3.477 \(\int \frac {(c-a^2 c x^2)^{3/2}}{\sin ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=206 \[ -\frac {4 \sqrt {2 \pi } c \sqrt {c-a^2 c x^2} C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\sin ^{-1}(a x)}\right )}{3 a \sqrt {1-a^2 x^2}}-\frac {8 \sqrt {\pi } c \sqrt {c-a^2 c x^2} C\left (\frac {2 \sqrt {\sin ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a \sqrt {1-a^2 x^2}}-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}+\frac {16 c x \left (1-a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 \sqrt {\sin ^{-1}(a x)}} \]

[Out]

-2/3*(-a^2*c*x^2+c)^(3/2)*(-a^2*x^2+1)^(1/2)/a/arcsin(a*x)^(3/2)-8/3*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)-4/3*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)+16/3*c*x*(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/arcsin(a*x)^(1
/2)

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Rubi [A]  time = 0.30, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4659, 4721, 4661, 3312, 3304, 3352, 4723, 4406} \[ -\frac {4 \sqrt {2 \pi } c \sqrt {c-a^2 c x^2} \text {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\sin ^{-1}(a x)}\right )}{3 a \sqrt {1-a^2 x^2}}-\frac {8 \sqrt {\pi } c \sqrt {c-a^2 c x^2} \text {FresnelC}\left (\frac {2 \sqrt {\sin ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a \sqrt {1-a^2 x^2}}-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}+\frac {16 c x \left (1-a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 \sqrt {\sin ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^(3/2)/ArcSin[a*x]^(5/2),x]

[Out]

(-2*Sqrt[1 - a^2*x^2]*(c - a^2*c*x^2)^(3/2))/(3*a*ArcSin[a*x]^(3/2)) + (16*c*x*(1 - a^2*x^2)*Sqrt[c - a^2*c*x^
2])/(3*Sqrt[ArcSin[a*x]]) - (4*c*Sqrt[2*Pi]*Sqrt[c - a^2*c*x^2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcSin[a*x]]])/(3*a
*Sqrt[1 - a^2*x^2]) - (8*c*Sqrt[Pi]*Sqrt[c - a^2*c*x^2]*FresnelC[(2*Sqrt[ArcSin[a*x]])/Sqrt[Pi]])/(3*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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4659

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*
(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(b*(n + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x],
x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]

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 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
((f*x)^m*Sqrt[1 - c^2*x^2]*(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(f*m*d^IntPar
t[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p - 1/
2)*(a + b*ArcSin[c*x])^(n + 1), x], x] + Dist[(c*(m + 2*p + 1)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(n +
1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x])
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1] && IGtQ[m, -3] && IGtQ[2*p, 0]

