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

Optimal. Leaf size=228 \[ -\frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )} \]

[Out]

-x*FresnelC((-1/b)^(1/2)*(a+b*arcsin(d*x^2-1))^(1/2)/Pi^(1/2))*(cos(1/2*a/b)-sin(1/2*a/b))*Pi^(1/2)/(cos(1/2*a
rcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))/(-1/b)^(1/2)+x*FresnelS((-1/b)^(1/2)*(a+b*arcsin(d*x^2-1))^(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)))/(-1/b)^(1/2)+x
*(a+b*arcsin(d*x^2-1))^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 228, normalized size of antiderivative = 1.00, 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 = {4811} \[ -\frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - b*ArcSin[1 - d*x^2]],x]

[Out]

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

Rule 4811

Int[Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[x*Sqrt[a + b*ArcSin[c + d*x^2]], x] + (
-Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqr
t[c/b]*(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[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqrt[c/b]*(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 \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )} \, dx &=x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}-\frac {\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 )}{\sqrt {-\frac {1}{b}} \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 {\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 )}{\sqrt {-\frac {1}{b}} \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 = 0.07, size = 225, normalized size = 0.99 \[ \frac {x \left (-\sqrt {\pi } \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )+\sqrt {\pi } \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )+\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right ) \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-(Sqrt[Pi]*FresnelC[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))
+ Sqrt[Pi]*FresnelS[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]) + S
qrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]]*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])))/(Sqrt[-b^(-
1)]*(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))^(1/2),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 \[ \int \sqrt {b \arcsin \left (d x^{2} - 1\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(d*x^2 - 1))^(1/2),x)

[Out]

int((a + b*asin(d*x^2 - 1))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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