\(\int \frac {1}{(a-b \arcsin (1-d x^2))^{5/2}} \, dx\) [429]

Optimal result

Integrand size = 18, antiderivative size = 281 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}} \, dx=-\frac {\sqrt {2 d x^2-d^2 x^4}}{3 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a-b \arcsin \left (1-d x^2\right )}}+\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{3 (-b)^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{3 (-b)^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \]

[Out]

1/3*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)/(-b)^(5/2
)/(cos(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))+1/3*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)/(-b)^(5/2)/(cos(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))-1/
3*(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2-1))^(3/2)+1/3*x/b^2/(a+b*arcsin(d*x^2-1))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4912, 4903} \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}} \, dx=\frac {x}{3 b^2 \sqrt {a-b \arcsin \left (1-d x^2\right )}}-\frac {\sqrt {2 d x^2-d^2 x^4}}{3 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}+\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right )}{3 (-b)^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right )}{3 (-b)^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \]

[In]

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

[Out]

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

Rule 4903

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[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] - 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]

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*} \text {integral}& = -\frac {\sqrt {2 d x^2-d^2 x^4}}{3 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a-b \arcsin \left (1-d x^2\right )}}-\frac {\int \frac {1}{\sqrt {a-b \arcsin \left (1-d x^2\right )}} \, dx}{3 b^2} \\ & = -\frac {\sqrt {2 d x^2-d^2 x^4}}{3 b d x \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a-b \arcsin \left (1-d x^2\right )}}+\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {a-b \arcsin \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 )}{3 (-b)^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a-b \arcsin \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 )}{3 (-b)^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}} \, dx=\frac {\frac {-\frac {b \sqrt {-d x^2 \left (-2+d x^2\right )}}{d}+x^2 \left (a-b \arcsin \left (1-d x^2\right )\right )}{x \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}+\frac {\sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\sqrt {-b} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {-b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\sqrt {-b} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}}{3 b^2} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(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)

Sympy [F]

\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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