3.5.14 \(\int \frac {1}{(a-b \text {ArcSin}(1-d x^2))^3} \, dx\) [414]

Optimal. Leaf size=240 \[ -\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )}-\frac {x \text {CosIntegral}\left (-\frac {a-b \text {ArcSin}\left (1-d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a}{2 b}-\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )\right )} \]

[Out]

1/8*x/b^2/(a+b*arcsin(d*x^2-1))+1/16*x*Si(1/2*a/b+1/2*arcsin(d*x^2-1))*(cos(1/2*a/b)-sin(1/2*a/b))/b^3/(cos(1/
2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))-1/16*x*Ci(1/2*(-a-b*arcsin(d*x^2-1))/b)*(cos(1/2*a/b)+sin(1/2*a/b
))/b^3/(cos(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))-1/4*(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^
2-1))^2

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Rubi [A]
time = 0.03, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4912, 4900} \begin {gather*} -\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {CosIntegral}\left (-\frac {a-b \text {ArcSin}\left (1-d x^2\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\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} \text {ArcSin}\left (1-d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )\right )}+\frac {x}{8 b^2 \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )}-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*ArcSin[1 - d*x^2])^(-3),x]

[Out]

-1/4*Sqrt[2*d*x^2 - d^2*x^4]/(b*d*x*(a - b*ArcSin[1 - d*x^2])^2) + x/(8*b^2*(a - b*ArcSin[1 - d*x^2])) - (x*Co
sIntegral[-1/2*(a - b*ArcSin[1 - d*x^2])/b]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(16*b^3*(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])/(16
*b^3*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]))

Rule 4900

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> Simp[(-x)*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*(
CosIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])]/(2*b*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))
), x] - Simp[x*(c*Cos[a/(2*b)] + Sin[a/(2*b)])*(SinIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])]/(2*b*(Cos[Arc
Sin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))), x] /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]

Rule 4912

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*((a + b*ArcSin[c + d*x^2])^(n + 2)/(
4*b^2*(n + 1)*(n + 2))), x] + (-Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcSin[c + d*x^2])^(n + 2), x], x]
+ Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]*((a + b*ArcSin[c + d*x^2])^(n + 1)/(2*b*d*(n + 1)*x)), x]) /; FreeQ[{a, b, c
, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]

Rubi steps

\begin {align*} \int \frac {1}{\left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3} \, dx &=-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}-\frac {\int \frac {1}{a-b \sin ^{-1}\left (1-d x^2\right )} \, dx}{8 b^2}\\ &=-\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}+\frac {x}{8 b^2 \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 )}{16 b^3 \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 )}{16 b^3 \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*}

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Mathematica [A]
time = 0.36, size = 195, normalized size = 0.81 \begin {gather*} -\frac {\frac {4 b^2 \sqrt {-d x^2 \left (-2+d x^2\right )}}{d \left (a-b \text {ArcSin}\left (1-d x^2\right )\right )^2}-\frac {2 b x^2}{a-b \text {ArcSin}\left (1-d x^2\right )}+\frac {\left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-d x^2\right )\right )\right ) \left (\text {CosIntegral}\left (\frac {1}{2} \left (-\frac {a}{b}+\text {ArcSin}\left (1-d x^2\right )\right )\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )+\left (-\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a-b \text {ArcSin}\left (1-d x^2\right )}{2 b}\right )\right )}{d}}{16 b^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((4*b^2*Sqrt[-(d*x^2*(-2 + d*x^2))])/(d*(a - b*ArcSin[1 - d*x^2])^2) - (2*b*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)])*SinIntegral[(a - b*ArcSin[1 - d*x^2])/(2*b)]))/d)/(b
^3*x)

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*d*x*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a*d*x - 2*sqrt(-d*x^2 + 2)*b*sqrt(d) - 8*(b^4*d*ar
ctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)^2 + 2*a*b^3*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a^
2*b^2*d)*integrate(1/8/(b^3*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a*b^2), x))/(b^4*d*arctan2(d*x^2
- 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)^2 + 2*a*b^3*d*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x) + a^2*b^2*d)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(d*x^2 - 1))^3,x)

[Out]

int(1/(a + b*asin(d*x^2 - 1))^3, x)

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