3.466 \(\int \frac {(c-a^2 c x^2)^{3/2}}{\sqrt {\sin ^{-1}(a x)}} \, dx\)

Optimal. Leaf size=170 \[ \frac {\sqrt {\frac {\pi }{2}} c \sqrt {c-a^2 c x^2} C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\sin ^{-1}(a x)}\right )}{8 a \sqrt {1-a^2 x^2}}+\frac {\sqrt {\pi } c \sqrt {c-a^2 c x^2} C\left (\frac {2 \sqrt {\sin ^{-1}(a x)}}{\sqrt {\pi }}\right )}{2 a \sqrt {1-a^2 x^2}}+\frac {3 c \sqrt {c-a^2 c x^2} \sqrt {\sin ^{-1}(a x)}}{4 a \sqrt {1-a^2 x^2}} \]

[Out]

1/16*c*FresnelC(2*2^(1/2)/Pi^(1/2)*arcsin(a*x)^(1/2))*2^(1/2)*Pi^(1/2)*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/
2)+1/2*c*FresnelC(2*arcsin(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)+3/4*c*(-a^2
*c*x^2+c)^(1/2)*arcsin(a*x)^(1/2)/a/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4663, 4661, 3312, 3304, 3352} \[ \frac {\sqrt {\frac {\pi }{2}} c \sqrt {c-a^2 c x^2} \text {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\sin ^{-1}(a x)}\right )}{8 a \sqrt {1-a^2 x^2}}+\frac {\sqrt {\pi } c \sqrt {c-a^2 c x^2} \text {FresnelC}\left (\frac {2 \sqrt {\sin ^{-1}(a x)}}{\sqrt {\pi }}\right )}{2 a \sqrt {1-a^2 x^2}}+\frac {3 c \sqrt {c-a^2 c x^2} \sqrt {\sin ^{-1}(a x)}}{4 a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^(3/2)/Sqrt[ArcSin[a*x]],x]

[Out]

(3*c*Sqrt[c - a^2*c*x^2]*Sqrt[ArcSin[a*x]])/(4*a*Sqrt[1 - a^2*x^2]) + (c*Sqrt[Pi/2]*Sqrt[c - a^2*c*x^2]*Fresne
lC[2*Sqrt[2/Pi]*Sqrt[ArcSin[a*x]]])/(8*a*Sqrt[1 - a^2*x^2]) + (c*Sqrt[Pi]*Sqrt[c - a^2*c*x^2]*FresnelC[(2*Sqrt
[ArcSin[a*x]])/Sqrt[Pi]])/(2*a*Sqrt[1 - a^2*x^2])

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4661

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[(
a + b*x)^n*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && I
GtQ[2*p, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 4663

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(d^(p - 1/2)*Sqrt[
d + e*x^2])/Sqrt[1 - c^2*x^2], Int[(1 - c^2*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IGtQ[2*p, 0] &&  !(IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {\left (c-a^2 c x^2\right )^{3/2}}{\sqrt {\sin ^{-1}(a x)}} \, dx &=\frac {\left (c \sqrt {c-a^2 c x^2}\right ) \int \frac {\left (1-a^2 x^2\right )^{3/2}}{\sqrt {\sin ^{-1}(a x)}} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\left (c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos ^4(x)}{\sqrt {x}} \, dx,x,\sin ^{-1}(a x)\right )}{a \sqrt {1-a^2 x^2}}\\ &=\frac {\left (c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {3}{8 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}+\frac {\cos (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a \sqrt {1-a^2 x^2}}\\ &=\frac {3 c \sqrt {c-a^2 c x^2} \sqrt {\sin ^{-1}(a x)}}{4 a \sqrt {1-a^2 x^2}}+\frac {\left (c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {x}} \, dx,x,\sin ^{-1}(a x)\right )}{8 a \sqrt {1-a^2 x^2}}+\frac {\left (c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\sin ^{-1}(a x)\right )}{2 a \sqrt {1-a^2 x^2}}\\ &=\frac {3 c \sqrt {c-a^2 c x^2} \sqrt {\sin ^{-1}(a x)}}{4 a \sqrt {1-a^2 x^2}}+\frac {\left (c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \cos \left (4 x^2\right ) \, dx,x,\sqrt {\sin ^{-1}(a x)}\right )}{4 a \sqrt {1-a^2 x^2}}+\frac {\left (c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\sin ^{-1}(a x)}\right )}{a \sqrt {1-a^2 x^2}}\\ &=\frac {3 c \sqrt {c-a^2 c x^2} \sqrt {\sin ^{-1}(a x)}}{4 a \sqrt {1-a^2 x^2}}+\frac {c \sqrt {\frac {\pi }{2}} \sqrt {c-a^2 c x^2} C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\sin ^{-1}(a x)}\right )}{8 a \sqrt {1-a^2 x^2}}+\frac {c \sqrt {\pi } \sqrt {c-a^2 c x^2} C\left (\frac {2 \sqrt {\sin ^{-1}(a x)}}{\sqrt {\pi }}\right )}{2 a \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.37, size = 182, normalized size = 1.07 \[ \frac {c \sqrt {c-a^2 c x^2} \sqrt {\sin ^{-1}(a x)} \left (24 \sqrt {\sin ^{-1}(a x)^2}-4 \sqrt {2} \sqrt {i \sin ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 i \sin ^{-1}(a x)\right )-4 \sqrt {2} \sqrt {-i \sin ^{-1}(a x)} \Gamma \left (\frac {1}{2},2 i \sin ^{-1}(a x)\right )-\sqrt {i \sin ^{-1}(a x)} \Gamma \left (\frac {1}{2},-4 i \sin ^{-1}(a x)\right )-\sqrt {-i \sin ^{-1}(a x)} \Gamma \left (\frac {1}{2},4 i \sin ^{-1}(a x)\right )\right )}{32 a \sqrt {1-a^2 x^2} \sqrt {\sin ^{-1}(a x)^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - a^2*c*x^2)^(3/2)/Sqrt[ArcSin[a*x]],x]

[Out]

(c*Sqrt[c - a^2*c*x^2]*Sqrt[ArcSin[a*x]]*(24*Sqrt[ArcSin[a*x]^2] - 4*Sqrt[2]*Sqrt[I*ArcSin[a*x]]*Gamma[1/2, (-
2*I)*ArcSin[a*x]] - 4*Sqrt[2]*Sqrt[(-I)*ArcSin[a*x]]*Gamma[1/2, (2*I)*ArcSin[a*x]] - Sqrt[I*ArcSin[a*x]]*Gamma
[1/2, (-4*I)*ArcSin[a*x]] - Sqrt[(-I)*ArcSin[a*x]]*Gamma[1/2, (4*I)*ArcSin[a*x]]))/(32*a*Sqrt[1 - a^2*x^2]*Sqr
t[ArcSin[a*x]^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^2*c*x^2+c)^(3/2)/arcsin(a*x)^(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 \[ \int \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{\sqrt {\arcsin \left (a x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-a^2*c*x^2 + c)^(3/2)/sqrt(arcsin(a*x)), x)

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{\sqrt {\arcsin \left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(3/2)/arcsin(a*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-a^2\,c\,x^2\right )}^{3/2}}{\sqrt {\mathrm {asin}\left (a\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a^2*c*x^2)^(3/2)/asin(a*x)^(1/2),x)

[Out]

int((c - a^2*c*x^2)^(3/2)/asin(a*x)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*c*x**2+c)**(3/2)/asin(a*x)**(1/2),x)

[Out]

Integral((-c*(a*x - 1)*(a*x + 1))**(3/2)/sqrt(asin(a*x)), x)

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