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

Optimal. Leaf size=101 \[ -\frac{a (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^2}-\frac{a \sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^2}+\frac{e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^2}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^2} \]

[Out]

-(a*E^ArcSin[a + b*x]*(a + b*x))/(2*b^2) - (a*E^ArcSin[a + b*x]*Sqrt[1 - (a + b*x)^2])/(2*b^2) - (E^ArcSin[a +
b*x]*Cos[2*ArcSin[a + b*x]])/(5*b^2) + (E^ArcSin[a + b*x]*Sin[2*ArcSin[a + b*x]])/(10*b^2)

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Rubi [A]  time = 0.183847, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {4836, 6741, 12, 6742, 4433, 4469, 4432} \[ -\frac{a (a+b x) e^{\sin ^{-1}(a+b x)}}{2 b^2}-\frac{a \sqrt{1-(a+b x)^2} e^{\sin ^{-1}(a+b x)}}{2 b^2}+\frac{e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^2}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSin[a + b*x]*x,x]

[Out]

-(a*E^ArcSin[a + b*x]*(a + b*x))/(2*b^2) - (a*E^ArcSin[a + b*x]*Sqrt[1 - (a + b*x)^2])/(2*b^2) - (E^ArcSin[a +
b*x]*Cos[2*ArcSin[a + b*x]])/(5*b^2) + (E^ArcSin[a + b*x]*Sin[2*ArcSin[a + b*x]])/(10*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 4433

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*C
os[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rule 4469

Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :
> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g
}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 4432

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*S
in[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] - Simp[(e*F^(c*(a + b*x))*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin{align*} \int e^{\sin ^{-1}(a+b x)} x \, dx &=\frac{\operatorname{Subst}\left (\int e^x \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 \cos (x) (-a+\sin (x))}{b} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) (-a+\sin (x)) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a e^x \cos (x)+e^x \cos (x) \sin (x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int e^x \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int e^x \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{a e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^2}-\frac{a e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{2} e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{a e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^2}-\frac{a e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^2}+\frac{\operatorname{Subst}\left (\int e^x \sin (2 x) \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac{a e^{\sin ^{-1}(a+b x)} (a+b x)}{2 b^2}-\frac{a e^{\sin ^{-1}(a+b x)} \sqrt{1-(a+b x)^2}}{2 b^2}-\frac{e^{\sin ^{-1}(a+b x)} \cos \left (2 \sin ^{-1}(a+b x)\right )}{5 b^2}+\frac{e^{\sin ^{-1}(a+b x)} \sin \left (2 \sin ^{-1}(a+b x)\right )}{10 b^2}\\ \end{align*}

Mathematica [A]  time = 0.145904, size = 59, normalized size = 0.58 \[ -\frac{e^{\sin ^{-1}(a+b x)} \left (\sqrt{1-(a+b x)^2} (3 a-2 b x)+5 a (a+b x)+2 \cos \left (2 \sin ^{-1}(a+b x)\right )\right )}{10 b^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSin[a + b*x]*x,x]

[Out]

-(E^ArcSin[a + b*x]*(5*a*(a + b*x) + (3*a - 2*b*x)*Sqrt[1 - (a + b*x)^2] + 2*Cos[2*ArcSin[a + b*x]]))/(10*b^2)

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Maple [F]  time = 0.007, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{\arcsin \left ( bx+a \right ) }}x\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(arcsin(b*x+a))*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 )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*e^(arcsin(b*x + a)), x)

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Fricas [A]  time = 2.05777, size = 153, normalized size = 1.51 \begin{align*} \frac{{\left (4 \, b^{2} x^{2} + 3 \, a b x - a^{2} + \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (2 \, b x - 3 \, a\right )} - 2\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{10 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(4*b^2*x^2 + 3*a*b*x - a^2 + sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(2*b*x - 3*a) - 2)*e^(arcsin(b*x + a))/b^
2

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Sympy [A]  time = 0.724474, size = 146, normalized size = 1.45 \begin{align*} \begin{cases} - \frac{a^{2} e^{\operatorname{asin}{\left (a + b x \right )}}}{10 b^{2}} + \frac{3 a x e^{\operatorname{asin}{\left (a + b x \right )}}}{10 b} - \frac{3 a \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a + b x \right )}}}{10 b^{2}} + \frac{2 x^{2} e^{\operatorname{asin}{\left (a + b x \right )}}}{5} + \frac{x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} e^{\operatorname{asin}{\left (a + b x \right )}}}{5 b} - \frac{e^{\operatorname{asin}{\left (a + b x \right )}}}{5 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} e^{\operatorname{asin}{\left (a \right )}}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*exp(asin(a + b*x))/(10*b**2) + 3*a*x*exp(asin(a + b*x))/(10*b) - 3*a*sqrt(-a**2 - 2*a*b*x - b
**2*x**2 + 1)*exp(asin(a + b*x))/(10*b**2) + 2*x**2*exp(asin(a + b*x))/5 + x*sqrt(-a**2 - 2*a*b*x - b**2*x**2
+ 1)*exp(asin(a + b*x))/(5*b) - exp(asin(a + b*x))/(5*b**2), Ne(b, 0)), (x**2*exp(asin(a))/2, True))

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Giac [A]  time = 1.26571, size = 146, normalized size = 1.45 \begin{align*} -\frac{{\left (b x + a\right )} a e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{2}} + \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{2}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1} a e^{\left (\arcsin \left (b x + a\right )\right )}}{2 \, b^{2}} + \frac{2 \,{\left ({\left (b x + a\right )}^{2} - 1\right )} e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{2}} + \frac{e^{\left (\arcsin \left (b x + a\right )\right )}}{5 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*x + a)*a*e^(arcsin(b*x + a))/b^2 + 1/5*sqrt(-(b*x + a)^2 + 1)*(b*x + a)*e^(arcsin(b*x + a))/b^2 - 1/2*
sqrt(-(b*x + a)^2 + 1)*a*e^(arcsin(b*x + a))/b^2 + 2/5*((b*x + a)^2 - 1)*e^(arcsin(b*x + a))/b^2 + 1/5*e^(arcs
in(b*x + a))/b^2