3.69 \(\int \frac{e^{c+b^2 x^2} \text{Erf}(b x)}{x^5} \, dx\)

Optimal. Leaf size=115 \[ \frac{b^5 e^c x \text{HypergeometricPFQ}\left (\left \{\frac{1}{2},1\right \},\left \{\frac{3}{2},\frac{3}{2}\right \},b^2 x^2\right )}{\sqrt{\pi }}-\frac{b^2 e^{b^2 x^2+c} \text{Erf}(b x)}{4 x^2}-\frac{e^{b^2 x^2+c} \text{Erf}(b x)}{4 x^4}-\frac{b^3 e^c}{2 \sqrt{\pi } x}-\frac{b e^c}{6 \sqrt{\pi } x^3} \]

[Out]

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

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Rubi [A]  time = 0.134677, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {6391, 6388, 12, 30} \[ \frac{b^5 e^c x \, _2F_2\left (\frac{1}{2},1;\frac{3}{2},\frac{3}{2};b^2 x^2\right )}{\sqrt{\pi }}-\frac{b^2 e^{b^2 x^2+c} \text{Erf}(b x)}{4 x^2}-\frac{e^{b^2 x^2+c} \text{Erf}(b x)}{4 x^4}-\frac{b^3 e^c}{2 \sqrt{\pi } x}-\frac{b e^c}{6 \sqrt{\pi } x^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^(c + b^2*x^2)*Erf[b*x])/x^5,x]

[Out]

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

Rule 6391

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])/(m + 1), x] + (-Dist[(2*d)/(m + 1), Int[x^(m + 2)*E^(c + d*x^2)*Erf[a + b*x], x], x] - Dist[(2*b)/(
(m + 1)*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] &&
ILtQ[m, -1]

Rule 6388

Int[(E^((c_.) + (d_.)*(x_)^2)*Erf[(b_.)*(x_)])/(x_), x_Symbol] :> Simp[(2*b*E^c*x*HypergeometricPFQ[{1/2, 1},
{3/2, 3/2}, 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 \frac{e^{c+b^2 x^2} \text{erf}(b x)}{x^5} \, dx &=-\frac{e^{c+b^2 x^2} \text{erf}(b x)}{4 x^4}+\frac{1}{2} b^2 \int \frac{e^{c+b^2 x^2} \text{erf}(b x)}{x^3} \, dx+\frac{b \int \frac{e^c}{x^4} \, dx}{2 \sqrt{\pi }}\\ &=-\frac{e^{c+b^2 x^2} \text{erf}(b x)}{4 x^4}-\frac{b^2 e^{c+b^2 x^2} \text{erf}(b x)}{4 x^2}+\frac{1}{2} b^4 \int \frac{e^{c+b^2 x^2} \text{erf}(b x)}{x} \, dx+\frac{b^3 \int \frac{e^c}{x^2} \, dx}{2 \sqrt{\pi }}+\frac{\left (b e^c\right ) \int \frac{1}{x^4} \, dx}{2 \sqrt{\pi }}\\ &=-\frac{b e^c}{6 \sqrt{\pi } x^3}-\frac{e^{c+b^2 x^2} \text{erf}(b x)}{4 x^4}-\frac{b^2 e^{c+b^2 x^2} \text{erf}(b x)}{4 x^2}+\frac{b^5 e^c x \, _2F_2\left (\frac{1}{2},1;\frac{3}{2},\frac{3}{2};b^2 x^2\right )}{\sqrt{\pi }}+\frac{\left (b^3 e^c\right ) \int \frac{1}{x^2} \, dx}{2 \sqrt{\pi }}\\ &=-\frac{b e^c}{6 \sqrt{\pi } x^3}-\frac{b^3 e^c}{2 \sqrt{\pi } x}-\frac{e^{c+b^2 x^2} \text{erf}(b x)}{4 x^4}-\frac{b^2 e^{c+b^2 x^2} \text{erf}(b x)}{4 x^2}+\frac{b^5 e^c x \, _2F_2\left (\frac{1}{2},1;\frac{3}{2},\frac{3}{2};b^2 x^2\right )}{\sqrt{\pi }}\\ \end{align*}

Mathematica [A]  time = 0.139924, size = 36, normalized size = 0.31 \[ -\frac{2 b e^c \text{HypergeometricPFQ}\left (\left \{-\frac{3}{2},1\right \},\left \{-\frac{1}{2},\frac{3}{2}\right \},b^2 x^2\right )}{3 \sqrt{\pi } x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(c + b^2*x^2)*Erf[b*x])/x^5,x]

[Out]

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

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Maple [F]  time = 0.279, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{{b}^{2}{x}^{2}+c}}{\it Erf} \left ( bx \right ) }{{x}^{5}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b^2*x^2+c)*erf(b*x)/x^5,x)

[Out]

int(exp(b^2*x^2+c)*erf(b*x)/x^5,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b^2*x^2+c)*erf(b*x)/x^5,x, algorithm="maxima")

[Out]

integrate(erf(b*x)*e^(b^2*x^2 + c)/x^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b^2*x^2+c)*erf(b*x)/x^5,x, algorithm="fricas")

[Out]

integral(erf(b*x)*e^(b^2*x^2 + c)/x^5, x)

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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)*erf(b*x)/x**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b^2*x^2+c)*erf(b*x)/x^5,x, algorithm="giac")

[Out]

integrate(erf(b*x)*e^(b^2*x^2 + c)/x^5, x)