3.96 \(\int \text{Erf}(b x) \sin (c+i b^2 x^2) \, dx\)

Optimal. Leaf size=66 \[ \frac{i b e^{-i c} x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{i \sqrt{\pi } e^{i c} \text{Erf}(b x)^2}{8 b} \]

[Out]

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

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Rubi [A]  time = 0.0630758, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6404, 6376, 6373, 30} \[ \frac{i b e^{-i c} x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{i \sqrt{\pi } e^{i c} \text{Erf}(b x)^2}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Erf[b*x]*Sin[c + I*b^2*x^2],x]

[Out]

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

Rule 6404

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

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 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]

Rubi steps

\begin{align*} \int \text{erf}(b x) \sin \left (c+i b^2 x^2\right ) \, dx &=-\left (\frac{1}{2} i \int e^{i c-b^2 x^2} \text{erf}(b x) \, dx\right )+\frac{1}{2} i \int e^{-i c+b^2 x^2} \text{erf}(b x) \, dx\\ &=\frac{i b e^{-i c} x^2 \, _2F_2\left (1,1;\frac{3}{2},2;b^2 x^2\right )}{2 \sqrt{\pi }}-\frac{\left (i e^{i c} \sqrt{\pi }\right ) \operatorname{Subst}(\int x \, dx,x,\text{erf}(b x))}{4 b}\\ &=-\frac{i e^{i c} \sqrt{\pi } \text{erf}(b x)^2}{8 b}+\frac{i b e^{-i 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 = 0.0690367, size = 69, normalized size = 1.05 \[ \frac{(\cos (c)-i \sin (c)) \left (4 i b^2 x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},b^2 x^2\right )+\pi \text{Erf}(b x)^2 (\sin (2 c)-i \cos (2 c))\right )}{8 \sqrt{\pi } b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.111, size = 0, normalized size = 0. \begin{align*} \int{\it Erf} \left ( bx \right ) \sin \left ( c+i{b}^{2}{x}^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{2} \,{\left (-i \, \operatorname{erf}\left (b x\right ) e^{\left (-2 \, b^{2} x^{2} + 2 i \, c\right )} + i \, \operatorname{erf}\left (b x\right )\right )} e^{\left (b^{2} x^{2} - i \, c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/2*(-I*erf(b*x)*e^(-2*b^2*x^2 + 2*I*c) + I*erf(b*x))*e^(b^2*x^2 - I*c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (i 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)*sin(c+I*b**2*x**2),x)

[Out]

Integral(sin(I*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 ) \sin \left (i \, 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)*sin(c+I*b^2*x^2),x, algorithm="giac")

[Out]

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