Integrand size = 19, antiderivative size = 134 \[ \int \frac {e^{c+b^2 x^2} \text {erfc}(b x)}{x^4} \, dx=\frac {b e^c}{3 \sqrt {\pi } x^2}-\frac {e^{c+b^2 x^2} \text {erfc}(b x)}{3 x^3}-\frac {2 b^2 e^{c+b^2 x^2} \text {erfc}(b x)}{3 x}+\frac {2}{3} b^3 e^c \sqrt {\pi } \text {erfi}(b x)-\frac {4 b^5 e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{3 \sqrt {\pi }}-\frac {4 b^3 e^c \log (x)}{3 \sqrt {\pi }} \]
-1/3*exp(b^2*x^2+c)*erfc(b*x)/x^3-2/3*b^2*exp(b^2*x^2+c)*erfc(b*x)/x+1/3*b *exp(c)/x^2/Pi^(1/2)-4/3*b^5*exp(c)*x^2*hypergeom([1, 1],[3/2, 2],b^2*x^2) /Pi^(1/2)-4/3*b^3*exp(c)*ln(x)/Pi^(1/2)+2/3*b^3*exp(c)*erfi(b*x)*Pi^(1/2)
Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.13 \[ \int \frac {e^{c+b^2 x^2} \text {erfc}(b x)}{x^4} \, dx=\frac {e^c \left (-e^{b^2 x^2} \sqrt {\pi }+b x-2 b^2 e^{b^2 x^2} \sqrt {\pi } x^2+e^{b^2 x^2} \sqrt {\pi } \left (1+2 b^2 x^2\right ) \text {erf}(b x)+2 b^3 \pi x^3 \text {erfi}(b x)-2 b^3 \pi x^3 \text {erf}(b x) \text {erfi}(b x)+4 b^5 x^5 \, _2F_2\left (1,1;\frac {3}{2},2;-b^2 x^2\right )-4 b^3 x^3 \log (x)\right )}{3 \sqrt {\pi } x^3} \]
(E^c*(-(E^(b^2*x^2)*Sqrt[Pi]) + b*x - 2*b^2*E^(b^2*x^2)*Sqrt[Pi]*x^2 + E^( b^2*x^2)*Sqrt[Pi]*(1 + 2*b^2*x^2)*Erf[b*x] + 2*b^3*Pi*x^3*Erfi[b*x] - 2*b^ 3*Pi*x^3*Erf[b*x]*Erfi[b*x] + 4*b^5*x^5*HypergeometricPFQ[{1, 1}, {3/2, 2} , -(b^2*x^2)] - 4*b^3*x^3*Log[x]))/(3*Sqrt[Pi]*x^3)
Time = 0.59 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6946, 15, 6946, 14, 6931, 2633, 6930}
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.
\(\displaystyle \int \frac {e^{b^2 x^2+c} \text {erfc}(b x)}{x^4} \, dx\) |
\(\Big \downarrow \) 6946 |
\(\displaystyle \frac {2}{3} b^2 \int \frac {e^{b^2 x^2+c} \text {erfc}(b x)}{x^2}dx-\frac {2 b \int \frac {e^c}{x^3}dx}{3 \sqrt {\pi }}-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{3 x^3}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2}{3} b^2 \int \frac {e^{b^2 x^2+c} \text {erfc}(b x)}{x^2}dx-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{3 x^3}+\frac {b e^c}{3 \sqrt {\pi } x^2}\) |
\(\Big \downarrow \) 6946 |
\(\displaystyle \frac {2}{3} b^2 \left (2 b^2 \int e^{b^2 x^2+c} \text {erfc}(b x)dx-\frac {2 b \int \frac {e^c}{x}dx}{\sqrt {\pi }}-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{x}\right )-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{3 x^3}+\frac {b e^c}{3 \sqrt {\pi } x^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {2}{3} b^2 \left (2 b^2 \int e^{b^2 x^2+c} \text {erfc}(b x)dx-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{x}-\frac {2 b e^c \log (x)}{\sqrt {\pi }}\right )-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{3 x^3}+\frac {b e^c}{3 \sqrt {\pi } x^2}\) |
\(\Big \downarrow \) 6931 |
