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

Optimal. Leaf size=305 \[ \frac{b c^3 d x^4 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}-\frac{5 b c d x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}+\frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{3}{8} d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{8 b c \sqrt{1-c^2 x^2}}+\frac{1}{32} b^2 c^2 d x^3 \sqrt{d-c^2 d x^2}-\frac{17}{64} b^2 d x \sqrt{d-c^2 d x^2}+\frac{17 b^2 d \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)}{64 c \sqrt{1-c^2 x^2}} \]

[Out]

(-17*b^2*d*x*Sqrt[d - c^2*d*x^2])/64 + (b^2*c^2*d*x^3*Sqrt[d - c^2*d*x^2])/32 + (17*b^2*d*Sqrt[d - c^2*d*x^2]*
ArcSin[c*x])/(64*c*Sqrt[1 - c^2*x^2]) - (5*b*c*d*x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*Sqrt[1 - c^2*
x^2]) + (b*c^3*d*x^4*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*Sqrt[1 - c^2*x^2]) + (3*d*x*Sqrt[d - c^2*d*x^
2]*(a + b*ArcSin[c*x])^2)/8 + (x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x])^2)/4 + (d*Sqrt[d - c^2*d*x^2]*(a +
b*ArcSin[c*x])^3)/(8*b*c*Sqrt[1 - c^2*x^2])

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Rubi [A]  time = 0.239648, antiderivative size = 307, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {4649, 4647, 4641, 4627, 321, 216, 4677, 195} \[ \frac{d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{8 b c \sqrt{1-c^2 x^2}}+\frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{3}{8} d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{b d \left (1-c^2 x^2\right )^{3/2} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}-\frac{3 b c d x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}-\frac{15}{64} b^2 d x \sqrt{d-c^2 d x^2}-\frac{1}{32} b^2 d x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}+\frac{9 b^2 d \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)}{64 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*b^2*d*x*Sqrt[d - c^2*d*x^2])/64 - (b^2*d*x*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2])/32 + (9*b^2*d*Sqrt[d - c^2*
d*x^2]*ArcSin[c*x])/(64*c*Sqrt[1 - c^2*x^2]) - (3*b*c*d*x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*Sqrt[1
- c^2*x^2]) + (b*d*(1 - c^2*x^2)^(3/2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*c) + (3*d*x*Sqrt[d - c^2*d
*x^2]*(a + b*ArcSin[c*x])^2)/8 + (x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x])^2)/4 + (d*Sqrt[d - c^2*d*x^2]*(a
+ b*ArcSin[c*x])^3)/(8*b*c*Sqrt[1 - c^2*x^2])

Rule 4649

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

Rule 4647

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

Rule 4641

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

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(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] && GtQ[n, 0] && NeQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rubi steps

\begin{align*} \int \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{1}{4} (3 d) \int \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx-\frac{\left (b c d \sqrt{d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx}{2 \sqrt{1-c^2 x^2}}\\ &=\frac{b d \left (1-c^2 x^2\right )^{3/2} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac{3}{8} d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{\left (3 d \sqrt{d-c^2 d x^2}\right ) \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{1-c^2 x^2}} \, dx}{8 \sqrt{1-c^2 x^2}}-\frac{\left (b^2 d \sqrt{d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \, dx}{8 \sqrt{1-c^2 x^2}}-\frac{\left (3 b c d \sqrt{d-c^2 d x^2}\right ) \int x \left (a+b \sin ^{-1}(c x)\right ) \, dx}{4 \sqrt{1-c^2 x^2}}\\ &=-\frac{1}{32} b^2 d x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}-\frac{3 b c d x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}+\frac{b d \left (1-c^2 x^2\right )^{3/2} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac{3}{8} d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{8 b c \sqrt{1-c^2 x^2}}-\frac{\left (3 b^2 d \sqrt{d-c^2 d x^2}\right ) \int \sqrt{1-c^2 x^2} \, dx}{32 \sqrt{1-c^2 x^2}}+\frac{\left (3 b^2 c^2 d \sqrt{d-c^2 d x^2}\right ) \int \frac{x^2}{\sqrt{1-c^2 x^2}} \, dx}{8 \sqrt{1-c^2 x^2}}\\ &=-\frac{15}{64} b^2 d x \sqrt{d-c^2 d x^2}-\frac{1}{32} b^2 d x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}-\frac{3 b c d x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}+\frac{b d \left (1-c^2 x^2\right )^{3/2} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac{3}{8} d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{8 b c \sqrt{1-c^2 x^2}}-\frac{\left (3 b^2 d \sqrt{d-c^2 d x^2}\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{64 \sqrt{1-c^2 x^2}}+\frac{\left (3 b^2 d \sqrt{d-c^2 d x^2}\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{16 \sqrt{1-c^2 x^2}}\\ &=-\frac{15}{64} b^2 d x \sqrt{d-c^2 d x^2}-\frac{1}{32} b^2 d x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}+\frac{9 b^2 d \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)}{64 c \sqrt{1-c^2 x^2}}-\frac{3 b c d x^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt{1-c^2 x^2}}+\frac{b d \left (1-c^2 x^2\right )^{3/2} \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac{3}{8} d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac{d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{8 b c \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.11664, size = 329, normalized size = 1.08 \[ \frac{d \sqrt{d-c^2 d x^2} \left (-64 a^2 c^3 x^3 \sqrt{1-c^2 x^2}+160 a^2 c x \sqrt{1-c^2 x^2}+64 a b \cos \left (2 \sin ^{-1}(c x)\right )+4 a b \cos \left (4 \sin ^{-1}(c x)\right )-32 b^2 \sin \left (2 \sin ^{-1}(c x)\right )-b^2 \sin \left (4 \sin ^{-1}(c x)\right )\right )-96 a^2 d^{3/2} \sqrt{1-c^2 x^2} \tan ^{-1}\left (\frac{c x \sqrt{d-c^2 d x^2}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )+8 b d \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)^2 \left (12 a+8 b \sin \left (2 \sin ^{-1}(c x)\right )+b \sin \left (4 \sin ^{-1}(c x)\right )\right )+4 b d \sqrt{d-c^2 d x^2} \sin ^{-1}(c x) \left (4 a \left (8 \sin \left (2 \sin ^{-1}(c x)\right )+\sin \left (4 \sin ^{-1}(c x)\right )\right )+16 b \cos \left (2 \sin ^{-1}(c x)\right )+b \cos \left (4 \sin ^{-1}(c x)\right )\right )+32 b^2 d \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)^3}{256 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^2*d*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^3 - 96*a^2*d^(3/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2]
)/(Sqrt[d]*(-1 + c^2*x^2))] + 8*b*d*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^2*(12*a + 8*b*Sin[2*ArcSin[c*x]] + b*Sin[4
*ArcSin[c*x]]) + d*Sqrt[d - c^2*d*x^2]*(160*a^2*c*x*Sqrt[1 - c^2*x^2] - 64*a^2*c^3*x^3*Sqrt[1 - c^2*x^2] + 64*
a*b*Cos[2*ArcSin[c*x]] + 4*a*b*Cos[4*ArcSin[c*x]] - 32*b^2*Sin[2*ArcSin[c*x]] - b^2*Sin[4*ArcSin[c*x]]) + 4*b*
d*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]*(16*b*Cos[2*ArcSin[c*x]] + b*Cos[4*ArcSin[c*x]] + 4*a*(8*Sin[2*ArcSin[c*x]]
+ Sin[4*ArcSin[c*x]])))/(256*c*Sqrt[1 - c^2*x^2])

