\(\int e^{c+d x^2} x^3 \text {erf}(b x) \, dx\) [54]

Optimal result

Integrand size = 17, antiderivative size = 155 \[ \int e^{c+d x^2} x^3 \text {erf}(b x) \, dx=\frac {b e^{c-\left (b^2-d\right ) x^2} x}{2 \left (b^2-d\right ) d \sqrt {\pi }}-\frac {e^{c+d x^2} \text {erf}(b x)}{2 d^2}+\frac {e^{c+d x^2} x^2 \text {erf}(b x)}{2 d}+\frac {b e^c \text {erf}\left (\sqrt {b^2-d} x\right )}{2 \sqrt {b^2-d} d^2}-\frac {b e^c \text {erf}\left (\sqrt {b^2-d} x\right )}{4 \left (b^2-d\right )^{3/2} d} \]

[Out]

-1/2*exp(d*x^2+c)*erf(b*x)/d^2+1/2*exp(d*x^2+c)*x^2*erf(b*x)/d-1/4*b*exp(c)*erf(x*(b^2-d)^(1/2))/(b^2-d)^(3/2)
/d+1/2*b*exp(c)*erf(x*(b^2-d)^(1/2))/d^2/(b^2-d)^(1/2)+1/2*b*exp(c-(b^2-d)*x^2)*x/(b^2-d)/d/Pi^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {6520, 6517, 2236, 2243} \[ \int e^{c+d x^2} x^3 \text {erf}(b x) \, dx=\frac {b e^c \text {erf}\left (x \sqrt {b^2-d}\right )}{2 d^2 \sqrt {b^2-d}}-\frac {b e^c \text {erf}\left (x \sqrt {b^2-d}\right )}{4 d \left (b^2-d\right )^{3/2}}+\frac {b x e^{c-x^2 \left (b^2-d\right )}}{2 \sqrt {\pi } d \left (b^2-d\right )}-\frac {\text {erf}(b x) e^{c+d x^2}}{2 d^2}+\frac {x^2 \text {erf}(b x) e^{c+d x^2}}{2 d} \]

[In]

Int[E^(c + d*x^2)*x^3*Erf[b*x],x]

[Out]

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

Rule 2236

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

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 6517

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.64 \[ \int e^{c+d x^2} x^3 \text {erf}(b x) \, dx=\frac {e^c \left (\frac {2 b d e^{\left (-b^2+d\right ) x^2} x}{\left (b^2-d\right ) \sqrt {\pi }}+2 e^{d x^2} \left (-1+d x^2\right ) \text {erf}(b x)+\frac {b \left (-2 b^2+3 d\right ) \text {erfi}\left (\sqrt {-b^2+d} x\right )}{\left (-b^2+d\right )^{3/2}}\right )}{4 d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {\frac {\operatorname {erf}\left (b x \right ) {\mathrm e}^{c} \left (\frac {b^{4} x^{2} {\mathrm e}^{d \,x^{2}}}{2 d}-\frac {b^{4} {\mathrm e}^{d \,x^{2}}}{2 d^{2}}\right )}{b^{3}}-\frac {{\mathrm e}^{c} \left (\frac {b^{2} \left (\frac {b x \,{\mathrm e}^{\left (-1+\frac {d}{b^{2}}\right ) b^{2} x^{2}}}{-2+\frac {2 d}{b^{2}}}-\frac {\sqrt {\pi }\, \operatorname {erf}\left (\sqrt {1-\frac {d}{b^{2}}}\, b x \right )}{4 \left (-1+\frac {d}{b^{2}}\right ) \sqrt {1-\frac {d}{b^{2}}}}\right )}{d}-\frac {b^{4} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {1-\frac {d}{b^{2}}}\, b x \right )}{2 d^{2} \sqrt {1-\frac {d}{b^{2}}}}\right )}{\sqrt {\pi }\, b^{3}}}{b}\)	\(168\)

[In]

int(exp(d*x^2+c)*x^3*erf(b*x),x,method=_RETURNVERBOSE)

[Out]

