\(\int \cosh (c-b^2 x^2) \text {erfc}(b x) \, dx\) [206]

Optimal result

Integrand size = 16, antiderivative size = 77 \[ \int \cosh \left (c-b^2 x^2\right ) \text {erfc}(b x) \, dx=-\frac {e^c \sqrt {\pi } \text {erfc}(b x)^2}{8 b}+\frac {e^{-c} \sqrt {\pi } \text {erfi}(b x)}{4 b}-\frac {b e^{-c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }} \]

[Out]

-1/2*b*x^2*hypergeom([1, 1],[3/2, 2],b^2*x^2)/exp(c)/Pi^(1/2)-1/8*exp(c)*erfc(b*x)^2*Pi^(1/2)/b+1/4*erfi(b*x)*
Pi^(1/2)/b/exp(c)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6549, 6509, 30, 6512, 2235, 6511} \[ \int \cosh \left (c-b^2 x^2\right ) \text {erfc}(b x) \, dx=-\frac {b e^{-c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }}-\frac {\sqrt {\pi } e^c \text {erfc}(b x)^2}{8 b}+\frac {\sqrt {\pi } e^{-c} \text {erfi}(b x)}{4 b} \]

[In]

Int[Cosh[c - b^2*x^2]*Erfc[b*x],x]

[Out]

-1/8*(E^c*Sqrt[Pi]*Erfc[b*x]^2)/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 30

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

Rule 2235

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 6509

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 6511

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]

Rule 6512

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 6549

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.31 \[ \int \cosh \left (c-b^2 x^2\right ) \text {erfc}(b x) \, dx=\frac {4 b^2 x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-b^2 x^2\right ) (\cosh (c)-\sinh (c))-\pi \left (2 \text {erf}(b x) (-\cosh (c)+\text {erfi}(b x) (\cosh (c)-\sinh (c))-\sinh (c))+2 \text {erfi}(b x) (-\cosh (c)+\sinh (c))+\text {erf}(b x)^2 (\cosh (c)+\sinh (c))\right )}{8 b \sqrt {\pi }} \]

[In]

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

[Out]

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

Maple [F]

\[\int \cosh \left (b^{2} x^{2}-c \right ) \operatorname {erfc}\left (b x \right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \cosh \left (c-b^2 x^2\right ) \text {erfc}(b x) \, dx=\int { \cosh \left (b^{2} x^{2} - c\right ) \operatorname {erfc}\left (b x\right ) \,d x } \]

[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)

Sympy [F]

\[ \int \cosh \left (c-b^2 x^2\right ) \text {erfc}(b x) \, dx=\int \cosh {\left (b^{2} x^{2} - c \right )} \operatorname {erfc}{\left (b x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \cosh \left (c-b^2 x^2\right ) \text {erfc}(b x) \, dx=\int { \cosh \left (b^{2} x^{2} - c\right ) \operatorname {erfc}\left (b x\right ) \,d x } \]

[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)

Giac [F]

\[ \int \cosh \left (c-b^2 x^2\right ) \text {erfc}(b x) \, dx=\int { \cosh \left (b^{2} x^{2} - c\right ) \operatorname {erfc}\left (b x\right ) \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \cosh \left (c-b^2 x^2\right ) \text {erfc}(b x) \, dx=\int \mathrm {cosh}\left (c-b^2\,x^2\right )\,\mathrm {erfc}\left (b\,x\right ) \,d x \]

[In]

int(cosh(c - b^2*x^2)*erfc(b*x),x)

[Out]

int(cosh(c - b^2*x^2)*erfc(b*x), x)