3.2.99 \(\int \text {Erfc}(b x) \sin (c+i b^2 x^2) \, dx\) [199]

Optimal. Leaf size=91 \[ \frac {i e^{i c} \sqrt {\pi } \text {Erfc}(b x)^2}{8 b}+\frac {i e^{-i c} \sqrt {\pi } \text {Erfi}(b x)}{4 b}-\frac {i b e^{-i c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6540, 6512, 2235, 6511, 6509, 30} \begin {gather*} -\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 {Erfc}(b x)^2}{8 b}+\frac {i \sqrt {\pi } e^{-i c} \text {Erfi}(b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Erfc[b*x]*Sin[c + I*b^2*x^2],x]

[Out]

((I/8)*E^(I*c)*Sqrt[Pi]*Erfc[b*x]^2)/b + ((I/4)*Sqrt[Pi]*Erfi[b*x])/(b*E^(I*c)) - ((I/2)*b*x^2*HypergeometricP
FQ[{1, 1}, {3/2, 2}, b^2*x^2])/(E^(I*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 6540

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

Rubi steps

\begin {align*} \int \text {erfc}(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 {erfc}(b x) \, dx\right )+\frac {1}{2} i \int e^{-i c+b^2 x^2} \text {erfc}(b x) \, dx\\ &=\frac {1}{2} i \int e^{-i c+b^2 x^2} \, dx-\frac {1}{2} i \int e^{-i c+b^2 x^2} \text {erf}(b x) \, dx+\frac {\left (i e^{i c} \sqrt {\pi }\right ) \text {Subst}(\int x \, dx,x,\text {erfc}(b x))}{4 b}\\ &=\frac {i e^{i c} \sqrt {\pi } \text {erfc}(b x)^2}{8 b}+\frac {i e^{-i c} \sqrt {\pi } \text {erfi}(b x)}{4 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 = 0.29, size = 94, normalized size = 1.03 \begin {gather*} \frac {(i \cos (c)+\sin (c)) \left (-4 b^2 x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )+\pi \left (2 \text {Erfi}(b x)-2 \text {Erf}(b x) (\cos (2 c)+i \sin (2 c))+\text {Erf}(b x)^2 (\cos (2 c)+i \sin (2 c))\right )\right )}{8 b \sqrt {\pi }} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \mathrm {erfc}\left (b x \right ) \sin \left (i b^{2} x^{2}+c \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(erfc(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin {\left (i b^{2} x^{2} + c \right )} \operatorname {erfc}{\left (b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(I*b**2*x**2 + c)*erfc(b*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sin \left (b^2\,x^2\,1{}\mathrm {i}+c\right )\,\mathrm {erfc}\left (b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + b^2*x^2*1i)*erfc(b*x),x)

[Out]

int(sin(c + b^2*x^2*1i)*erfc(b*x), x)

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