\(\int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx\) [70]

Optimal result

Integrand size = 19, antiderivative size = 119 \[ \int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx=\frac {3 e^c x^2}{4 b^3 \sqrt {\pi }}-\frac {e^c x^4}{4 b \sqrt {\pi }}-\frac {3 e^{c+b^2 x^2} x \text {erf}(b x)}{4 b^4}+\frac {e^{c+b^2 x^2} x^3 \text {erf}(b x)}{2 b^2}+\frac {3 e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{4 b^3 \sqrt {\pi }} \]

[Out]

-3/4*exp(b^2*x^2+c)*x*erf(b*x)/b^4+1/2*exp(b^2*x^2+c)*x^3*erf(b*x)/b^2+3/4*exp(c)*x^2/b^3/Pi^(1/2)-1/4*exp(c)*
x^4/b/Pi^(1/2)+3/4*exp(c)*x^2*hypergeom([1, 1],[3/2, 2],b^2*x^2)/b^3/Pi^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {6520, 6511, 12, 30} \[ \int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx=\frac {3 e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{4 \sqrt {\pi } b^3}+\frac {3 e^c x^2}{4 \sqrt {\pi } b^3}+\frac {x^3 e^{b^2 x^2+c} \text {erf}(b x)}{2 b^2}-\frac {3 x e^{b^2 x^2+c} \text {erf}(b x)}{4 b^4}-\frac {e^c x^4}{4 \sqrt {\pi } b} \]

[In]

Int[E^(c + b^2*x^2)*x^4*Erf[b*x],x]

[Out]

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

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]

Rule 6511

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

Rule 6520

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]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{c+b^2 x^2} x^3 \text {erf}(b x)}{2 b^2}-\frac {3 \int e^{c+b^2 x^2} x^2 \text {erf}(b x) \, dx}{2 b^2}-\frac {\int e^c x^3 \, dx}{b \sqrt {\pi }} \\ & = -\frac {3 e^{c+b^2 x^2} x \text {erf}(b x)}{4 b^4}+\frac {e^{c+b^2 x^2} x^3 \text {erf}(b x)}{2 b^2}+\frac {3 \int e^{c+b^2 x^2} \text {erf}(b x) \, dx}{4 b^4}+\frac {3 \int e^c x \, dx}{2 b^3 \sqrt {\pi }}-\frac {e^c \int x^3 \, dx}{b \sqrt {\pi }} \\ & = -\frac {e^c x^4}{4 b \sqrt {\pi }}-\frac {3 e^{c+b^2 x^2} x \text {erf}(b x)}{4 b^4}+\frac {e^{c+b^2 x^2} x^3 \text {erf}(b x)}{2 b^2}+\frac {3 e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{4 b^3 \sqrt {\pi }}+\frac {\left (3 e^c\right ) \int x \, dx}{2 b^3 \sqrt {\pi }} \\ & = \frac {3 e^c x^2}{4 b^3 \sqrt {\pi }}-\frac {e^c x^4}{4 b \sqrt {\pi }}-\frac {3 e^{c+b^2 x^2} x \text {erf}(b x)}{4 b^4}+\frac {e^{c+b^2 x^2} x^3 \text {erf}(b x)}{2 b^2}+\frac {3 e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{4 b^3 \sqrt {\pi }} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.84 \[ \int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx=\frac {e^c \left (6 b^2 x^2-2 b^4 x^4+2 b e^{b^2 x^2} \sqrt {\pi } x \left (-3+2 b^2 x^2\right ) \text {erf}(b x)+3 \pi \text {erf}(b x) \text {erfi}(b x)-6 b^2 x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-b^2 x^2\right )\right )}{8 b^5 \sqrt {\pi }} \]

[In]

Integrate[E^(c + b^2*x^2)*x^4*Erf[b*x],x]

[Out]

(E^c*(6*b^2*x^2 - 2*b^4*x^4 + 2*b*E^(b^2*x^2)*Sqrt[Pi]*x*(-3 + 2*b^2*x^2)*Erf[b*x] + 3*Pi*Erf[b*x]*Erfi[b*x] -
6*b^2*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, -(b^2*x^2)]))/(8*b^5*Sqrt[Pi])

Maple [F]

\[\int {\mathrm e}^{b^{2} x^{2}+c} x^{4} \operatorname {erf}\left (b x \right )d x\]

[In]

int(exp(b^2*x^2+c)*x^4*erf(b*x),x)

[Out]

int(exp(b^2*x^2+c)*x^4*erf(b*x),x)

Fricas [F]

\[ \int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx=\int { x^{4} \operatorname {erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )} \,d x } \]

[In]

integrate(exp(b^2*x^2+c)*x^4*erf(b*x),x, algorithm="fricas")

[Out]

integral(x^4*erf(b*x)*e^(b^2*x^2 + c), x)

Sympy [A] (verification not implemented)

Time = 171.60 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.20 \[ \int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx=\frac {b x^{6} e^{c} {{}_{2}F_{2}\left (\begin {matrix} 1, 3 \\ \frac {3}{2}, 4 \end {matrix}\middle | {b^{2} x^{2}} \right )}}{3 \sqrt {\pi }} \]

[In]

integrate(exp(b**2*x**2+c)*x**4*erf(b*x),x)

[Out]

b*x**6*exp(c)*hyper((1, 3), (3/2, 4), b**2*x**2)/(3*sqrt(pi))

Maxima [F]

\[ \int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx=\int { x^{4} \operatorname {erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )} \,d x } \]

[In]

integrate(exp(b^2*x^2+c)*x^4*erf(b*x),x, algorithm="maxima")

[Out]

integrate(x^4*erf(b*x)*e^(b^2*x^2 + c), x)

Giac [F]

\[ \int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx=\int { x^{4} \operatorname {erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )} \,d x } \]

[In]

integrate(exp(b^2*x^2+c)*x^4*erf(b*x),x, algorithm="giac")

[Out]

integrate(x^4*erf(b*x)*e^(b^2*x^2 + c), x)

Mupad [F(-1)]

Timed out. \[ \int e^{c+b^2 x^2} x^4 \text {erf}(b x) \, dx=\int x^4\,{\mathrm {e}}^{b^2\,x^2+c}\,\mathrm {erf}\left (b\,x\right ) \,d x \]

[In]

int(x^4*exp(c + b^2*x^2)*erf(b*x),x)

[Out]

int(x^4*exp(c + b^2*x^2)*erf(b*x), x)