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

Optimal. Leaf size=251 \[ \frac {\sqrt {\pi } a^2 \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } a^2 \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^3}-\frac {\sqrt {\pi } a \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{4 e b^3}-\frac {\sqrt {\pi } a \text {erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{4 e b^3}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-3\right )\right )}{16 e^{9/4} b^3}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{16 \sqrt [4]{e} b^3}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{16 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+3\right )\right )}{16 e^{9/4} b^3} \]

[Out]

1/4*a*erfi(-1+arcsinh(b*x+a))*Pi^(1/2)/b^3/E-1/4*a*erfi(1+arcsinh(b*x+a))*Pi^(1/2)/b^3/E+1/16*erfi(-3/2+arcsin
h(b*x+a))*Pi^(1/2)/b^3/exp(9/4)-1/16*erfi(-1/2+arcsinh(b*x+a))*Pi^(1/2)/b^3/exp(1/4)+1/4*a^2*erfi(-1/2+arcsinh
(b*x+a))*Pi^(1/2)/b^3/exp(1/4)-1/16*erfi(1/2+arcsinh(b*x+a))*Pi^(1/2)/b^3/exp(1/4)+1/4*a^2*erfi(1/2+arcsinh(b*
x+a))*Pi^(1/2)/b^3/exp(1/4)+1/16*erfi(3/2+arcsinh(b*x+a))*Pi^(1/2)/b^3/exp(9/4)

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Rubi [A]  time = 0.52, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5898, 6741, 12, 6742, 5513, 2234, 2204, 5514} \[ \frac {\sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^3}-\frac {\sqrt {\pi } a \text {Erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{4 e b^3}-\frac {\sqrt {\pi } a \text {Erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{4 e b^3}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-3\right )\right )}{16 e^{9/4} b^3}-\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{16 \sqrt [4]{e} b^3}-\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{16 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+3\right )\right )}{16 e^{9/4} b^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]^2*x^2,x]

[Out]

-(a*Sqrt[Pi]*Erfi[1 - ArcSinh[a + b*x]])/(4*b^3*E) - (a*Sqrt[Pi]*Erfi[1 + ArcSinh[a + b*x]])/(4*b^3*E) + (Sqrt
[Pi]*Erfi[(-3 + 2*ArcSinh[a + b*x])/2])/(16*b^3*E^(9/4)) - (Sqrt[Pi]*Erfi[(-1 + 2*ArcSinh[a + b*x])/2])/(16*b^
3*E^(1/4)) + (a^2*Sqrt[Pi]*Erfi[(-1 + 2*ArcSinh[a + b*x])/2])/(4*b^3*E^(1/4)) - (Sqrt[Pi]*Erfi[(1 + 2*ArcSinh[
a + b*x])/2])/(16*b^3*E^(1/4)) + (a^2*Sqrt[Pi]*Erfi[(1 + 2*ArcSinh[a + b*x])/2])/(4*b^3*E^(1/4)) + (Sqrt[Pi]*E
rfi[(3 + 2*ArcSinh[a + b*x])/2])/(16*b^3*E^(9/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 5514

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

Rule 5898

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

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int e^{\sinh ^{-1}(a+b x)^2} x^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh (x) \left (-\frac {a}{b}+\frac {\sinh (x)}{b}\right )^2 \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e^{x^2} \cosh (x) (a-\sinh (x))^2}{b^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh (x) (a-\sinh (x))^2 \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 e^{x^2} \cosh (x)-2 a e^{x^2} \cosh (x) \sinh (x)+e^{x^2} \cosh (x) \sinh ^2(x)\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh (x) \sinh ^2(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int e^{x^2} \cosh (x) \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^{x^2} \cosh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{8} e^{-3 x+x^2}-\frac {1}{8} e^{-x+x^2}-\frac {e^{x+x^2}}{8}+\frac {1}{8} e^{3 x+x^2}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \left (-\frac {1}{4} e^{-2 x+x^2}+\frac {1}{4} e^{2 x+x^2}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \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^3}\\ &=\frac {\operatorname {Subst}\left (\int e^{-3 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3}-\frac {\operatorname {Subst}\left (\int e^{-x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3}-\frac {\operatorname {Subst}\left (\int e^{x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3}+\frac {\operatorname {Subst}\left (\int e^{3 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3}+\frac {a \operatorname {Subst}\left (\int e^{-2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}-\frac {a \operatorname {Subst}\left (\int e^{2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^{-x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^{x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}\\ &=\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (-3+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3 e^{9/4}}+\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (3+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3 e^{9/4}}+\frac {a \operatorname {Subst}\left (\int e^{\frac {1}{4} (-2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3 e}-\frac {a \operatorname {Subst}\left (\int e^{\frac {1}{4} (2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3 e}-\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3 \sqrt [4]{e}}-\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3 \sqrt [4]{e}}+\frac {a^2 \operatorname {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3 \sqrt [4]{e}}+\frac {a^2 \operatorname {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3 \sqrt [4]{e}}\\ &=-\frac {a \sqrt {\pi } \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{4 b^3 e}-\frac {a \sqrt {\pi } \text {erfi}\left (1+\sinh ^{-1}(a+b x)\right )}{4 b^3 e}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-3+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 \sqrt [4]{e}}+\frac {a^2 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^3 \sqrt [4]{e}}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 \sqrt [4]{e}}+\frac {a^2 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^3 \sqrt [4]{e}}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (3+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 138, normalized size = 0.55 \[ -\frac {\sqrt {\pi } \left (-4 e^2 a^2 \text {erfi}\left (\sinh ^{-1}(a+b x)+\frac {1}{2}\right )+e^2 \left (4 a^2-1\right ) \text {erfi}\left (\frac {1}{2}-\sinh ^{-1}(a+b x)\right )+4 e^{5/4} a \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )+4 e^{5/4} a \text {erfi}\left (\sinh ^{-1}(a+b x)+1\right )+\text {erfi}\left (\frac {3}{2}-\sinh ^{-1}(a+b x)\right )+e^2 \text {erfi}\left (\sinh ^{-1}(a+b x)+\frac {1}{2}\right )-\text {erfi}\left (\sinh ^{-1}(a+b x)+\frac {3}{2}\right )\right )}{16 e^{9/4} b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSinh[a + b*x]^2*x^2,x]

[Out]

-1/16*(Sqrt[Pi]*((-1 + 4*a^2)*E^2*Erfi[1/2 - ArcSinh[a + b*x]] + 4*a*E^(5/4)*Erfi[1 - ArcSinh[a + b*x]] + Erfi
[3/2 - ArcSinh[a + b*x]] + E^2*Erfi[1/2 + ArcSinh[a + b*x]] - 4*a^2*E^2*Erfi[1/2 + ArcSinh[a + b*x]] + 4*a*E^(
5/4)*Erfi[1 + ArcSinh[a + b*x]] - Erfi[3/2 + ArcSinh[a + b*x]]))/(b^3*E^(9/4))

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} e^{\left (\operatorname {arsinh}\left (b x + a\right )^{2}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*e^(arcsinh(b*x + a)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*e^(arcsinh(b*x + a)^2), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\arcsinh \left (b x +a \right )^{2}} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*e^(arcsinh(b*x + a)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*exp(asinh(a + b*x)^2),x)

[Out]

int(x^2*exp(asinh(a + b*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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