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]

-(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))

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Rubi [A]  time = 0.516042, antiderivative size = 251, normalized size of antiderivative = 1., 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 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 12

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

Rule 6742

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

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 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 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 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]

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*}

Mathematica [A]  time = 0.183048, 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]

-(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]]))/(16*b^3*E^(9/4))

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Maple [F]  time = 0.01, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}}{x}^{2}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int x^{2} e^{\left (\operatorname{arsinh}\left (b x + a\right )^{2}\right )}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} e^{\left (\operatorname{arsinh}\left (b x + a\right )^{2}\right )}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} e^{\operatorname{asinh}^{2}{\left (a + b x \right )}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} e^{\left (\operatorname{arsinh}\left (b x + a\right )^{2}\right )}\,{d x} \end{align*}

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)