3.358 \(\int e^{\sinh ^{-1}(a+b x)^2} x^3 \, dx\)

Optimal. Leaf size=359 \[ -\frac {\sqrt {\pi } a^3 \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^4}-\frac {\sqrt {\pi } a^3 \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^4}+\frac {3 \sqrt {\pi } a^2 \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{8 e b^4}+\frac {3 \sqrt {\pi } a^2 \text {erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{8 e b^4}-\frac {3 \sqrt {\pi } a \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-3\right )\right )}{16 e^{9/4} b^4}+\frac {3 \sqrt {\pi } a \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{16 \sqrt [4]{e} b^4}+\frac {3 \sqrt {\pi } a \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{16 \sqrt [4]{e} b^4}-\frac {3 \sqrt {\pi } a \text {erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+3\right )\right )}{16 e^{9/4} b^4}-\frac {\sqrt {\pi } \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{16 e b^4}+\frac {\sqrt {\pi } \text {erfi}\left (2-\sinh ^{-1}(a+b x)\right )}{32 e^4 b^4}-\frac {\sqrt {\pi } \text {erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{16 e b^4}+\frac {\sqrt {\pi } \text {erfi}\left (\sinh ^{-1}(a+b x)+2\right )}{32 e^4 b^4} \]

[Out]

-1/32*erfi(-2+arcsinh(b*x+a))*Pi^(1/2)/b^4/exp(4)+1/16*erfi(-1+arcsinh(b*x+a))*Pi^(1/2)/b^4/E-3/8*a^2*erfi(-1+
arcsinh(b*x+a))*Pi^(1/2)/b^4/E-1/16*erfi(1+arcsinh(b*x+a))*Pi^(1/2)/b^4/E+3/8*a^2*erfi(1+arcsinh(b*x+a))*Pi^(1
/2)/b^4/E+1/32*erfi(2+arcsinh(b*x+a))*Pi^(1/2)/b^4/exp(4)-3/16*a*erfi(-3/2+arcsinh(b*x+a))*Pi^(1/2)/b^4/exp(9/
4)+3/16*a*erfi(-1/2+arcsinh(b*x+a))*Pi^(1/2)/b^4/exp(1/4)-1/4*a^3*erfi(-1/2+arcsinh(b*x+a))*Pi^(1/2)/b^4/exp(1
/4)+3/16*a*erfi(1/2+arcsinh(b*x+a))*Pi^(1/2)/b^4/exp(1/4)-1/4*a^3*erfi(1/2+arcsinh(b*x+a))*Pi^(1/2)/b^4/exp(1/
4)-3/16*a*erfi(3/2+arcsinh(b*x+a))*Pi^(1/2)/b^4/exp(9/4)

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Rubi [A]  time = 0.74, antiderivative size = 359, normalized size of antiderivative = 1.00, number of steps used = 37, 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} \[ -\frac {\sqrt {\pi } a^3 \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^4}-\frac {\sqrt {\pi } a^3 \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^4}+\frac {3 \sqrt {\pi } a^2 \text {Erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{8 e b^4}+\frac {3 \sqrt {\pi } a^2 \text {Erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{8 e b^4}-\frac {3 \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-3\right )\right )}{16 e^{9/4} b^4}+\frac {3 \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{16 \sqrt [4]{e} b^4}+\frac {3 \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{16 \sqrt [4]{e} b^4}-\frac {3 \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+3\right )\right )}{16 e^{9/4} b^4}-\frac {\sqrt {\pi } \text {Erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{16 e b^4}+\frac {\sqrt {\pi } \text {Erfi}\left (2-\sinh ^{-1}(a+b x)\right )}{32 e^4 b^4}-\frac {\sqrt {\pi } \text {Erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{16 e b^4}+\frac {\sqrt {\pi } \text {Erfi}\left (\sinh ^{-1}(a+b x)+2\right )}{32 e^4 b^4} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]^2*x^3,x]

[Out]

