3.97 \(\int \text {erf}(b x) \sin (c-i b^2 x^2) \, dx\)

Optimal. Leaf size=66 \[ \frac {i \sqrt {\pi } e^{-i c} \text {erf}(b x)^2}{8 b}-\frac {i b e^{i c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, 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, 6373, 30, 6376} \[ \frac {i \sqrt {\pi } e^{-i c} \text {Erf}(b x)^2}{8 b}-\frac {i b e^{i c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }} \]

Antiderivative was successfully verified.

[In]

Int[Erf[b*x]*Sin[c - I*b^2*x^2],x]

[Out]

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

Rubi steps

\begin {align*} \int \text {erf}(b x) \sin \left (c-i b^2 x^2\right ) \, dx &=\frac {1}{2} i \int e^{-i c-b^2 x^2} \text {erf}(b x) \, dx-\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*}

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Mathematica [A]  time = 0.09, size = 67, normalized size = 1.02 \[ \frac {(\sin (c)+i \cos (c)) \left (\pi \text {erf}(b x)^2-4 b^2 x^2 (\cos (2 c)+i \sin (2 c)) \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )\right )}{8 \sqrt {\pi } b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{2} \, {\left (i \, \operatorname {erf}\left (b x\right ) e^{\left (-2 \, b^{2} x^{2} - 2 i \, c\right )} - i \, \operatorname {erf}\left (b x\right )\right )} e^{\left (b^{2} x^{2} + i \, c\right )}, x\right ) \]

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

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)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int -\erf \left (b x \right ) \sin \left (i b^{2} x^{2}-c \right )\, dx \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c - b^2*x^2*1i)*erf(b*x),x)

[Out]

int(sin(c - b^2*x^2*1i)*erf(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-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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