3.308 \(\int \text {erfi}(b x) \sinh (c+b^2 x^2) \, dx\)

Optimal. Leaf size=57 \[ \frac {\sqrt {\pi } e^c \text {erfi}(b x)^2}{8 b}-\frac {b e^{-c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-b^2 x^2\right )}{2 \sqrt {\pi }} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6412, 6375, 30, 6378} \[ \frac {\sqrt {\pi } e^c \text {Erfi}(b x)^2}{8 b}-\frac {b e^{-c} x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-b^2 x^2\right )}{2 \sqrt {\pi }} \]

Antiderivative was successfully verified.

[In]

Int[Erfi[b*x]*Sinh[c + b^2*x^2],x]

[Out]

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

Rule 30

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

Rule 6375

Int[E^((c_.) + (d_.)*(x_)^2)*Erfi[(b_.)*(x_)]^(n_.), x_Symbol] :> Dist[(E^c*Sqrt[Pi])/(2*b), Subst[Int[x^n, x]
, x, Erfi[b*x]], x] /; FreeQ[{b, c, d, n}, x] && EqQ[d, b^2]

Rule 6378

Int[E^((c_.) + (d_.)*(x_)^2)*Erfi[(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 6412

Int[Erfi[(b_.)*(x_)]*Sinh[(c_.) + (d_.)*(x_)^2], x_Symbol] :> Dist[1/2, Int[E^(c + d*x^2)*Erfi[b*x], x], x] -
Dist[1/2, Int[E^(-c - d*x^2)*Erfi[b*x], x], x] /; FreeQ[{b, c, d}, x] && EqQ[d^2, b^4]

Rubi steps

\begin {align*} \int \text {erfi}(b x) \sinh \left (c+b^2 x^2\right ) \, dx &=-\left (\frac {1}{2} \int e^{-c-b^2 x^2} \text {erfi}(b x) \, dx\right )+\frac {1}{2} \int e^{c+b^2 x^2} \text {erfi}(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 {erfi}(b x))}{4 b}\\ &=\frac {e^c \sqrt {\pi } \text {erfi}(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*}

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Mathematica [A]  time = 1.56, size = 74, normalized size = 1.30 \[ \frac {4 b^2 x^2 (\cosh (c)-\sinh (c)) \, _2F_2\left (1,1;\frac {3}{2},2;b^2 x^2\right )+\pi \text {erfi}(b x) (\text {erfi}(b x) (\sinh (c)+\cosh (c))-2 \text {erf}(b x) (\cosh (c)-\sinh (c)))}{8 \sqrt {\pi } b} \]

Antiderivative was successfully verified.

[In]

Integrate[Erfi[b*x]*Sinh[c + b^2*x^2],x]

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {erfi}\left (b x\right ) \sinh \left (b^{2} x^{2} + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(erfi(b*x)*sinh(b^2*x^2+c),x, algorithm="fricas")

[Out]

integral(erfi(b*x)*sinh(b^2*x^2 + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {erfi}\left (b x\right ) \sinh \left (b^{2} x^{2} + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(erfi(b*x)*sinh(b^2*x^2+c),x, algorithm="giac")

[Out]

integrate(erfi(b*x)*sinh(b^2*x^2 + c), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \erfi \left (b x \right ) \sinh \left (b^{2} x^{2}+c \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(erfi(b*x)*sinh(b^2*x^2+c),x)

[Out]

int(erfi(b*x)*sinh(b^2*x^2+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {erfi}\left (b x\right ) \sinh \left (b^{2} x^{2} + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(erfi(b*x)*sinh(b^2*x^2+c),x, algorithm="maxima")

[Out]

integrate(erfi(b*x)*sinh(b^2*x^2 + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + b^2*x^2)*erfi(b*x),x)

[Out]

int(sinh(c + b^2*x^2)*erfi(b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\left (b^{2} x^{2} + c \right )} \operatorname {erfi}{\left (b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(erfi(b*x)*sinh(b**2*x**2+c),x)

[Out]

Integral(sinh(b**2*x**2 + c)*erfi(b*x), x)

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