3.101 \(\int \text{Erf}(b x) \sinh (c-b^2 x^2) \, dx\)

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

[Out]

(E^c*Sqrt[Pi]*Erf[b*x]^2)/(8*b) - (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 = 0.0561107, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6410, 6373, 30, 6376} \[ \frac{\sqrt{\pi } e^c \text{Erf}(b x)^2}{8 b}-\frac{b e^{-c} x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi }} \]

Antiderivative was successfully verified.

[In]

Int[Erf[b*x]*Sinh[c - b^2*x^2],x]

[Out]

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

Rule 6410

Int[Erf[(b_.)*(x_)]*Sinh[(c_.) + (d_.)*(x_)^2], 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]

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 30

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

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 \text{erf}(b x) \sinh \left (c-b^2 x^2\right ) \, 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*}

Mathematica [A]  time = 0.0394838, size = 61, normalized size = 1.09 \[ \frac{(\cosh (c)-\sinh (c)) \left (\pi \text{Erf}(b x)^2 (\sinh (2 c)+\cosh (2 c))-4 b^2 x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )\right )}{8 \sqrt{\pi } b} \]

Antiderivative was successfully verified.

[In]

Integrate[Erf[b*x]*Sinh[c - b^2*x^2],x]

[Out]

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

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Maple [F]  time = 0.065, size = 0, normalized size = 0. \begin{align*} \int -{\it Erf} \left ( bx \right ) \sinh \left ({b}^{2}{x}^{2}-c \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-erf(b*x)*sinh(b^2*x^2-c),x)

[Out]

int(-erf(b*x)*sinh(b^2*x^2-c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{\pi } \operatorname{erf}\left (b x\right )^{2} e^{c}}{8 \, b} - \frac{1}{2} \, \int \operatorname{erf}\left (b x\right ) e^{\left (b^{2} x^{2} - c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-erf(b*x)*sinh(b^2*x^2-c),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-erf(b*x)*sinh(b**2*x**2-c),x)

[Out]

-Integral(sinh(b**2*x**2 - c)*erf(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-erf(b*x)*sinh(b^2*x^2-c),x, algorithm="giac")

[Out]

integrate(-erf(b*x)*sinh(b^2*x^2 - c), x)