3.391 \(\int \frac {x^2 (1-c^2 x^2)^{3/2}}{(a+b \sin ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=220 \[ \frac {\sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}-\frac {3 \sin \left (\frac {6 a}{b}\right ) \text {Ci}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}+\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.64, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4721, 4723, 4406, 3303, 3299, 3302} \[ \frac {\sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {3 \sin \left (\frac {6 a}{b}\right ) \text {CosIntegral}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{16 b^2 c^3}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{4 b^2 c^3}+\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{16 b^2 c^3}-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(1 - c^2*x^2)^(3/2))/(a + b*ArcSin[c*x])^2,x]

[Out]

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

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 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 {x^2 \left (1-c^2 x^2\right )^{3/2}}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \int \frac {x \left (1-c^2 x^2\right )}{a+b \sin ^{-1}(c x)} \, dx}{b c}-\frac {(6 c) \int \frac {x^3 \left (1-c^2 x^2\right )}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {\cos ^3(x) \sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}-\frac {6 \operatorname {Subst}\left (\int \frac {\cos ^3(x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \operatorname {Subst}\left (\int \left (\frac {\sin (2 x)}{4 (a+b x)}+\frac {\sin (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}-\frac {6 \operatorname {Subst}\left (\int \left (\frac {3 \sin (2 x)}{32 (a+b x)}-\frac {\sin (6 x)}{32 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^2}{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^3}+\frac {\operatorname {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}+\frac {\operatorname {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c^3}-\frac {9 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\cos \left (\frac {2 a}{b}\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 )}{2 b c^3}-\frac {\left (9 \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^3}+\frac {\cos \left (\frac {4 a}{b}\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^3}+\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^3}-\frac {\sin \left (\frac {2 a}{b}\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 )}{2 b c^3}+\frac {\left (9 \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^3}-\frac {\sin \left (\frac {4 a}{b}\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^3}-\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^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right ) \sin \left (\frac {2 a}{b}\right )}{16 b^2 c^3}-\frac {\text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right ) \sin \left (\frac {4 a}{b}\right )}{4 b^2 c^3}-\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^3}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{4 b^2 c^3}+\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^3}\\ \end {align*}

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Mathematica [A]  time = 0.86, size = 306, normalized size = 1.39 \[ -\frac {-\sin \left (\frac {2 a}{b}\right ) \left (a+b \sin ^{-1}(c x)\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+4 \sin \left (\frac {4 a}{b}\right ) \left (a+b \sin ^{-1}(c x)\right ) \text {Ci}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+3 a \sin \left (\frac {6 a}{b}\right ) \text {Ci}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+3 b \sin \left (\frac {6 a}{b}\right ) \sin ^{-1}(c x) \text {Ci}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+a \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+b \cos \left (\frac {2 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-4 a \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-4 b \cos \left (\frac {4 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-3 a \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-3 b \cos \left (\frac {6 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+16 b c^6 x^6-32 b c^4 x^4+16 b c^2 x^2}{16 b^2 c^3 \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(1 - c^2*x^2)^(3/2))/(a + b*ArcSin[c*x])^2,x]

[Out]

