3.430 \(\int \frac {1}{(a-b \sin ^{-1}(1-d x^2))^{7/2}} \, dx\)

Optimal. Leaf size=339 \[ \frac {\sqrt {2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}+\frac {x}{15 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}-\frac {\sqrt {2 d x^2-d^2 x^4}}{5 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}+\frac {\sqrt {\pi } \left (-\frac {1}{b}\right )^{7/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{15 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac {\sqrt {\pi } \left (-\frac {1}{b}\right )^{7/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{15 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )} \]

[Out]

1/15*x/b^2/(a+b*arcsin(d*x^2-1))^(3/2)+1/15*(-1/b)^(7/2)*x*FresnelC((-1/b)^(1/2)*(a+b*arcsin(d*x^2-1))^(1/2)/P
i^(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)))-1/15*(-1/b)^
(7/2)*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)))-1/5*(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2-1))^(5/2)+
1/15*(-d^2*x^4+2*d*x^2)^(1/2)/b^3/d/x/(a+b*arcsin(d*x^2-1))^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 339, 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 = {4828, 4822} \[ \frac {\sqrt {2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}+\frac {x}{15 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}-\frac {\sqrt {2 d x^2-d^2 x^4}}{5 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}+\frac {\sqrt {\pi } \left (-\frac {1}{b}\right )^{7/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{15 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac {\sqrt {\pi } \left (-\frac {1}{b}\right )^{7/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{15 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*ArcSin[1 - d*x^2])^(-7/2),x]

[Out]

-Sqrt[2*d*x^2 - d^2*x^4]/(5*b*d*x*(a - b*ArcSin[1 - d*x^2])^(5/2)) + x/(15*b^2*(a - b*ArcSin[1 - d*x^2])^(3/2)
) + Sqrt[2*d*x^2 - d^2*x^4]/(15*b^3*d*x*Sqrt[a - b*ArcSin[1 - d*x^2]]) + ((-b^(-1))^(7/2)*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)]))/(15*(Cos[ArcSin[1 - d*x
^2]/2] - Sin[ArcSin[1 - d*x^2]/2])) - ((-b^(-1))^(7/2)*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)]))/(15*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]))

Rule 4822

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> -Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]/(b*d*x*S
qrt[a + b*ArcSin[c + d*x^2]]), x] + (-Simp[((c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sq
rt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2]), x] +
Simp[((c/b)^(3/2)*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]]])/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2]), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2,
1]

Rule 4828

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 )^{7/2}} \, dx &=-\frac {\sqrt {2 d x^2-d^2 x^4}}{5 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}-\frac {\int \frac {1}{\left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}} \, dx}{15 b^2}\\ &=-\frac {\sqrt {2 d x^2-d^2 x^4}}{5 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}+\frac {\sqrt {2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}+\frac {\left (-\frac {1}{b}\right )^{7/2} \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 )}{15 \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 {\left (-\frac {1}{b}\right )^{7/2} \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 )}{15 \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 = 1.07, size = 319, normalized size = 0.94 \[ \frac {\frac {\sqrt {\pi } \left (-\frac {1}{b}\right )^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac {\sqrt {\pi } b \left (-\frac {1}{b}\right )^{5/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac {x^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )+\frac {\sqrt {d x^2 \left (2-d x^2\right )} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{b d}-\frac {3 b \sqrt {d x^2 \left (2-d x^2\right )}}{d}}{x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}}{15 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*ArcSin[1 - d*x^2])^(-7/2),x]

[Out]

(((-3*b*Sqrt[d*x^2*(2 - d*x^2)])/d + x^2*(a - b*ArcSin[1 - d*x^2]) + (Sqrt[d*x^2*(2 - d*x^2)]*(a - b*ArcSin[1
- d*x^2])^2)/(b*d))/(x*(a - b*ArcSin[1 - d*x^2])^(5/2)) + ((-b^(-1))^(3/2)*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)]))/(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin
[1 - d*x^2]/2]) + ((-b^(-1))^(5/2)*b*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)]))/(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]))/(15*b^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(1/(a+b*arcsin(d*x^2-1))^(7/2),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 \[ \int \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x^2-1))^(7/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 - 1) + a)^(-7/2), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsin(d*x^2-1))^(7/2),x)

[Out]

int(1/(a+b*arcsin(d*x^2-1))^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x^2-1))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*arcsin(d*x^2 - 1) + a)^(-7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(d*x^2 - 1))^(7/2),x)

[Out]

int(1/(a + b*asin(d*x^2 - 1))^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asin(d*x**2-1))**(7/2),x)

[Out]

Integral((a + b*asin(d*x**2 - 1))**(-7/2), x)

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