3.402 \(\int \frac {(1-c^2 x^2)^{5/2}}{(a+b \sin ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=217 \[ \frac {15 \sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c}+\frac {3 \sin \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2 c}+\frac {3 \sin \left (\frac {6 a}{b}\right ) \text {Ci}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c}-\frac {15 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c}-\frac {3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2 c}-\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c}-\frac {\left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

-(-c^2*x^2+1)^3/b/c/(a+b*arcsin(c*x))-15/16*cos(2*a/b)*Si(2*(a+b*arcsin(c*x))/b)/b^2/c-3/4*cos(4*a/b)*Si(4*(a+
b*arcsin(c*x))/b)/b^2/c-3/16*cos(6*a/b)*Si(6*(a+b*arcsin(c*x))/b)/b^2/c+15/16*Ci(2*(a+b*arcsin(c*x))/b)*sin(2*
a/b)/b^2/c+3/4*Ci(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b^2/c+3/16*Ci(6*(a+b*arcsin(c*x))/b)*sin(6*a/b)/b^2/c

________________________________________________________________________________________

Rubi [A]  time = 0.40, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4659, 4723, 4406, 3303, 3299, 3302} \[ \frac {15 \sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c}+\frac {3 \sin \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{4 b^2 c}+\frac {3 \sin \left (\frac {6 a}{b}\right ) \text {CosIntegral}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{16 b^2 c}-\frac {15 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c}-\frac {3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{4 b^2 c}-\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{16 b^2 c}-\frac {\left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - c^2*x^2)^3/(b*c*(a + b*ArcSin[c*x]))) + (15*CosIntegral[(2*a)/b + 2*ArcSin[c*x]]*Sin[(2*a)/b])/(16*b^2*
c) + (3*CosIntegral[(4*a)/b + 4*ArcSin[c*x]]*Sin[(4*a)/b])/(4*b^2*c) + (3*CosIntegral[(6*a)/b + 6*ArcSin[c*x]]
*Sin[(6*a)/b])/(16*b^2*c) - (15*Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]])/(16*b^2*c) - (3*Cos[(4*a)/b
]*SinIntegral[(4*a)/b + 4*ArcSin[c*x]])/(4*b^2*c) - (3*Cos[(6*a)/b]*SinIntegral[(6*a)/b + 6*ArcSin[c*x]])/(16*
b^2*c)

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

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

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

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 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 (1-c^2 x^2\right )^{5/2}}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {\left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {(6 c) \int \frac {x \left (1-c^2 x^2\right )^2}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {6 \operatorname {Subst}\left (\int \frac {\cos ^5(x) \sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {6 \operatorname {Subst}\left (\int \left (\frac {5 \sin (2 x)}{32 (a+b x)}+\frac {\sin (4 x)}{8 (a+b x)}+\frac {\sin (6 x)}{32 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {3 \operatorname {Subst}\left (\int \frac {\sin (6 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c}-\frac {3 \operatorname {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c}-\frac {15 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\left (15 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c}-\frac {\left (3 \cos \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c}-\frac {\left (3 \cos \left (\frac {6 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c}+\frac {\left (15 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c}+\frac {\left (3 \sin \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c}+\frac {\left (3 \sin \left (\frac {6 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c}\\ &=-\frac {\left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {15 \text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right ) \sin \left (\frac {2 a}{b}\right )}{16 b^2 c}+\frac {3 \text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right ) \sin \left (\frac {4 a}{b}\right )}{4 b^2 c}+\frac {3 \text {Ci}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right ) \sin \left (\frac {6 a}{b}\right )}{16 b^2 c}-\frac {15 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c}-\frac {3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{4 b^2 c}-\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{16 b^2 c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.92, size = 311, normalized size = 1.43 \[ -\frac {-15 \sin \left (\frac {2 a}{b}\right ) \left (a+b \sin ^{-1}(c x)\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-12 \sin \left (\frac {4 a}{b}\right ) \left (a+b \sin ^{-1}(c x)\right ) \text {Ci}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-3 a \sin \left (\frac {6 a}{b}\right ) \text {Ci}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-3 b \sin \left (\frac {6 a}{b}\right ) \sin ^{-1}(c x) \text {Ci}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+15 a \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+15 b \cos \left (\frac {2 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+12 a \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+12 b \cos \left (\frac {4 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+3 a \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+3 b \cos \left (\frac {6 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-16 b c^6 x^6+48 b c^4 x^4-48 b c^2 x^2+16 b}{16 b^2 c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(16*b - 48*b*c^2*x^2 + 48*b*c^4*x^4 - 16*b*c^6*x^6 - 15*(a + b*ArcSin[c*x])*CosIntegral[2*(a/b + ArcSin[
c*x])]*Sin[(2*a)/b] - 12*(a + b*ArcSin[c*x])*CosIntegral[4*(a/b + ArcSin[c*x])]*Sin[(4*a)/b] - 3*a*CosIntegral
[6*(a/b + ArcSin[c*x])]*Sin[(6*a)/b] - 3*b*ArcSin[c*x]*CosIntegral[6*(a/b + ArcSin[c*x])]*Sin[(6*a)/b] + 15*a*
Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + 15*b*ArcSin[c*x]*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*
x])] + 12*a*Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] + 12*b*ArcSin[c*x]*Cos[(4*a)/b]*SinIntegral[4*(a/b
+ ArcSin[c*x])] + 3*a*Cos[(6*a)/b]*SinIntegral[6*(a/b + ArcSin[c*x])] + 3*b*ArcSin[c*x]*Cos[(6*a)/b]*SinInteg
ral[6*(a/b + ArcSin[c*x])])/(b^2*c*(a + b*ArcSin[c*x]))

