3.424 \(\int (a-b \sin ^{-1}(1-d x^2))^{5/2} \, dx\)

Optimal. Leaf size=299 \[ -15 b^2 x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}{d x}+\frac {15 \sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\left (-\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac {15 \sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\left (-\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2} \]

[Out]

x*(a+b*arcsin(d*x^2-1))^(5/2)+15*x*FresnelC((-1/b)^(1/2)*(a+b*arcsin(d*x^2-1))^(1/2)/Pi^(1/2))*(cos(1/2*a/b)-s
in(1/2*a/b))*Pi^(1/2)/(-1/b)^(5/2)/(cos(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))-15*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)/(-1/b)^(5/2)/(cos(1/2*arcsin(d*
x^2-1))+sin(1/2*arcsin(d*x^2-1)))+5*b*(a+b*arcsin(d*x^2-1))^(3/2)*(-d^2*x^4+2*d*x^2)^(1/2)/d/x-15*b^2*x*(a+b*a
rcsin(d*x^2-1))^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 299, 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, 4811} \[ -15 b^2 x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}{d x}+\frac {15 \sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\left (-\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac {15 \sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\left (-\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*ArcSin[1 - d*x^2])^(5/2),x]

[Out]

-15*b^2*x*Sqrt[a - b*ArcSin[1 - d*x^2]] + (5*b*Sqrt[2*d*x^2 - d^2*x^4]*(a - b*ArcSin[1 - d*x^2])^(3/2))/(d*x)
+ x*(a - b*ArcSin[1 - d*x^2])^(5/2) + (15*Sqrt[Pi]*x*FresnelC[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sq
rt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/((-b^(-1))^(5/2)*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]))
- (15*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)
]))/((-b^(-1))^(5/2)*(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]

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]

Rubi steps

\begin {align*} \int \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2} \, dx &=\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}-\left (15 b^2\right ) \int \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )} \, dx\\ &=-15 b^2 x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}+\frac {15 \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 )}{\left (-\frac {1}{b}\right )^{5/2} \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 {15 \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 )}{\left (-\frac {1}{b}\right )^{5/2} \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*}

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Mathematica [A]  time = 0.34, size = 292, normalized size = 0.98 \[ \frac {15 b x \left (-\sqrt {\pi } \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )+\sqrt {\pi } \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )+\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right ) \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}\right )}{\left (-\frac {1}{b}\right )^{3/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}+\frac {5 b \sqrt {-d x^2 \left (d x^2-2\right )} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}{d x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*ArcSin[1 - d*x^2])^(5/2),x]

[Out]

(5*b*Sqrt[-(d*x^2*(-2 + d*x^2))]*(a - b*ArcSin[1 - d*x^2])^(3/2))/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^(5/2) +
(15*b*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)]
) + Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]]*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])))/((-b^(
-1))^(3/2)*(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))^(5/2),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 \[ \int {\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2-1))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 - 1) + a)^(5/2), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsin \left (d \,x^{2}-1\right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x^2-1))^(5/2),x)

[Out]

int((a+b*arcsin(d*x^2-1))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2-1))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arcsin(d*x^2 - 1) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(d*x^2 - 1))^(5/2),x)

[Out]

int((a + b*asin(d*x^2 - 1))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x**2-1))**(5/2),x)

[Out]

Integral((a + b*asin(d*x**2 - 1))**(5/2), x)

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