3.102 \(\int \cosh (c+b^2 x^2) \text {erf}(b x) \, dx\)

Optimal. Leaf size=56 \[ \frac {b e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }}+\frac {\sqrt {\pi } e^{-c} \text {erf}(b x)^2}{8 b} \]

[Out]

1/2*b*exp(c)*x^2*HypergeometricPFQ([1, 1],[3/2, 2],b^2*x^2)/Pi^(1/2)+1/8*erf(b*x)^2*Pi^(1/2)/b/exp(c)

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Rubi [A]  time = 0.05, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6413, 6376, 6373, 30} \[ \frac {b e^c x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }}+\frac {\sqrt {\pi } e^{-c} \text {Erf}(b x)^2}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + b^2*x^2]*Erf[b*x],x]

[Out]

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

Rule 30

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

Rule 6373

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

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]

Rule 6413

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

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 93, normalized size = 1.66 \[ \frac {4 b^2 x^2 \sinh (c) \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )-4 b^2 x^2 \cosh (c) \, _2F_2\left (1,1;\frac {3}{2},2;-b^2 x^2\right )+\pi \text {erf}(b x) (\text {erf}(b x) (\cosh (c)-\sinh (c))+2 \cosh (c) \text {erfi}(b x))}{8 \sqrt {\pi } b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + b^2*x^2]*Erf[b*x],x]

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cosh \left (b^{2} x^{2} + c\right ) \operatorname {erf}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cosh \left (b^{2} x^{2} + c\right ) \operatorname {erf}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \cosh \left (b^{2} x^{2}+c \right ) \erf \left (b x \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\sqrt {\pi } \operatorname {erf}\left (b x\right )^{2} e^{\left (-c\right )}}{8 \, b} + \frac {1}{2} \, \int \operatorname {erf}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(pi)*erf(b*x)^2*e^(-c)/b + 1/2*integrate(erf(b*x)*e^(b^2*x^2 + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {cosh}\left (b^2\,x^2+c\right )\,\mathrm {erf}\left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + b^2*x^2)*erf(b*x),x)

[Out]

int(cosh(c + b^2*x^2)*erf(b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cosh {\left (b^{2} x^{2} + c \right )} \operatorname {erf}{\left (b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b**2*x**2+c)*erf(b*x),x)

[Out]

Integral(cosh(b**2*x**2 + c)*erf(b*x), x)

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