3.1.95 \(\int (\frac {e^{-b^2 x^2} \text {Erf}(b x)}{x^3}+\frac {b^2 e^{-b^2 x^2} \text {Erf}(b x)}{x}) \, dx\) [95]

Optimal. Leaf size=62 \[ -\frac {b e^{-2 b^2 x^2}}{\sqrt {\pi } x}-\frac {e^{-b^2 x^2} \text {Erf}(b x)}{2 x^2}-\sqrt {2} b^2 \text {Erf}\left (\sqrt {2} b x\right ) \]

[Out]

-1/2*erf(b*x)/exp(b^2*x^2)/x^2-b^2*erf(b*x*2^(1/2))*2^(1/2)-b/exp(2*b^2*x^2)/x/Pi^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {6526, 2245, 2236} \begin {gather*} -\frac {e^{-b^2 x^2} \text {Erf}(b x)}{2 x^2}-\sqrt {2} b^2 \text {Erf}\left (\sqrt {2} b x\right )-\frac {b e^{-2 b^2 x^2}}{\sqrt {\pi } x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Erf[b*x]/(E^(b^2*x^2)*x^3) + (b^2*Erf[b*x])/(E^(b^2*x^2)*x),x]

[Out]

-(b/(E^(2*b^2*x^2)*Sqrt[Pi]*x)) - Erf[b*x]/(2*E^(b^2*x^2)*x^2) - Sqrt[2]*b^2*Erf[Sqrt[2]*b*x]

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 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 6526

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]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 62, normalized size = 1.00 \begin {gather*} -\frac {b e^{-2 b^2 x^2}}{\sqrt {\pi } x}-\frac {e^{-b^2 x^2} \text {Erf}(b x)}{2 x^2}-\sqrt {2} b^2 \text {Erf}\left (\sqrt {2} b x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(b/(E^(2*b^2*x^2)*Sqrt[Pi]*x)) - Erf[b*x]/(2*E^(b^2*x^2)*x^2) - Sqrt[2]*b^2*Erf[Sqrt[2]*b*x]

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Maple [A]
time = 0.70, size = 67, normalized size = 1.08

method	result	size
default	\(\frac {-\frac {\erf \left (b x \right ) b \,{\mathrm e}^{-b^{2} x^{2}}}{2 x^{2}}+\frac {b^{3} \left (-\frac {{\mathrm e}^{-2 b^{2} x^{2}}}{b x}-\sqrt {2}\, \sqrt {\pi }\, \erf \left (b x \sqrt {2}\right )\right )}{\sqrt {\pi }}}{b}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 66, normalized size = 1.06 \begin {gather*} -\frac {2 \, \sqrt {2} \pi \sqrt {b^{2}} b x^{2} \operatorname {erf}\left (\sqrt {2} \sqrt {b^{2}} x\right ) + 2 \, \sqrt {\pi } b x e^{\left (-2 \, b^{2} x^{2}\right )} + \pi \operatorname {erf}\left (b x\right ) e^{\left (-b^{2} x^{2}\right )}}{2 \, \pi x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*sqrt(2)*pi*sqrt(b^2)*b*x^2*erf(sqrt(2)*sqrt(b^2)*x) + 2*sqrt(pi)*b*x*e^(-2*b^2*x^2) + pi*erf(b*x)*e^(-
b^2*x^2))/(pi*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b^{2} x^{2} + 1\right ) e^{- b^{2} x^{2}} \operatorname {erf}{\left (b x \right )}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b**2*x**2 + 1)*exp(-b**2*x**2)*erf(b*x)/x**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(b^2*erf(b*x)*e^(-b^2*x^2)/x + erf(b*x)*e^(-b^2*x^2)/x^3, x)

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Mupad [B]
time = 0.22, size = 52, normalized size = 0.84 \begin {gather*} -\frac {\frac {{\mathrm {e}}^{-b^2\,x^2}\,\mathrm {erf}\left (b\,x\right )}{2}+\frac {b\,x\,{\mathrm {e}}^{-2\,b^2\,x^2}}{\sqrt {\pi }}}{x^2}-\sqrt {2}\,b^2\,\mathrm {erf}\left (\sqrt {2}\,b\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-b^2*x^2)*erf(b*x))/x^3 + (b^2*exp(-b^2*x^2)*erf(b*x))/x,x)

[Out]

- ((exp(-b^2*x^2)*erf(b*x))/2 + (b*x*exp(-2*b^2*x^2))/pi^(1/2))/x^2 - 2^(1/2)*b^2*erf(2^(1/2)*b*x)

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