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

Optimal. Leaf size=267 \[ \frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{d x}+\frac {3 \sqrt {\pi } (-b)^{3/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac {3 \sqrt {\pi } (-b)^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2} \]

[Out]

x*(a+b*arcsin(d*x^2-1))^(3/2)+3*(-b)^(3/2)*x*FresnelS((a+b*arcsin(d*x^2-1))^(1/2)/(-b)^(1/2)/Pi^(1/2))*(cos(1/
2*a/b)-sin(1/2*a/b))*Pi^(1/2)/(cos(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))+3*(-b)^(3/2)*x*FresnelC((a+b
*arcsin(d*x^2-1))^(1/2)/(-b)^(1/2)/Pi^(1/2))*(cos(1/2*a/b)+sin(1/2*a/b))*Pi^(1/2)/(cos(1/2*arcsin(d*x^2-1))+si
n(1/2*arcsin(d*x^2-1)))+3*b*(-d^2*x^4+2*d*x^2)^(1/2)*(a+b*arcsin(d*x^2-1))^(1/2)/d/x

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Rubi [A]  time = 0.10, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4814, 4819} \[ \frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{d x}+\frac {3 \sqrt {\pi } (-b)^{3/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi } \sqrt {-b}}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac {3 \sqrt {\pi } (-b)^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*Sqrt[2*d*x^2 - d^2*x^4]*Sqrt[a - b*ArcSin[1 - d*x^2]])/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^(3/2) + (3*(-b
)^(3/2)*Sqrt[Pi]*x*FresnelS[Sqrt[a - b*ArcSin[1 - d*x^2]]/(Sqrt[-b]*Sqrt[Pi])]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/
(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]) + (3*(-b)^(3/2)*Sqrt[Pi]*x*FresnelC[Sqrt[a - b*ArcSin[1
- d*x^2]]/(Sqrt[-b]*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 4814

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

Rule 4819

Int[1/Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> -Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] - c*Sin[a/
(2*b)])*FresnelC[(1*Sqrt[a + b*ArcSin[c + d*x^2]])/(Sqrt[b*c]*Sqrt[Pi])])/(Sqrt[b*c]*(Cos[ArcSin[c + d*x^2]/2]
- c*Sin[ArcSin[c + d*x^2]/2])), x] - Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelS[(1/(Sqrt[b*c]*
Sqrt[Pi]))*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqrt[b*c]*(Cos[ArcSin[c + d*x^2]/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 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2} \, dx &=\frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}-\left (3 b^2\right ) \int \frac {1}{\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}} \, dx\\ &=\frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}+\frac {3 (-b)^{3/2} \sqrt {\pi } x S\left (\frac {\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {-b} \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 {3 (-b)^{3/2} \sqrt {\pi } x C\left (\frac {\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {-b} \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*}

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Mathematica [A]  time = 0.55, size = 265, normalized size = 0.99 \[ \frac {3 \sqrt {\pi } (-b)^{3/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac {3 \sqrt {\pi } (-b)^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}+\frac {3 b \sqrt {-d x^2 \left (d x^2-2\right )} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*Sqrt[-(d*x^2*(-2 + d*x^2))]*Sqrt[a - b*ArcSin[1 - d*x^2]])/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^(3/2) + (3
*(-b)^(3/2)*Sqrt[Pi]*x*FresnelS[Sqrt[a - b*ArcSin[1 - d*x^2]]/(Sqrt[-b]*Sqrt[Pi])]*(Cos[a/(2*b)] - Sin[a/(2*b)
]))/(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]) + (3*(-b)^(3/2)*Sqrt[Pi]*x*FresnelC[Sqrt[a - b*ArcSi
n[1 - d*x^2]]/(Sqrt[-b]*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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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsin \left (d \,x^{2}-1\right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(d*x^2 - 1))^(3/2),x)

[Out]

int((a + b*asin(d*x^2 - 1))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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