3.411 \(\int (a-b \sin ^{-1}(1-d x^2)) \, dx\)

Optimal. Leaf size=45 \[ a x+\frac{2 b \sqrt{2 d x^2-d^2 x^4}}{d x}+b (-x) \sin ^{-1}\left (1-d x^2\right ) \]

[Out]

a*x + (2*b*Sqrt[2*d*x^2 - d^2*x^4])/(d*x) - b*x*ArcSin[1 - d*x^2]

________________________________________________________________________________________

Rubi [A]  time = 0.0392835, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4840, 12, 1588} \[ a x+\frac{2 b \sqrt{2 d x^2-d^2 x^4}}{d x}+b (-x) \sin ^{-1}\left (1-d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[a - b*ArcSin[1 - d*x^2],x]

[Out]

a*x + (2*b*Sqrt[2*d*x^2 - d^2*x^4])/(d*x) - b*x*ArcSin[1 - d*x^2]

Rule 4840

Int[ArcSin[u_], x_Symbol] :> Simp[x*ArcSin[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/Sqrt[1 - u^2], x], x] /;
 InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

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

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \left (a-b \sin ^{-1}\left (1-d x^2\right )\right ) \, dx &=a x-b \int \sin ^{-1}\left (1-d x^2\right ) \, dx\\ &=a x-b x \sin ^{-1}\left (1-d x^2\right )+b \int -\frac{2 d x^2}{\sqrt{2 d x^2-d^2 x^4}} \, dx\\ &=a x-b x \sin ^{-1}\left (1-d x^2\right )-(2 b d) \int \frac{x^2}{\sqrt{2 d x^2-d^2 x^4}} \, dx\\ &=a x+\frac{2 b \sqrt{2 d x^2-d^2 x^4}}{d x}-b x \sin ^{-1}\left (1-d x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0268153, size = 43, normalized size = 0.96 \[ a x+\frac{2 b \sqrt{-d x^2 \left (d x^2-2\right )}}{d x}+b (-x) \sin ^{-1}\left (1-d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[a - b*ArcSin[1 - d*x^2],x]

[Out]

a*x + (2*b*Sqrt[-(d*x^2*(-2 + d*x^2))])/(d*x) - b*x*ArcSin[1 - d*x^2]

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 45, normalized size = 1. \begin{align*} ax+b \left ( x\arcsin \left ( d{x}^{2}-1 \right ) -2\,{\frac{x \left ( d{x}^{2}-2 \right ) }{\sqrt{-{d}^{2}{x}^{4}+2\,d{x}^{2}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a+b*arcsin(d*x^2-1),x)

[Out]

a*x+b*(x*arcsin(d*x^2-1)-2/(-d^2*x^4+2*d*x^2)^(1/2)*x*(d*x^2-2))

________________________________________________________________________________________

Maxima [A]  time = 1.51336, size = 61, normalized size = 1.36 \begin{align*}{\left (x \arcsin \left (d x^{2} - 1\right ) - \frac{2 \,{\left (d^{\frac{3}{2}} x^{2} - 2 \, \sqrt{d}\right )}}{\sqrt{-d x^{2} + 2} d}\right )} b + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arcsin(d*x^2-1),x, algorithm="maxima")

[Out]

(x*arcsin(d*x^2 - 1) - 2*(d^(3/2)*x^2 - 2*sqrt(d))/(sqrt(-d*x^2 + 2)*d))*b + a*x

________________________________________________________________________________________

Fricas [A]  time = 2.24798, size = 103, normalized size = 2.29 \begin{align*} \frac{b d x^{2} \arcsin \left (d x^{2} - 1\right ) + a d x^{2} + 2 \, \sqrt{-d^{2} x^{4} + 2 \, d x^{2}} b}{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arcsin(d*x^2-1),x, algorithm="fricas")

[Out]

(b*d*x^2*arcsin(d*x^2 - 1) + a*d*x^2 + 2*sqrt(-d^2*x^4 + 2*d*x^2)*b)/(d*x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{asin}{\left (d x^{2} - 1 \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*asin(d*x**2-1),x)

[Out]

Integral(a + b*asin(d*x**2 - 1), x)

________________________________________________________________________________________

Giac [A]  time = 1.1674, size = 74, normalized size = 1.64 \begin{align*} -{\left (2 \, d{\left (\frac{\sqrt{2} \mathrm{sgn}\left (x\right )}{d^{\frac{3}{2}}} - \frac{\sqrt{-d^{2} x^{2} + 2 \, d}}{d^{2} \mathrm{sgn}\left (x\right )}\right )} - x \arcsin \left (d x^{2} - 1\right )\right )} b + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arcsin(d*x^2-1),x, algorithm="giac")

[Out]

-(2*d*(sqrt(2)*sgn(x)/d^(3/2) - sqrt(-d^2*x^2 + 2*d)/(d^2*sgn(x))) - x*arcsin(d*x^2 - 1))*b + a*x