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

Optimal. Leaf size=265 \[ \frac {\sqrt [4]{e} \sqrt {\pi } a^2 \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } a^2 \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{4 b^3}-\frac {e \sqrt {\pi } a \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {e \sqrt {\pi } a \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{16 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-3 i\right )\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+3 i\right )\right )}{16 b^3} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 265, 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 = {4836, 6741, 12, 6742, 4473, 2234, 2204, 4474} \[ \frac {\sqrt [4]{e} \sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{4 b^3}-\frac {e \sqrt {\pi } a \text {Erf}\left (1-i \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {e \sqrt {\pi } a \text {Erf}\left (1+i \sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{16 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-3 i\right )\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+3 i\right )\right )}{16 b^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a + b*x]^2*x^2,x]

[Out]

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

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 4473

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

Rule 4474

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

Rule 4836

Int[(u_.)*(f_)^(ArcSin[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[(u /. x -> -(a/b) +
Sin[x]/b)*f^(c*x^n)*Cos[x], x], x, ArcSin[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && 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^{\sin ^{-1}(a+b x)^2} x^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cos (x) \left (-\frac {a}{b}+\frac {\sin (x)}{b}\right )^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e^{x^2} \cos (x) (a-\sin (x))^2}{b^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cos (x) (a-\sin (x))^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 e^{x^2} \cos (x)-2 a e^{x^2} \cos (x) \sin (x)+e^{x^2} \cos (x) \sin ^2(x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int e^{x^2} \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^{x^2} \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{8} e^{-i x+x^2}+\frac {1}{8} e^{i x+x^2}-\frac {1}{8} e^{-3 i x+x^2}-\frac {1}{8} e^{3 i x+x^2}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \operatorname {Subst}\left (\int \left (\frac {1}{4} i e^{-2 i x+x^2}-\frac {1}{4} i e^{2 i x+x^2}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \operatorname {Subst}\left (\int \left (\frac {1}{2} e^{-i x+x^2}+\frac {1}{2} e^{i x+x^2}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {\operatorname {Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {\operatorname {Subst}\left (\int e^{-3 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {\operatorname {Subst}\left (\int e^{3 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {(i a) \operatorname {Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {(i a) \operatorname {Subst}\left (\int e^{2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \operatorname {Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}\\ &=\frac {\sqrt [4]{e} \operatorname {Subst}\left (\int e^{\frac {1}{4} (-i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {\sqrt [4]{e} \operatorname {Subst}\left (\int e^{\frac {1}{4} (i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {\left (a^2 \sqrt [4]{e}\right ) \operatorname {Subst}\left (\int e^{\frac {1}{4} (-i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {\left (a^2 \sqrt [4]{e}\right ) \operatorname {Subst}\left (\int e^{\frac {1}{4} (i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}-\frac {(i a e) \operatorname {Subst}\left (\int e^{\frac {1}{4} (-2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {(i a e) \operatorname {Subst}\left (\int e^{\frac {1}{4} (2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}-\frac {e^{9/4} \operatorname {Subst}\left (\int e^{\frac {1}{4} (-3 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {e^{9/4} \operatorname {Subst}\left (\int e^{\frac {1}{4} (3 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}\\ &=-\frac {a e \sqrt {\pi } \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a e \sqrt {\pi } \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^3}+\frac {a^2 \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^3}+\frac {a^2 \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-3 i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (3 i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 161, normalized size = 0.61 \[ -\frac {\sqrt {\pi } \left (i \sqrt [4]{e} \left (4 a^2 \text {erf}\left (\frac {1}{2}+i \sin ^{-1}(a+b x)\right )-\left (4 a^2+1\right ) \text {erf}\left (\frac {1}{2}-i \sin ^{-1}(a+b x)\right )-4 i e^{3/4} a \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )+e^2 \text {erf}\left (\frac {3}{2}-i \sin ^{-1}(a+b x)\right )+\text {erf}\left (\frac {1}{2}+i \sin ^{-1}(a+b x)\right )-e^2 \text {erf}\left (\frac {3}{2}+i \sin ^{-1}(a+b x)\right )\right )+4 e a \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )\right )}{16 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSin[a + b*x]^2*x^2,x]

[Out]

-1/16*(Sqrt[Pi]*(4*a*E*Erf[1 - I*ArcSin[a + b*x]] + I*E^(1/4)*(-((1 + 4*a^2)*Erf[1/2 - I*ArcSin[a + b*x]]) + E
^2*Erf[3/2 - I*ArcSin[a + b*x]] + Erf[1/2 + I*ArcSin[a + b*x]] + 4*a^2*Erf[1/2 + I*ArcSin[a + b*x]] - (4*I)*a*
E^(3/4)*Erf[1 + I*ArcSin[a + b*x]] - E^2*Erf[3/2 + I*ArcSin[a + b*x]])))/b^3

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} e^{\left (\arcsin \left (b x + a\right )^{2}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*exp(asin(a + b*x)^2),x)

[Out]

int(x^2*exp(asin(a + b*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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