\(\displaystyle \frac {2}{3} b^2 \left (2 b^2 \left (\int e^{b^2 x^2+c}dx-\int e^{b^2 x^2+c} \text {erf}(b x)dx\right )-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{x}-\frac {2 b e^c \log (x)}{\sqrt {\pi }}\right )-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{3 x^3}+\frac {b e^c}{3 \sqrt {\pi } x^2}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {2}{3} b^2 \left (2 b^2 \left (\frac {\sqrt {\pi } e^c \text {erfi}(b x)}{2 b}-\int e^{b^2 x^2+c} \text {erf}(b x)dx\right )-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{x}-\frac {2 b e^c \log (x)}{\sqrt {\pi }}\right )-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{3 x^3}+\frac {b e^c}{3 \sqrt {\pi } x^2}\) |
\(\Big \downarrow \) 6930 |
\(\displaystyle \frac {2}{3} b^2 \left (2 b^2 \left (\frac {\sqrt {\pi } e^c \text {erfi}(b x)}{2 b}-\frac {b e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{\sqrt {\pi }}\right )-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{x}-\frac {2 b e^c \log (x)}{\sqrt {\pi }}\right )-\frac {e^{b^2 x^2+c} \text {erfc}(b x)}{3 x^3}+\frac {b e^c}{3 \sqrt {\pi } x^2}\) |
(b*E^c)/(3*Sqrt[Pi]*x^2) - (E^(c + b^2*x^2)*Erfc[b*x])/(3*x^3) + (2*b^2*(- ((E^(c + b^2*x^2)*Erfc[b*x])/x) + 2*b^2*((E^c*Sqrt[Pi]*Erfi[b*x])/(2*b) - (b*E^c*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, b^2*x^2])/Sqrt[Pi]) - (2*b* E^c*Log[x])/Sqrt[Pi]))/3
3.2.77.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
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]
Int[E^((c_.) + (d_.)*(x_)^2)*Erfc[(b_.)*(x_)], x_Symbol] :> Int[E^(c + d*x^ 2), x] - Int[E^(c + d*x^2)*Erf[b*x], x] /; FreeQ[{b, c, d}, x] && EqQ[d, b^ 2]
Int[E^((c_.) + (d_.)*(x_)^2)*Erfc[(a_.) + (b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[x^(m + 1)*E^(c + d*x^2)*(Erfc[a + b*x]/(m + 1)), x] + (-Simp[2*(d/( m + 1)) Int[x^(m + 2)*E^(c + d*x^2)*Erfc[a + b*x], x], x] + Simp[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]
\[\int \frac {{\mathrm e}^{b^{2} x^{2}+c} \operatorname {erfc}\left (b x \right )}{x^{4}}d x\]
\[ \int \frac {e^{c+b^2 x^2} \text {erfc}(b x)}{x^4} \, dx=\int { \frac {\operatorname {erfc}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}}{x^{4}} \,d x } \]
Time = 80.53 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.16 \[ \int \frac {e^{c+b^2 x^2} \text {erfc}(b x)}{x^4} \, dx=- \frac {b^{3} {G_{3, 2}^{1, 3}\left (\begin {matrix} 2, \frac {5}{2}, 1 & \\2 & 0 \end {matrix} \middle | {\frac {1}{b^{2} x^{2}}} \right )} e^{c}}{2 \pi } \]
\[ \int \frac {e^{c+b^2 x^2} \text {erfc}(b x)}{x^4} \, dx=\int { \frac {\operatorname {erfc}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}}{x^{4}} \,d x } \]
\[ \int \frac {e^{c+b^2 x^2} \text {erfc}(b x)}{x^4} \, dx=\int { \frac {\operatorname {erfc}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {e^{c+b^2 x^2} \text {erfc}(b x)}{x^4} \, dx=\int \frac {{\mathrm {e}}^{b^2\,x^2+c}\,\mathrm {erfc}\left (b\,x\right )}{x^4} \,d x \]