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

Optimal. Leaf size=62 \[ \frac {b e^{i c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }}+\frac {\sqrt {\pi } e^{-i c} \text {erf}(b x)^2}{8 b} \]

[Out]

1/2*b*exp(I*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(I*c)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 62, 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 = {6407, 6373, 30, 6376} \[ \frac {b e^{i c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )}{2 \sqrt {\pi }}+\frac {\sqrt {\pi } e^{-i c} \text {Erf}(b x)^2}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c - I*b^2*x^2]*Erf[b*x],x]

[Out]

(Sqrt[Pi]*Erf[b*x]^2)/(8*b*E^(I*c)) + (b*E^(I*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 6407

Int[Cos[(c_.) + (d_.)*(x_)^2]*Erf[(b_.)*(x_)], x_Symbol] :> Dist[1/2, Int[E^(-(I*c) - I*d*x^2)*Erf[b*x], x], x
] + Dist[1/2, 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 \cos \left (c-i b^2 x^2\right ) \text {erf}(b x) \, dx &=\frac {1}{2} \int e^{-i c-b^2 x^2} \text {erf}(b x) \, dx+\frac {1}{2} \int e^{i c+b^2 x^2} \text {erf}(b x) \, dx\\ &=\frac {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 (e^{-i c} \sqrt {\pi }\right ) \operatorname {Subst}(\int x \, dx,x,\text {erf}(b x))}{4 b}\\ &=\frac {e^{-i c} \sqrt {\pi } \text {erf}(b x)^2}{8 b}+\frac {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 [F]  time = 0.77, size = 0, normalized size = 0.00 \[ \int \cos \left (c-i b^2 x^2\right ) \text {erf}(b x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{2} \, {\left (\operatorname {erf}\left (b x\right ) e^{\left (-2 \, b^{2} x^{2} - 2 i \, c\right )} + \operatorname {erf}\left (b x\right )\right )} e^{\left (b^{2} x^{2} + i \, c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c - b^2*x^2*1i)*erf(b*x),x)

[Out]

int(cos(c - b^2*x^2*1i)*erf(b*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(-c+I*b**2*x**2)*erf(b*x),x)

[Out]

Integral(cos(I*b**2*x**2 - c)*erf(b*x), x)

________________________________________________________________________________________