\(\int e^{c+b^2 x^2} x^4 \text {erfi}(b x) \, dx\) [289]

Optimal result

Integrand size = 19, antiderivative size = 121 \[ \int e^{c+b^2 x^2} x^4 \text {erfi}(b x) \, dx=\frac {e^{c+2 b^2 x^2}}{2 b^5 \sqrt {\pi }}-\frac {e^{c+2 b^2 x^2} x^2}{4 b^3 \sqrt {\pi }}-\frac {3 e^{c+b^2 x^2} x \text {erfi}(b x)}{4 b^4}+\frac {e^{c+b^2 x^2} x^3 \text {erfi}(b x)}{2 b^2}+\frac {3 e^c \sqrt {\pi } \text {erfi}(b x)^2}{16 b^5} \] Output:

1/2*exp(2*b^2*x^2+c)/b^5/Pi^(1/2)-1/4*exp(2*b^2*x^2+c)*x^2/b^3/Pi^(1/2)-3/ 
4*exp(b^2*x^2+c)*x*erfi(b*x)/b^4+1/2*exp(b^2*x^2+c)*x^3*erfi(b*x)/b^2+3/16 
*exp(c)*Pi^(1/2)*erfi(b*x)^2/b^5
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.64 \[ \int e^{c+b^2 x^2} x^4 \text {erfi}(b x) \, dx=\frac {e^c \left (-4 e^{2 b^2 x^2} \left (-2+b^2 x^2\right )+4 b e^{b^2 x^2} \sqrt {\pi } x \left (-3+2 b^2 x^2\right ) \text {erfi}(b x)+3 \pi \text {erfi}(b x)^2\right )}{16 b^5 \sqrt {\pi }} \] Input:

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

Output:

(E^c*(-4*E^(2*b^2*x^2)*(-2 + b^2*x^2) + 4*b*E^(b^2*x^2)*Sqrt[Pi]*x*(-3 + 2 
*b^2*x^2)*Erfi[b*x] + 3*Pi*Erfi[b*x]^2))/(16*b^5*Sqrt[Pi])
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.27, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6941, 2641, 2638, 6941, 2638, 6929, 15}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^4 e^{b^2 x^2+c} \text {erfi}(b x) \, dx\)

\(\Big \downarrow \) 6941

\(\displaystyle -\frac {3 \int e^{b^2 x^2+c} x^2 \text {erfi}(b x)dx}{2 b^2}-\frac {\int e^{2 b^2 x^2+c} x^3dx}{\sqrt {\pi } b}+\frac {x^3 e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}\)

\(\Big \downarrow \) 2641

\(\displaystyle -\frac {3 \int e^{b^2 x^2+c} x^2 \text {erfi}(b x)dx}{2 b^2}-\frac {\frac {x^2 e^{2 b^2 x^2+c}}{4 b^2}-\frac {\int e^{2 b^2 x^2+c} xdx}{2 b^2}}{\sqrt {\pi } b}+\frac {x^3 e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}\)

\(\Big \downarrow \) 2638

\(\displaystyle -\frac {3 \int e^{b^2 x^2+c} x^2 \text {erfi}(b x)dx}{2 b^2}+\frac {x^3 e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}-\frac {\frac {x^2 e^{2 b^2 x^2+c}}{4 b^2}-\frac {e^{2 b^2 x^2+c}}{8 b^4}}{\sqrt {\pi } b}\)

\(\Big \downarrow \) 6941

\(\displaystyle -\frac {3 \left (-\frac {\int e^{b^2 x^2+c} \text {erfi}(b x)dx}{2 b^2}-\frac {\int e^{2 b^2 x^2+c} xdx}{\sqrt {\pi } b}+\frac {x e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}\right )}{2 b^2}+\frac {x^3 e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}-\frac {\frac {x^2 e^{2 b^2 x^2+c}}{4 b^2}-\frac {e^{2 b^2 x^2+c}}{8 b^4}}{\sqrt {\pi } b}\)

