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

Optimal. Leaf size=123 \[ \frac{e \sqrt{\pi } \text{Erf}\left (1-i \sin ^{-1}(a+b x)\right )}{8 b^2}+\frac{e \sqrt{\pi } \text{Erf}\left (1+i \sin ^{-1}(a+b x)\right )}{8 b^2}-\frac{\sqrt [4]{e} \sqrt{\pi } a \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{4 b^2}-\frac{\sqrt [4]{e} \sqrt{\pi } a \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{4 b^2} \]

[Out]

(E*Sqrt[Pi]*Erf[1 - I*ArcSin[a + b*x]])/(8*b^2) + (E*Sqrt[Pi]*Erf[1 + I*ArcSin[a + b*x]])/(8*b^2) - (a*E^(1/4)
*Sqrt[Pi]*Erfi[(-I + 2*ArcSin[a + b*x])/2])/(4*b^2) - (a*E^(1/4)*Sqrt[Pi]*Erfi[(I + 2*ArcSin[a + b*x])/2])/(4*
b^2)

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Rubi [A]  time = 0.272831, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4836, 6741, 12, 6742, 4473, 2234, 2204, 4474} \[ \frac{e \sqrt{\pi } \text{Erf}\left (1-i \sin ^{-1}(a+b x)\right )}{8 b^2}+\frac{e \sqrt{\pi } \text{Erf}\left (1+i \sin ^{-1}(a+b x)\right )}{8 b^2}-\frac{\sqrt [4]{e} \sqrt{\pi } a \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{4 b^2}-\frac{\sqrt [4]{e} \sqrt{\pi } a \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a + b*x]^2*x,x]

[Out]

(E*Sqrt[Pi]*Erf[1 - I*ArcSin[a + b*x]])/(8*b^2) + (E*Sqrt[Pi]*Erf[1 + I*ArcSin[a + b*x]])/(8*b^2) - (a*E^(1/4)
*Sqrt[Pi]*Erfi[(-I + 2*ArcSin[a + b*x])/2])/(4*b^2) - (a*E^(1/4)*Sqrt[Pi]*Erfi[(I + 2*ArcSin[a + b*x])/2])/(4*
b^2)

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 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 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 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 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]

Rubi steps

\begin{align*} \int e^{\sin ^{-1}(a+b x)^2} x \, dx &=\frac{\operatorname{Subst}\left (\int e^{x^2} \cos (x) \left (-\frac{a}{b}+\frac{\sin (x)}{b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{e^{x^2} \cos (x) (-a+\sin (x))}{b} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int e^{x^2} \cos (x) (-a+\sin (x)) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a e^{x^2} \cos (x)+e^{x^2} \cos (x) \sin (x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int e^{x^2} \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int e^{x^2} \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\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^2}-\frac{a \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^2}\\ &=\frac{i \operatorname{Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}-\frac{i \operatorname{Subst}\left (\int e^{2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}-\frac{a \operatorname{Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \operatorname{Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac{\left (a \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^2}-\frac{\left (a \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^2}+\frac{(i e) \operatorname{Subst}\left (\int e^{\frac{1}{4} (-2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}-\frac{(i e) \operatorname{Subst}\left (\int e^{\frac{1}{4} (2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}\\ &=\frac{e \sqrt{\pi } \text{erf}\left (1-i \sin ^{-1}(a+b x)\right )}{8 b^2}+\frac{e \sqrt{\pi } \text{erf}\left (1+i \sin ^{-1}(a+b x)\right )}{8 b^2}-\frac{a \sqrt [4]{e} \sqrt{\pi } \text{erfi}\left (\frac{1}{2} \left (-i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^2}-\frac{a \sqrt [4]{e} \sqrt{\pi } \text{erfi}\left (\frac{1}{2} \left (i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^2}\\ \end{align*}

Mathematica [A]  time = 0.1061, size = 93, normalized size = 0.76 \[ \frac{\sqrt{\pi } \left (e \text{Erf}\left (1-i \sin ^{-1}(a+b x)\right )+e \text{Erf}\left (1+i \sin ^{-1}(a+b x)\right )-2 \sqrt [4]{e} a \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )-2 \sqrt [4]{e} a \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )\right )}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSin[a + b*x]^2*x,x]

[Out]

(Sqrt[Pi]*(E*Erf[1 - I*ArcSin[a + b*x]] + E*Erf[1 + I*ArcSin[a + b*x]] - 2*a*E^(1/4)*Erfi[(-I + 2*ArcSin[a + b
*x])/2] - 2*a*E^(1/4)*Erfi[(I + 2*ArcSin[a + b*x])/2]))/(8*b^2)

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Maple [F]  time = 0.009, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}}x\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*e^(arcsin(b*x + a)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*e^(arcsin(b*x + a)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*exp(asin(a + b*x)**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*e^(arcsin(b*x + a)^2), x)