3.428 \(\int \frac{1}{(a-b \sin ^{-1}(1-d x^2))^{3/2}} \, dx\)

Optimal. Leaf size=256 \[ -\frac{\sqrt{2 d x^2-d^2 x^4}}{b d x \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}-\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{3/2} x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{3/2} x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )} \]

[Out]

-(Sqrt[2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a - b*ArcSin[1 - d*x^2]])) - ((-b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelC[(Sqrt[
-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Cos[ArcSin[1 - d*x^2]/2] - S
in[ArcSin[1 - d*x^2]/2]) + ((-b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelS[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/
Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])

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Rubi [A]  time = 0.05705, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {4822} \[ -\frac{\sqrt{2 d x^2-d^2 x^4}}{b d x \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}-\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{3/2} x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{3/2} x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*ArcSin[1 - d*x^2])^(-3/2),x]

[Out]

-(Sqrt[2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a - b*ArcSin[1 - d*x^2]])) - ((-b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelC[(Sqrt[
-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Cos[ArcSin[1 - d*x^2]/2] - S
in[ArcSin[1 - d*x^2]/2]) + ((-b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelS[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/
Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])

Rule 4822

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> -Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]/(b*d*x*S
qrt[a + b*ArcSin[c + d*x^2]]), x] + (-Simp[((c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sq
rt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2]), x] +
Simp[((c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] - c*Sin[a/(2*b)])*FresnelS[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*
x^2]]])/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2]), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2,
1]

Rubi steps

\begin{align*} \int \frac{1}{\left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}} \, dx &=-\frac{\sqrt{2 d x^2-d^2 x^4}}{b d x \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}-\frac{\left (-\frac{1}{b}\right )^{3/2} \sqrt{\pi } x C\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right ) \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac{\left (-\frac{1}{b}\right )^{3/2} \sqrt{\pi } x S\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right ) \left (\cos \left (\frac{a}{2 b}\right )+\sin \left (\frac{a}{2 b}\right )\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.435859, size = 256, normalized size = 1. \[ -\frac{\sqrt{2 d x^2-d^2 x^4}}{b d x \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}-\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{3/2} x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{3/2} x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*ArcSin[1 - d*x^2])^(-3/2),x]

[Out]

-(Sqrt[2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a - b*ArcSin[1 - d*x^2]])) - ((-b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelC[(Sqrt[
-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Cos[ArcSin[1 - d*x^2]/2] - S
in[ArcSin[1 - d*x^2]/2]) + ((-b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelS[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/
Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])

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Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arcsin \left ( d{x}^{2}-1 \right ) \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x^2-1))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x^2-1))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{asin}{\left (d x^{2} - 1 \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asin(d*x**2-1))**(3/2),x)

[Out]

Integral((a + b*asin(d*x**2 - 1))**(-3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x^2-1))^(3/2),x, algorithm="giac")

[Out]

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