\(\Big \downarrow \) 2638

\(\displaystyle -\frac {3 \left (-\frac {\int e^{b^2 x^2+c} \text {erfi}(b x)dx}{2 b^2}+\frac {x e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}-\frac {e^{2 b^2 x^2+c}}{4 \sqrt {\pi } b^3}\right )}{2 b^2}+\frac {x^3 e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}-\frac {\frac {x^2 e^{2 b^2 x^2+c}}{4 b^2}-\frac {e^{2 b^2 x^2+c}}{8 b^4}}{\sqrt {\pi } b}\)

\(\Big \downarrow \) 6929

\(\displaystyle -\frac {3 \left (-\frac {\sqrt {\pi } e^c \int \text {erfi}(b x)d\text {erfi}(b x)}{4 b^3}+\frac {x e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}-\frac {e^{2 b^2 x^2+c}}{4 \sqrt {\pi } b^3}\right )}{2 b^2}+\frac {x^3 e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}-\frac {\frac {x^2 e^{2 b^2 x^2+c}}{4 b^2}-\frac {e^{2 b^2 x^2+c}}{8 b^4}}{\sqrt {\pi } b}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {x^3 e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}-\frac {\frac {x^2 e^{2 b^2 x^2+c}}{4 b^2}-\frac {e^{2 b^2 x^2+c}}{8 b^4}}{\sqrt {\pi } b}-\frac {3 \left (-\frac {\sqrt {\pi } e^c \text {erfi}(b x)^2}{8 b^3}+\frac {x e^{b^2 x^2+c} \text {erfi}(b x)}{2 b^2}-\frac {e^{2 b^2 x^2+c}}{4 \sqrt {\pi } b^3}\right )}{2 b^2}\)

Input:

Int[E^(c + b^2*x^2)*x^4*Erfi[b*x],x]
 

Output:

-((-1/8*E^(c + 2*b^2*x^2)/b^4 + (E^(c + 2*b^2*x^2)*x^2)/(4*b^2))/(b*Sqrt[P 
i])) + (E^(c + b^2*x^2)*x^3*Erfi[b*x])/(2*b^2) - (3*(-1/4*E^(c + 2*b^2*x^2 
)/(b^3*Sqrt[Pi]) + (E^(c + b^2*x^2)*x*Erfi[b*x])/(2*b^2) - (E^c*Sqrt[Pi]*E 
rfi[b*x]^2)/(8*b^3)))/(2*b^2)
 

Defintions of rubi rules used

rule 15

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

rule 2638

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 2641

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*L 
og[F])), x] - Simp[(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 6929

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

rule 6941

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

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.61 \[ \int e^{c+b^2 x^2} x^4 \text {erfi}(b x) \, dx=\frac {{\left (4 \, {\left (2 \, \pi b^{3} x^{3} - 3 \, \pi b x\right )} \operatorname {erfi}\left (b x\right ) e^{\left (b^{2} x^{2}\right )} + \sqrt {\pi } {\left (3 \, \pi \operatorname {erfi}\left (b x\right )^{2} - 4 \, {\left (b^{2} x^{2} - 2\right )} e^{\left (2 \, b^{2} x^{2}\right )}\right )}\right )} e^{c}}{16 \, \pi b^{5}} \] Input:

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

Output:

1/16*(4*(2*pi*b^3*x^3 - 3*pi*b*x)*erfi(b*x)*e^(b^2*x^2) + sqrt(pi)*(3*pi*e 
rfi(b*x)^2 - 4*(b^2*x^2 - 2)*e^(2*b^2*x^2)))*e^c/(pi*b^5)
 

