\frac{e^c x^2}{2 b \sqrt{\pi }}+\frac{e^{c+b^2 x^2} x \text{erf}(b x)}{2 b^2}-\frac{e^c x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 b \sqrt{\pi }}\\ \end{align*}

Mathematica [A]  time = 0.214236, size = 80, normalized size = 1.05 \[ \frac{e^c \left (2 b^2 x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-b^2 x^2\right )+\text{Erf}(b x) \left (2 \sqrt{\pi } b x e^{b^2 x^2}-\pi \text{Erfi}(b x)\right )-2 b^2 x^2\right )}{4 \sqrt{\pi } b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^c*(-2*b^2*x^2 + Erf[b*x]*(2*b*E^(b^2*x^2)*Sqrt[Pi]*x - Pi*Erfi[b*x]) + 2*b^2*x^2*HypergeometricPFQ[{1, 1},
{3/2, 2}, -(b^2*x^2)]))/(4*b^3*Sqrt[Pi])

________________________________________________________________________________________

Maple [F]  time = 0.207, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{{b}^{2}{x}^{2}+c}}{x}^{2}{\it Erf} \left ( bx \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*erf(b*x)*e^(b^2*x^2 + c), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*erf(b*x)*e^(b^2*x^2 + c), x)

________________________________________________________________________________________

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**2*erf(b*x),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*erf(b*x)*e^(b^2*x^2 + c), x)