-1/16*(16*b*c^2*x^2 - 32*b*c^4*x^4 + 16*b*c^6*x^6 - (a + b*ArcSin[c*x])*CosIntegral[2*(a/b + ArcSin[c*x])]*Sin
[(2*a)/b] + 4*(a + b*ArcSin[c*x])*CosIntegral[4*(a/b + ArcSin[c*x])]*Sin[(4*a)/b] + 3*a*CosIntegral[6*(a/b + A
rcSin[c*x])]*Sin[(6*a)/b] + 3*b*ArcSin[c*x]*CosIntegral[6*(a/b + ArcSin[c*x])]*Sin[(6*a)/b] + a*Cos[(2*a)/b]*S
inIntegral[2*(a/b + ArcSin[c*x])] + b*ArcSin[c*x]*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] - 4*a*Cos[(4
*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] - 4*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]*SinIntegral[6*(a/b + ArcSin
[c*x])])/(b^2*c^3*(a + b*ArcSin[c*x]))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (c^{2} x^{4} - x^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.69, size = 1553, normalized size = 7.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*x^2+1)^(3/2)/(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^3*arcsin(c*x) + a*b^2*c^3) + 6
*b*arcsin(c*x)*cos(a/b)^6*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 6*a*cos(a/b)
^5*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 6*a*cos(a/b)^6*sin_integra
l(6*a/b + 6*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 6*b*arcsin(c*x)*cos(a/b)^3*cos_integral(6*a/b + 6
*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 2*b*arcsin(c*x)*cos(a/b)^3*cos_integral(4*a/b + 4*a
rcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 9*b*arcsin(c*x)*cos(a/b)^4*sin_integral(6*a/b + 6*arc
sin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 2*b*arcsin(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(
b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 6*a*cos(a/b)^3*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin
(c*x) + a*b^2*c^3) - 2*a*cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*
c^3) - 9*a*cos(a/b)^4*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 2*a*cos(a/b)^4*s
in_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 9/8*b*arcsin(c*x)*cos(a/b)*cos_integral
(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c^3*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^3*arcsin(c*x) + a*b^2*c^3) + 1/8*b*arcsin(c*x)*cos(a/b)*cos_integral(2*a/b +
2*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 27/8*b*arcsin(c*x)*cos(a/b)^2*sin_integral(6*a/b
+ 6*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 2*b*arcsin(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(
c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 1/8*b*arcsin(c*x)*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^
3*c^3*arcsin(c*x) + a*b^2*c^3) - (c^2*x^2 - 1)^3*b/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 9/8*a*cos(a/b)*cos_inte
gral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + a*cos(a/b)*cos_integral(4*a/b + 4*arc
sin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/8*a*cos(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(
a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 27/8*a*cos(a/b)^2*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c^3*arcsin
(c*x) + a*b^2*c^3) - 2*a*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 1/
8*a*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - (c^2*x^2 - 1)^2*b/(b^3*
c^3*arcsin(c*x) + a*b^2*c^3) - 3/16*b*arcsin(c*x)*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a
*b^2*c^3) + 1/4*b*arcsin(c*x)*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/16*b*a
rcsin(c*x)*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 3/16*a*sin_integral(6*a/b +
6*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/4*a*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(
c*x) + a*b^2*c^3) + 1/16*a*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3)

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maple [A]  time = 0.10, size = 364, normalized size = 1.65 \[ \frac {8 \arcsin \left (c x \right ) \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -8 \arcsin \left (c x \right ) \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b -2 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b +2 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b +6 \arcsin \left (c x \right ) \Si \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) b -6 \arcsin \left (c x \right ) \Ci \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) b +8 \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -8 \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a -2 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a +2 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +6 \Si \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) a -6 \Ci \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) a +2 \cos \left (4 \arcsin \left (c x \right )\right ) b -\cos \left (2 \arcsin \left (c x \right )\right ) b +\cos \left (6 \arcsin \left (c x \right )\right ) b -2 b}{32 c^{3} \left (a +b \arcsin \left (c x \right )\right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/c^3*(8*arcsin(c*x)*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*b-8*arcsin(c*x)*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*
b-2*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*arcsin(c*x)*b+2*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*arcsin(c*x)*b+6*arcs
in(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+8*Si(4*arcsin(
c*x)+4*a/b)*cos(4*a/b)*a-8*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*a-2*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*a+2*Ci(2*
arcsin(c*x)+2*a/b)*sin(2*a/b)*a+6*Si(6*arcsin(c*x)+6*a/b)*cos(6*a/b)*a-6*Ci(6*arcsin(c*x)+6*a/b)*sin(6*a/b)*a+
2*cos(4*arcsin(c*x))*b-cos(2*arcsin(c*x))*b+cos(6*arcsin(c*x))*b-2*b)/(a+b*arcsin(c*x))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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