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*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] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {\left (c-a^2 c x^2\right )^{3/2}}{\sin ^{-1}(a x)^{5/2}} \, dx &=-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}-\frac {\left (8 a c \sqrt {c-a^2 c x^2}\right ) \int \frac {x \left (1-a^2 x^2\right )}{\sin ^{-1}(a x)^{3/2}} \, dx}{3 \sqrt {1-a^2 x^2}}\\ &=-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}+\frac {16 c x \left (1-a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 \sqrt {\sin ^{-1}(a x)}}-\frac {\left (16 c \sqrt {c-a^2 c x^2}\right ) \int \frac {\sqrt {1-a^2 x^2}}{\sqrt {\sin ^{-1}(a x)}} \, dx}{3 \sqrt {1-a^2 x^2}}+\frac {\left (64 a^2 c \sqrt {c-a^2 c x^2}\right ) \int \frac {x^2 \sqrt {1-a^2 x^2}}{\sqrt {\sin ^{-1}(a x)}} \, dx}{3 \sqrt {1-a^2 x^2}}\\ &=-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}+\frac {16 c x \left (1-a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 \sqrt {\sin ^{-1}(a x)}}-\frac {\left (16 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 )}{3 a \sqrt {1-a^2 x^2}}+\frac {\left (64 c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos ^2(x) \sin ^2(x)}{\sqrt {x}} \, dx,x,\sin ^{-1}(a x)\right )}{3 a \sqrt {1-a^2 x^2}}\\ &=-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}+\frac {16 c x \left (1-a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 \sqrt {\sin ^{-1}(a x)}}-\frac {\left (16 c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{3 a \sqrt {1-a^2 x^2}}+\frac {\left (64 c \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{8 \sqrt {x}}-\frac {\cos (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{3 a \sqrt {1-a^2 x^2}}\\ &=-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}+\frac {16 c x \left (1-a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 \sqrt {\sin ^{-1}(a x)}}-\frac {\left (8 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 )}{3 a \sqrt {1-a^2 x^2}}-\frac {\left (8 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 )}{3 a \sqrt {1-a^2 x^2}}\\ &=-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}+\frac {16 c x \left (1-a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 \sqrt {\sin ^{-1}(a x)}}-\frac {\left (16 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 )}{3 a \sqrt {1-a^2 x^2}}-\frac {\left (16 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 )}{3 a \sqrt {1-a^2 x^2}}\\ &=-\frac {2 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^{3/2}}{3 a \sin ^{-1}(a x)^{3/2}}+\frac {16 c x \left (1-a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 \sqrt {\sin ^{-1}(a x)}}-\frac {4 c \sqrt {2 \pi } \sqrt {c-a^2 c x^2} C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\sin ^{-1}(a x)}\right )}{3 a \sqrt {1-a^2 x^2}}-\frac {8 c \sqrt {\pi } \sqrt {c-a^2 c x^2} C\left (\frac {2 \sqrt {\sin ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 1.60, size = 251, normalized size = 1.22 \[ \frac {c \sqrt {c-a^2 c x^2} \left (16 a^2 x^2+64 a x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)-e^{-4 i \sin ^{-1}(a x)}-e^{4 i \sin ^{-1}(a x)}+8 i e^{-4 i \sin ^{-1}(a x)} \sin ^{-1}(a x)-8 i e^{4 i \sin ^{-1}(a x)} \sin ^{-1}(a x)-16 \sqrt {2} \left (-i \sin ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-2 i \sin ^{-1}(a x)\right )-16 \sqrt {2} \left (i \sin ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},2 i \sin ^{-1}(a x)\right )-16 \left (-i \sin ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-4 i \sin ^{-1}(a x)\right )-16 \left (i \sin ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},4 i \sin ^{-1}(a x)\right )-14\right )}{24 a \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - a^2*c*x^2)^(3/2)/ArcSin[a*x]^(5/2),x]

[Out]

(c*Sqrt[c - a^2*c*x^2]*(-14 - E^((-4*I)*ArcSin[a*x]) - E^((4*I)*ArcSin[a*x]) + 16*a^2*x^2 + ((8*I)*ArcSin[a*x]
)/E^((4*I)*ArcSin[a*x]) - (8*I)*E^((4*I)*ArcSin[a*x])*ArcSin[a*x] + 64*a*x*Sqrt[1 - a^2*x^2]*ArcSin[a*x] - 16*
Sqrt[2]*((-I)*ArcSin[a*x])^(3/2)*Gamma[1/2, (-2*I)*ArcSin[a*x]] - 16*Sqrt[2]*(I*ArcSin[a*x])^(3/2)*Gamma[1/2,
(2*I)*ArcSin[a*x]] - 16*((-I)*ArcSin[a*x])^(3/2)*Gamma[1/2, (-4*I)*ArcSin[a*x]] - 16*(I*ArcSin[a*x])^(3/2)*Gam
ma[1/2, (4*I)*ArcSin[a*x]]))/(24*a*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^(3/2))

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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)^(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 \[ \int \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{\arcsin \left (a x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(3/2)/arcsin(a*x)^(5/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(3/2)/arcsin(a*x)^(5/2), x)

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{\arcsin \left (a x \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*c*x^2+c)^(3/2)/arcsin(a*x)^(5/2),x)

[Out]

int((-a^2*c*x^2+c)^(3/2)/arcsin(a*x)^(5/2),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)^(5/2),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 = -0.00 \[ \int \frac {{\left (c-a^2\,c\,x^2\right )}^{3/2}}{{\mathrm {asin}\left (a\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a^2*c*x^2)^(3/2)/asin(a*x)^(5/2),x)

[Out]

int((c - a^2*c*x^2)^(3/2)/asin(a*x)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*c*x**2+c)**(3/2)/asin(a*x)**(5/2),x)

[Out]

Timed out

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