3.361 \(\int e^{\sinh ^{-1}(a+b x)^2} \, dx\)

Optimal. Leaf size=65 \[ \frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b} \]

[Out]

1/4*erfi(-1/2+arcsinh(b*x+a))*Pi^(1/2)/b/exp(1/4)+1/4*erfi(1/2+arcsinh(b*x+a))*Pi^(1/2)/b/exp(1/4)

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Rubi [A]  time = 0.06, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5896, 5513, 2234, 2204} \[ \frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]^2,x]

[Out]

(Sqrt[Pi]*Erfi[(-1 + 2*ArcSinh[a + b*x])/2])/(4*b*E^(1/4)) + (Sqrt[Pi]*Erfi[(1 + 2*ArcSinh[a + b*x])/2])/(4*b*
E^(1/4))

Rule 2204

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 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 5513

Int[Cosh[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cosh[v]^n, x], x] /; FreeQ[F, x] && (Linea
rQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]

Rule 5896

Int[(f_)^(ArcSinh[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[f^(c*x^n)*Cosh[x], x], x,
ArcSinh[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int e^{\sinh ^{-1}(a+b x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{2} e^{-x+x^2}+\frac {e^{x+x^2}}{2}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^{-x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b}+\frac {\operatorname {Subst}\left (\int e^{x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b}\\ &=\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b \sqrt [4]{e}}+\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b \sqrt [4]{e}}\\ &=\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b \sqrt [4]{e}}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b \sqrt [4]{e}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 44, normalized size = 0.68 \[ \frac {\sqrt {\pi } \left (\text {erfi}\left (\sinh ^{-1}(a+b x)+\frac {1}{2}\right )+\text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )\right )}{4 \sqrt [4]{e} b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSinh[a + b*x]^2,x]

[Out]

(Sqrt[Pi]*(Erfi[1/2 + ArcSinh[a + b*x]] + Erfi[(-1 + 2*ArcSinh[a + b*x])/2]))/(4*b*E^(1/4))

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (e^{\left (\operatorname {arsinh}\left (b x + a\right )^{2}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsinh(b*x+a)^2),x, algorithm="fricas")

[Out]

integral(e^(arcsinh(b*x + a)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsinh(b*x+a)^2),x, algorithm="giac")

[Out]

integrate(e^(arcsinh(b*x + a)^2), x)

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\arcsinh \left (b x +a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arcsinh(b*x+a)^2),x)

[Out]

int(exp(arcsinh(b*x+a)^2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsinh(b*x+a)^2),x, algorithm="maxima")

[Out]

integrate(e^(arcsinh(b*x + a)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(asinh(a + b*x)^2),x)

[Out]

int(exp(asinh(a + b*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(asinh(b*x+a)**2),x)

[Out]

Integral(exp(asinh(a + b*x)**2), x)

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