________________________________________________________________________________________

fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt {-c^{2} x^{2} + 1}}{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(-c^2*x^2 + 1)/(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2), x)

________________________________________________________________________________________

giac [B]  time = 0.75, size = 1394, normalized size = 6.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*b*arcsin(c*x)*cos(a/b)^5*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 6*b*ar
csin(c*x)*cos(a/b)^6*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 6*a*cos(a/b)^5*cos_in
tegral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 6*a*cos(a/b)^6*sin_integral(6*a/b + 6*a
rcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 6*b*arcsin(c*x)*cos(a/b)^3*cos_integral(6*a/b + 6*arcsin(c*x))*sin
(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 6*b*arcsin(c*x)*cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/
(b^3*c*arcsin(c*x) + a*b^2*c) + 9*b*arcsin(c*x)*cos(a/b)^4*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c*arcsin(c
*x) + a*b^2*c) - 6*b*arcsin(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c)
- 6*a*cos(a/b)^3*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 6*a*cos(a/b)^3*c
os_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 9*a*cos(a/b)^4*sin_integral(6*a/b
+ 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 6*a*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arc
sin(c*x) + a*b^2*c) + 9/8*b*arcsin(c*x)*cos(a/b)*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*
x) + a*b^2*c) - 3*b*arcsin(c*x)*cos(a/b)*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b
^2*c) + 15/8*b*arcsin(c*x)*cos(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c)
- 27/8*b*arcsin(c*x)*cos(a/b)^2*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 6*b*arcsi
n(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 15/8*b*arcsin(c*x)*cos(a
/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + (c^2*x^2 - 1)^3*b/(b^3*c*arcsin(c*x)
+ a*b^2*c) + 9/8*a*cos(a/b)*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 3*a*
cos(a/b)*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 15/8*a*cos(a/b)*cos_inte
gral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 27/8*a*cos(a/b)^2*sin_integral(6*a/b + 6*
arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 6*a*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(
c*x) + a*b^2*c) - 15/8*a*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 3/16*b
*arcsin(c*x)*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 3/4*b*arcsin(c*x)*sin_integra
l(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 15/16*b*arcsin(c*x)*sin_integral(2*a/b + 2*arcsin(c*x
))/(b^3*c*arcsin(c*x) + a*b^2*c) + 3/16*a*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) -
3/4*a*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 15/16*a*sin_integral(2*a/b + 2*arcsi
n(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c)

________________________________________________________________________________________

maple [A]  time = 0.10, size = 364, normalized size = 1.68 \[ -\frac {6 \arcsin \left (c x \right ) \Si \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) b -6 \arcsin \left (c x \right ) \Ci \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) b +24 \arcsin \left (c x \right ) \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -24 \arcsin \left (c x \right ) \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b +30 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b -30 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b +6 \Si \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) a -6 \Ci \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) a +24 \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -24 \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a +30 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -30 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +\cos \left (6 \arcsin \left (c x \right )\right ) b +6 \cos \left (4 \arcsin \left (c x \right )\right ) b +15 \cos \left (2 \arcsin \left (c x \right )\right ) b +10 b}{32 c \left (a +b \arcsin \left (c x \right )\right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/c*(6*arcsin(c*x)*Si(6*arcsin(c*x)+6*a/b)*cos(6*a/b)*b-6*arcsin(c*x)*Ci(6*arcsin(c*x)+6*a/b)*sin(6*a/b)*b
+24*arcsin(c*x)*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*b-24*arcsin(c*x)*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*b+30*Si
(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*arcsin(c*x)*b-30*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*arcsin(c*x)*b+6*Si(6*arcs
in(c*x)+6*a/b)*cos(6*a/b)*a-6*Ci(6*arcsin(c*x)+6*a/b)*sin(6*a/b)*a+24*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*a-24*
Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*a+30*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*a-30*Ci(2*arcsin(c*x)+2*a/b)*sin(2*
a/b)*a+cos(6*arcsin(c*x))*b+6*cos(4*arcsin(c*x))*b+15*cos(2*arcsin(c*x))*b+10*b)/(a+b*arcsin(c*x))/b^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 6 \, {\left (b^{2} c \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c\right )} \int \frac {c^{5} x^{5} - 2 \, c^{3} x^{3} + c x}{b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b}\,{d x} - 1}{b^{2} c \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^6*x^6 - 3*c^4*x^4 + 3*c^2*x^2 - (b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integrate(6*(c^5
*x^5 - 2*c^3*x^3 + c*x)/(b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b), x) - 1)/(b^2*c*arctan2(c*x, sq
rt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________