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

Optimal. Leaf size=216 \[ -\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (-\frac{a-b \sin ^{-1}\left (1-d x^2\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac{\sqrt{2 d x^2-d^2 x^4}}{2 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.024594, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {4825} \[ -\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (-\frac{a-b \sin ^{-1}\left (1-d x^2\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac{\sqrt{2 d x^2-d^2 x^4}}{2 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*ArcSin[1 - d*x^2])^(-2),x]

[Out]

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

Rule 4825

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-2), x_Symbol] :> -Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]/(2*b*d*x*(
a + b*ArcSin[c + d*x^2])), x] + (-Simp[(x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*CosIntegral[(c/(2*b))*(a + b*ArcSin[
c + d*x^2])])/(4*b^2*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2])), x] + Simp[(x*(Cos[a/(2*b)] - c*
Sin[a/(2*b)])*SinIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])])/(4*b^2*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSi
n[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 )^2} \, dx &=-\frac{\sqrt{2 d x^2-d^2 x^4}}{2 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}-\frac{x \text{Ci}\left (-\frac{a-b \sin ^{-1}\left (1-d x^2\right )}{2 b}\right ) \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right )}{4 b^2 \left (\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 )\right )}-\frac{x \left (\cos \left (\frac{a}{2 b}\right )+\sin \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}{4 b^2 \left (\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 )\right )}\\ \end{align*}

Mathematica [A]  time = 0.359269, size = 183, normalized size = 0.85 \[ \frac{\left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right ) \left (a-b \sin ^{-1}\left (1-d x^2\right )\right ) \left (\left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\sin ^{-1}\left (1-d x^2\right )-\frac{a}{b}\right )\right )+\left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a-b \sin ^{-1}\left (1-d x^2\right )}{2 b}\right )\right )+2 b \sqrt{d x^2 \left (2-d x^2\right )}}{4 b^2 d x \left (b \sin ^{-1}\left (1-d x^2\right )-a\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*b*Sqrt[d*x^2*(2 - d*x^2)] + (a - b*ArcSin[1 - d*x^2])*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])
*(CosIntegral[(-(a/b) + ArcSin[1 - d*x^2])/2]*(Cos[a/(2*b)] - Sin[a/(2*b)]) + (Cos[a/(2*b)] + Sin[a/(2*b)])*Si
nIntegral[(a - b*ArcSin[1 - d*x^2])/(2*b)]))/(4*b^2*d*x*(-a + b*ArcSin[1 - d*x^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 ) ^{-2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b^2*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a*b*d)*sqrt(d)*integrate(1/2*sqrt(-d*x^2 + 2)*x
/(a*b*d*x^2 - 2*a*b + (b^2*d*x^2 - 2*b^2)*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)), x) - sqrt(-d*x^2 +
2)*sqrt(d))/(b^2*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a*b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(d*x**2 - 1))**(-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 )}^{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))^2,x, algorithm="giac")

[Out]

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