Sympy [A] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int e^{c+b^2 x^2} x^4 \text {erfi}(b x) \, dx=\begin {cases} \frac {x^{3} e^{c} e^{b^{2} x^{2}} \operatorname {erfi}{\left (b x \right )}}{2 b^{2}} - \frac {x^{2} e^{c} e^{2 b^{2} x^{2}}}{4 \sqrt {\pi } b^{3}} - \frac {3 x e^{c} e^{b^{2} x^{2}} \operatorname {erfi}{\left (b x \right )}}{4 b^{4}} + \frac {e^{c} e^{2 b^{2} x^{2}}}{2 \sqrt {\pi } b^{5}} + \frac {3 \sqrt {\pi } e^{c} \operatorname {erfi}^{2}{\left (b x \right )}}{16 b^{5}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((x**3*exp(c)*exp(b**2*x**2)*erfi(b*x)/(2*b**2) - x**2*exp(c)*exp 
(2*b**2*x**2)/(4*sqrt(pi)*b**3) - 3*x*exp(c)*exp(b**2*x**2)*erfi(b*x)/(4*b 
**4) + exp(c)*exp(2*b**2*x**2)/(2*sqrt(pi)*b**5) + 3*sqrt(pi)*exp(c)*erfi( 
b*x)**2/(16*b**5), Ne(b, 0)), (0, True))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.04 \[ \int e^{c+b^2 x^2} x^4 \text {erfi}(b x) \, dx=\mathrm {erfi}\left (b\,x\right )\,\left (\frac {x^3\,{\mathrm {e}}^{b^2\,x^2+c}}{2\,b^2}-\frac {3\,x\,{\mathrm {e}}^{b^2\,x^2+c}}{4\,b^4}+\frac {3\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b^2\,x}{\sqrt {b^2}}\right )\,{\mathrm {e}}^c}{8\,{\left (b^2\right )}^{5/2}}\right )+\frac {8\,{\mathrm {e}}^{2\,b^2\,x^2+c}-3\,\pi \,{\mathrm {erfi}\left (\frac {b^2\,x}{\sqrt {b^2}}\right )}^2\,{\mathrm {e}}^c}{16\,b^5\,\sqrt {\pi }}-\frac {x^2\,{\mathrm {e}}^{2\,b^2\,x^2+c}}{4\,b^3\,\sqrt {\pi }} \] Input:

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

Output:

erfi(b*x)*((x^3*exp(c + b^2*x^2))/(2*b^2) - (3*x*exp(c + b^2*x^2))/(4*b^4) 
 + (3*pi^(1/2)*erfi((b^2*x)/(b^2)^(1/2))*exp(c))/(8*(b^2)^(5/2))) + (8*exp 
(c + 2*b^2*x^2) - 3*pi*erfi((b^2*x)/(b^2)^(1/2))^2*exp(c))/(16*b^5*pi^(1/2 
)) - (x^2*exp(c + 2*b^2*x^2))/(4*b^3*pi^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84 \[ \int e^{c+b^2 x^2} x^4 \text {erfi}(b x) \, dx=\frac {e^{c} \left (-3 \sqrt {\pi }\, \mathrm {erf}\left (b i x \right )^{2} \pi -8 e^{b^{2} x^{2}} \mathrm {erf}\left (b i x \right ) b^{3} i \pi \,x^{3}+12 e^{b^{2} x^{2}} \mathrm {erf}\left (b i x \right ) b i \pi x -4 \sqrt {\pi }\, e^{2 b^{2} x^{2}} b^{2} x^{2}+8 \sqrt {\pi }\, e^{2 b^{2} x^{2}}\right )}{16 b^{5} \pi } \] Input:

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

Output:

(e**c*( - 3*sqrt(pi)*erf(b*i*x)**2*pi - 8*e**(b**2*x**2)*erf(b*i*x)*b**3*i 
*pi*x**3 + 12*e**(b**2*x**2)*erf(b*i*x)*b*i*pi*x - 4*sqrt(pi)*e**(2*b**2*x 
**2)*b**2*x**2 + 8*sqrt(pi)*e**(2*b**2*x**2)))/(16*b**5*pi)