(erf(b*x)/b^3*exp(c)*(1/2/d*b^4*x^2*exp(d*x^2)-1/2/d^2*b^4*exp(d*x^2))-1/Pi^(1/2)/b^3*exp(c)*(1/d*b^2*(1/2/(-1
+d/b^2)*b*x*exp((-1+d/b^2)*b^2*x^2)-1/4/(-1+d/b^2)*Pi^(1/2)/(1-d/b^2)^(1/2)*erf((1-d/b^2)^(1/2)*b*x))-1/2/d^2*
b^4*Pi^(1/2)/(1-d/b^2)^(1/2)*erf((1-d/b^2)^(1/2)*b*x)))/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96 \[ \int e^{c+d x^2} x^3 \text {erf}(b x) \, dx=\frac {\pi {\left (2 \, b^{3} - 3 \, b d\right )} \sqrt {b^{2} - d} \operatorname {erf}\left (\sqrt {b^{2} - d} x\right ) e^{c} + 2 \, \sqrt {\pi } {\left (b^{3} d - b d^{2}\right )} x e^{\left (-b^{2} x^{2} + d x^{2} + c\right )} + 2 \, {\left (\pi {\left (b^{4} d - 2 \, b^{2} d^{2} + d^{3}\right )} x^{2} - \pi {\left (b^{4} - 2 \, b^{2} d + d^{2}\right )}\right )} \operatorname {erf}\left (b x\right ) e^{\left (d x^{2} + c\right )}}{4 \, \pi {\left (b^{4} d^{2} - 2 \, b^{2} d^{3} + d^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int e^{c+d x^2} x^3 \text {erf}(b x) \, dx=e^{c} \int x^{3} e^{d x^{2}} \operatorname {erf}{\left (b x \right )}\, dx \]

[In]

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

[Out]

exp(c)*Integral(x**3*exp(d*x**2)*erf(b*x), x)

Maxima [F]

\[ \int e^{c+d x^2} x^3 \text {erf}(b x) \, dx=\int { x^{3} \operatorname {erf}\left (b x\right ) e^{\left (d x^{2} + c\right )} \,d x } \]

[In]

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

[Out]

1/2*(d*x^2*e^c - e^c)*erf(b*x)*e^(d*x^2)/d^2 - integrate((b*d*x^2*e^c - b*e^c)*e^(-b^2*x^2 + d*x^2), x)/(sqrt(
pi)*d^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.91 \[ \int e^{c+d x^2} x^3 \text {erf}(b x) \, dx=\frac {1}{2} \, {\left (\frac {{\left (d x^{2} + c - 1\right )} e^{\left (d x^{2} + c\right )}}{d^{2}} - \frac {c e^{\left (d x^{2} + c\right )}}{d^{2}}\right )} \operatorname {erf}\left (b x\right ) + \frac {b d {\left (\frac {2 \, x e^{\left (-b^{2} x^{2} + d x^{2} + c\right )}}{b^{2} - d} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {b^{2} - d} x\right ) e^{c}}{{\left (b^{2} - d\right )}^{\frac {3}{2}}}\right )} - \frac {2 \, \sqrt {\pi } b \operatorname {erf}\left (-\sqrt {b^{2} - d} x\right ) e^{c}}{\sqrt {b^{2} - d}}}{4 \, \sqrt {\pi } d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.85 \[ \int e^{c+d x^2} x^3 \text {erf}(b x) \, dx=\frac {b\,x\,{\mathrm {e}}^{-b^2\,x^2+d\,x^2+c}}{2\,\sqrt {\pi }\,\left (b^2\,d-d^2\right )}-\mathrm {erf}\left (b\,x\right )\,\left (\frac {{\mathrm {e}}^{d\,x^2+c}}{2\,d^2}-\frac {x^2\,{\mathrm {e}}^{d\,x^2+c}}{2\,d}\right )+\frac {b\,{\mathrm {e}}^c\,\mathrm {erf}\left (x\,\sqrt {b^2-d}\right )}{2\,d^2\,\sqrt {b^2-d}}+\frac {b\,{\mathrm {e}}^c\,\mathrm {erfi}\left (x\,\sqrt {d-b^2}\right )}{4\,d\,{\left (d-b^2\right )}^{3/2}} \]

[In]

int(x^3*exp(c + d*x^2)*erf(b*x),x)

[Out]

(b*x*exp(c + d*x^2 - b^2*x^2))/(2*pi^(1/2)*(b^2*d - d^2)) - erf(b*x)*(exp(c + d*x^2)/(2*d^2) - (x^2*exp(c + d*
x^2))/(2*d)) + (b*exp(c)*erf(x*(b^2 - d)^(1/2)))/(2*d^2*(b^2 - d)^(1/2)) + (b*exp(c)*erfi(x*(d - b^2)^(1/2)))/
(4*d*(d - b^2)^(3/2))