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Maple [B]  time = 0.243, size = 820, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(-c^2*d*x^2+d)^(3/2)*a^2+3/8*a^2*d*x*(-c^2*d*x^2+d)^(1/2)+3/8*a^2*d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)
*x/(-c^2*d*x^2+d)^(1/2))-17/64*b^2*(-d*(c^2*x^2-1))^(1/2)*d/c/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-1/8*b
^2*(-d*(c^2*x^2-1))^(1/2)*d*c^3/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^4+5/8*b^2*(-d*(c^2*x^2-1))^(1/2)*
d*c/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2-1/8*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/(c^2*x^
2-1)*arcsin(c*x)^3*d+1/32*b^2*(-d*(c^2*x^2-1))^(1/2)*d*c^4/(c^2*x^2-1)*x^5-19/64*b^2*(-d*(c^2*x^2-1))^(1/2)*d*
c^2/(c^2*x^2-1)*x^3+17/64*b^2*(-d*(c^2*x^2-1))^(1/2)*d/(c^2*x^2-1)*x-1/4*b^2*(-d*(c^2*x^2-1))^(1/2)*d*c^4/(c^2
*x^2-1)*arcsin(c*x)^2*x^5+7/8*b^2*(-d*(c^2*x^2-1))^(1/2)*d*c^2/(c^2*x^2-1)*arcsin(c*x)^2*x^3-5/8*b^2*(-d*(c^2*
x^2-1))^(1/2)*d/(c^2*x^2-1)*arcsin(c*x)^2*x-1/8*a*b*(-d*(c^2*x^2-1))^(1/2)*d*c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2
)*x^4+5/8*a*b*(-d*(c^2*x^2-1))^(1/2)*d*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-1/2*a*b*(-d*(c^2*x^2-1))^(1/2)*d*c
^4/(c^2*x^2-1)*arcsin(c*x)*x^5+7/4*a*b*(-d*(c^2*x^2-1))^(1/2)*d*c^2/(c^2*x^2-1)*arcsin(c*x)*x^3-17/64*a*b*(-d*
(c^2*x^2-1))^(1/2)*d/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-5/4*a*b*(-d*(c^2*x^2-1))^(1/2)*d/(c^2*x^2-1)*arcsin(c*x)
*x-3/8*a*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/(c^2*x^2-1)*arcsin(c*x)^2*d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (a^{2} c^{2} d x^{2} - a^{2} d +{\left (b^{2} c^{2} d x^{2} - b^{2} d\right )} \arcsin \left (c x\right )^{2} + 2 \,{\left (a b c^{2} d x^{2} - a b d\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a^2*c^2*d*x^2 - a^2*d + (b^2*c^2*d*x^2 - b^2*d)*arcsin(c*x)^2 + 2*(a*b*c^2*d*x^2 - a*b*d)*arcsin(c*
x))*sqrt(-c^2*d*x^2 + d), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- d \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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