-(Sqrt[Pi]*Erfi[1 - ArcSinh[a + b*x]])/(16*b^4*E) + (3*a^2*Sqrt[Pi]*Erfi[1 - ArcSinh[a + b*x]])/(8*b^4*E) + (S
qrt[Pi]*Erfi[2 - ArcSinh[a + b*x]])/(32*b^4*E^4) - (Sqrt[Pi]*Erfi[1 + ArcSinh[a + b*x]])/(16*b^4*E) + (3*a^2*S
qrt[Pi]*Erfi[1 + ArcSinh[a + b*x]])/(8*b^4*E) + (Sqrt[Pi]*Erfi[2 + ArcSinh[a + b*x]])/(32*b^4*E^4) - (3*a*Sqrt
[Pi]*Erfi[(-3 + 2*ArcSinh[a + b*x])/2])/(16*b^4*E^(9/4)) + (3*a*Sqrt[Pi]*Erfi[(-1 + 2*ArcSinh[a + b*x])/2])/(1
6*b^4*E^(1/4)) - (a^3*Sqrt[Pi]*Erfi[(-1 + 2*ArcSinh[a + b*x])/2])/(4*b^4*E^(1/4)) + (3*a*Sqrt[Pi]*Erfi[(1 + 2*
ArcSinh[a + b*x])/2])/(16*b^4*E^(1/4)) - (a^3*Sqrt[Pi]*Erfi[(1 + 2*ArcSinh[a + b*x])/2])/(4*b^4*E^(1/4)) - (3*
a*Sqrt[Pi]*Erfi[(3 + 2*ArcSinh[a + b*x])/2])/(16*b^4*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^3 \, dx &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh (x) \left (-\frac {a}{b}+\frac {\sinh (x)}{b}\right )^3 \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e^{x^2} \cosh (x) (-a+\sinh (x))^3}{b^3} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh (x) (-a+\sinh (x))^3 \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^4}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a^3 e^{x^2} \cosh (x)+3 a^2 e^{x^2} \cosh (x) \sinh (x)-3 a e^{x^2} \cosh (x) \sinh ^2(x)+e^{x^2} \cosh (x) \sinh ^3(x)\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^4}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh (x) \sinh ^3(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^4}-\frac {(3 a) \operatorname {Subst}\left (\int e^{x^2} \cosh (x) \sinh ^2(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^4}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int e^{x^2} \cosh (x) \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^4}-\frac {a^3 \operatorname {Subst}\left (\int e^{x^2} \cosh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^4}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{16} e^{-4 x+x^2}+\frac {1}{8} e^{-2 x+x^2}-\frac {1}{8} e^{2 x+x^2}+\frac {1}{16} e^{4 x+x^2}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^4}-\frac {(3 a) \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^4}+\frac {\left (3 a^2\right ) \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^4}-\frac {a^3 \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^4}\\ &=-\frac {\operatorname {Subst}\left (\int e^{-4 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{16 b^4}+\frac {\operatorname {Subst}\left (\int e^{4 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{16 b^4}+\frac {\operatorname {Subst}\left (\int e^{-2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4}-\frac {\operatorname {Subst}\left (\int e^{2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4}-\frac {(3 a) \operatorname {Subst}\left (\int e^{-3 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4}+\frac {(3 a) \operatorname {Subst}\left (\int e^{-x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4}+\frac {(3 a) \operatorname {Subst}\left (\int e^{x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4}-\frac {(3 a) \operatorname {Subst}\left (\int e^{3 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int e^{-2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^4}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int e^{2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^4}-\frac {a^3 \operatorname {Subst}\left (\int e^{-x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^4}-\frac {a^3 \operatorname {Subst}\left (\int e^{x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^4}\\ &=-\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (-4+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{16 b^4 e^4}+\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (4+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{16 b^4 e^4}-\frac {(3 a) \operatorname {Subst}\left (\int e^{\frac {1}{4} (-3+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4 e^{9/4}}-\frac {(3 a) \operatorname {Subst}\left (\int e^{\frac {1}{4} (3+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4 e^{9/4}}+\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (-2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4 e}-\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4 e}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int e^{\frac {1}{4} (-2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^4 e}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int e^{\frac {1}{4} (2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^4 e}+\frac {(3 a) \operatorname {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4 \sqrt [4]{e}}+\frac {(3 a) \operatorname {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^4 \sqrt [4]{e}}-\frac {a^3 \operatorname {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^4 \sqrt [4]{e}}-\frac {a^3 \operatorname {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^4 \sqrt [4]{e}}\\ &=-\frac {\sqrt {\pi } \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{16 b^4 e}+\frac {3 a^2 \sqrt {\pi } \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{8 b^4 e}+\frac {\sqrt {\pi } \text {erfi}\left (2-\sinh ^{-1}(a+b x)\right )}{32 b^4 e^4}-\frac {\sqrt {\pi } \text {erfi}\left (1+\sinh ^{-1}(a+b x)\right )}{16 b^4 e}+\frac {3 a^2 \sqrt {\pi } \text {erfi}\left (1+\sinh ^{-1}(a+b x)\right )}{8 b^4 e}+\frac {\sqrt {\pi } \text {erfi}\left (2+\sinh ^{-1}(a+b x)\right )}{32 b^4 e^4}-\frac {3 a \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-3+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^4 e^{9/4}}+\frac {3 a \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^4 \sqrt [4]{e}}-\frac {a^3 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^4 \sqrt [4]{e}}+\frac {3 a \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^4 \sqrt [4]{e}}-\frac {a^3 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^4 \sqrt [4]{e}}-\frac {3 a \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (3+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^4 e^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 198, normalized size = 0.55 \[ \frac {\sqrt {\pi } \left (-8 e^{15/4} a^3 \text {erfi}\left (\sinh ^{-1}(a+b x)+\frac {1}{2}\right )+12 e^3 a^2 \text {erfi}\left (\sinh ^{-1}(a+b x)+1\right )+2 e^{15/4} \left (4 a^2-3\right ) a \text {erfi}\left (\frac {1}{2}-\sinh ^{-1}(a+b x)\right )+2 e^3 \left (6 a^2-1\right ) \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )+6 e^{7/4} a \text {erfi}\left (\frac {3}{2}-\sinh ^{-1}(a+b x)\right )+6 e^{15/4} a \text {erfi}\left (\sinh ^{-1}(a+b x)+\frac {1}{2}\right )-6 e^{7/4} a \text {erfi}\left (\sinh ^{-1}(a+b x)+\frac {3}{2}\right )+\text {erfi}\left (2-\sinh ^{-1}(a+b x)\right )-2 e^3 \text {erfi}\left (\sinh ^{-1}(a+b x)+1\right )+\text {erfi}\left (\sinh ^{-1}(a+b x)+2\right )\right )}{32 e^4 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSinh[a + b*x]^2*x^3,x]

