3.206 \(\int \cosh (c-b^2 x^2) \text{Erfc}(b x) \, dx\)

Optimal. Leaf size=77 \[ -\frac{b e^{-c} x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{\sqrt{\pi } e^c \text{Erfc}(b x)^2}{8 b}+\frac{\sqrt{\pi } e^{-c} \text{Erfi}(b x)}{4 b} \]

[Out]

-(E^c*Sqrt[Pi]*Erfc[b*x]^2)/(8*b) + (Sqrt[Pi]*Erfi[b*x])/(4*b*E^c) - (b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}
, b^2*x^2])/(2*E^c*Sqrt[Pi])

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Rubi [A]  time = 0.0714556, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6414, 6374, 30, 6377, 2204, 6376} \[ -\frac{b e^{-c} x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{\sqrt{\pi } e^c \text{Erfc}(b x)^2}{8 b}+\frac{\sqrt{\pi } e^{-c} \text{Erfi}(b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c - b^2*x^2]*Erfc[b*x],x]

[Out]

-(E^c*Sqrt[Pi]*Erfc[b*x]^2)/(8*b) + (Sqrt[Pi]*Erfi[b*x])/(4*b*E^c) - (b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}
, b^2*x^2])/(2*E^c*Sqrt[Pi])

Rule 6414

Int[Cosh[(c_.) + (d_.)*(x_)^2]*Erfc[(b_.)*(x_)], x_Symbol] :> Dist[1/2, Int[E^(c + d*x^2)*Erfc[b*x], x], x] +
Dist[1/2, Int[E^(-c - d*x^2)*Erfc[b*x], x], x] /; FreeQ[{b, c, d}, x] && EqQ[d^2, b^4]

Rule 6374

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

Rule 30

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

Rule 6377

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]

Rule 2204

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]

Rule 6376

Int[E^((c_.) + (d_.)*(x_)^2)*Erf[(b_.)*(x_)], x_Symbol] :> Simp[(b*E^c*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2},
b^2*x^2])/Sqrt[Pi], x] /; FreeQ[{b, c, d}, x] && EqQ[d, b^2]

Rubi steps

\begin{align*} \int \cosh \left (c-b^2 x^2\right ) \text{erfc}(b x) \, dx &=\frac{1}{2} \int e^{c-b^2 x^2} \text{erfc}(b x) \, dx+\frac{1}{2} \int e^{-c+b^2 x^2} \text{erfc}(b x) \, dx\\ &=\frac{1}{2} \int e^{-c+b^2 x^2} \, dx-\frac{1}{2} \int e^{-c+b^2 x^2} \text{erf}(b x) \, dx-\frac{\left (e^c \sqrt{\pi }\right ) \operatorname{Subst}(\int x \, dx,x,\text{erfc}(b x))}{4 b}\\ &=-\frac{e^c \sqrt{\pi } \text{erfc}(b x)^2}{8 b}+\frac{e^{-c} \sqrt{\pi } \text{erfi}(b x)}{4 b}-\frac{b e^{-c} x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi }}\\ \end{align*}

Mathematica [A]  time = 0.11046, size = 117, normalized size = 1.52 \[ \frac{4 b^2 x^2 \sinh (c) \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )+4 b^2 x^2 \cosh (c) \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-b^2 x^2\right )-\pi \left (2 \text{Erf}(b x) (\cosh (c) \text{Erfi}(b x)-\sinh (c)-\cosh (c))+\text{Erf}(b x)^2 (\sinh (c)+\cosh (c))+2 \text{Erfi}(b x) (\sinh (c)-\cosh (c))\right )}{8 \sqrt{\pi } b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cosh[c - b^2*x^2]*Erfc[b*x],x]

[Out]

(4*b^2*x^2*Cosh[c]*HypergeometricPFQ[{1, 1}, {3/2, 2}, -(b^2*x^2)] + 4*b^2*x^2*HypergeometricPFQ[{1, 1}, {3/2,
2}, b^2*x^2]*Sinh[c] - Pi*(2*Erf[b*x]*(-Cosh[c] + Cosh[c]*Erfi[b*x] - Sinh[c]) + 2*Erfi[b*x]*(-Cosh[c] + Sinh
[c]) + Erf[b*x]^2*(Cosh[c] + Sinh[c])))/(8*b*Sqrt[Pi])

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Maple [F]  time = 0.342, size = 0, normalized size = 0. \begin{align*} \int \cosh \left ({b}^{2}{x}^{2}-c \right ){\it erfc} \left ( bx \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b^2*x^2-c)*erfc(b*x),x)

[Out]

int(cosh(b^2*x^2-c)*erfc(b*x),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh \left (b^{2} x^{2} - c\right ) \operatorname{erfc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b^2*x^2-c)*erfc(b*x),x, algorithm="maxima")

[Out]

integrate(cosh(b^2*x^2 - c)*erfc(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b^2*x^2-c)*erfc(b*x),x, algorithm="fricas")

[Out]

integral(-cosh(b^2*x^2 - c)*erf(b*x) + cosh(b^2*x^2 - c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh{\left (b^{2} x^{2} - c \right )} \operatorname{erfc}{\left (b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b**2*x**2-c)*erfc(b*x),x)

[Out]

Integral(cosh(b**2*x**2 - c)*erfc(b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh \left (b^{2} x^{2} - c\right ) \operatorname{erfc}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b^2*x^2-c)*erfc(b*x),x, algorithm="giac")

[Out]

integrate(cosh(b^2*x^2 - c)*erfc(b*x), x)