[Out]

(Sqrt[Pi]*(2*a*(-3 + 4*a^2)*E^(15/4)*Erfi[1/2 - ArcSinh[a + b*x]] + 2*(-1 + 6*a^2)*E^3*Erfi[1 - ArcSinh[a + b*
x]] + 6*a*E^(7/4)*Erfi[3/2 - ArcSinh[a + b*x]] + Erfi[2 - ArcSinh[a + b*x]] + 6*a*E^(15/4)*Erfi[1/2 + ArcSinh[
a + b*x]] - 8*a^3*E^(15/4)*Erfi[1/2 + ArcSinh[a + b*x]] - 2*E^3*Erfi[1 + ArcSinh[a + b*x]] + 12*a^2*E^3*Erfi[1
+ ArcSinh[a + b*x]] - 6*a*E^(7/4)*Erfi[3/2 + ArcSinh[a + b*x]] + Erfi[2 + ArcSinh[a + b*x]]))/(32*b^4*E^4)

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} e^{\left (\operatorname {arsinh}\left (b x + a\right )^{2}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsinh(b*x+a)^2)*x^3,x, algorithm="fricas")

[Out]

integral(x^3*e^(arcsinh(b*x + a)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} e^{\left (\operatorname {arsinh}\left (b x + a\right )^{2}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsinh(b*x+a)^2)*x^3,x, algorithm="giac")

[Out]

integrate(x^3*e^(arcsinh(b*x + a)^2), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\arcsinh \left (b x +a \right )^{2}} x^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arcsinh(b*x+a)^2)*x^3,x)

[Out]

int(exp(arcsinh(b*x+a)^2)*x^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} e^{\left (\operatorname {arsinh}\left (b x + a\right )^{2}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsinh(b*x+a)^2)*x^3,x, algorithm="maxima")

[Out]

integrate(x^3*e^(arcsinh(b*x + a)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\mathrm {e}}^{{\mathrm {asinh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*exp(asinh(a + b*x)^2),x)

[Out]

int(x^3*exp(asinh(a + b*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} e^{\operatorname {asinh}^{2}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(asinh(b*x+a)**2)*x**3,x)

[Out]

Integral(x**3*exp(asinh(a + b*x)**2), x)

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