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

Optimal. Leaf size=172 \[ \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{b x^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d^3}+\frac{b^2}{12 c^4 d^3 \left (1-c^2 x^2\right )}+\frac{b^2 \log \left (1-c^2 x^2\right )}{3 c^4 d^3} \]

[Out]

b^2/(12*c^4*d^3*(1 - c^2*x^2)) - (b*x^3*(a + b*ArcSin[c*x]))/(6*c*d^3*(1 - c^2*x^2)^(3/2)) + (b*x*(a + b*ArcSi
n[c*x]))/(2*c^3*d^3*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])^2/(4*c^4*d^3) + (x^4*(a + b*ArcSin[c*x])^2)/(4*d^
3*(1 - c^2*x^2)^2) + (b^2*Log[1 - c^2*x^2])/(3*c^4*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.33432, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4681, 4703, 4641, 260, 266, 43} \[ \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{b x^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d^3}+\frac{b^2}{12 c^4 d^3 \left (1-c^2 x^2\right )}+\frac{b^2 \log \left (1-c^2 x^2\right )}{3 c^4 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^2/(12*c^4*d^3*(1 - c^2*x^2)) - (b*x^3*(a + b*ArcSin[c*x]))/(6*c*d^3*(1 - c^2*x^2)^(3/2)) + (b*x*(a + b*ArcSi
n[c*x]))/(2*c^3*d^3*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])^2/(4*c^4*d^3) + (x^4*(a + b*ArcSin[c*x])^2)/(4*d^
3*(1 - c^2*x^2)^2) + (b^2*Log[1 - c^2*x^2])/(3*c^4*d^3)

Rule 4681

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

Rule 4703

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

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac{x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{(b c) \int \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{2 d^3}\\ &=-\frac{b x^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{b^2 \int \frac{x^3}{\left (1-c^2 x^2\right )^2} \, dx}{6 d^3}+\frac{b \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 c d^3}\\ &=-\frac{b x^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d^3 \sqrt{1-c^2 x^2}}+\frac{x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{x}{\left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{12 d^3}-\frac{b \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{2 c^3 d^3}-\frac{b^2 \int \frac{x}{1-c^2 x^2} \, dx}{2 c^2 d^3}\\ &=-\frac{b x^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d^3}+\frac{x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{b^2 \log \left (1-c^2 x^2\right )}{4 c^4 d^3}+\frac{b^2 \operatorname{Subst}\left (\int \left (\frac{1}{c^2 \left (-1+c^2 x\right )^2}+\frac{1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{12 d^3}\\ &=\frac{b^2}{12 c^4 d^3 \left (1-c^2 x^2\right )}-\frac{b x^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d^3}+\frac{x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{b^2 \log \left (1-c^2 x^2\right )}{3 c^4 d^3}\\ \end{align*}

Mathematica [A]  time = 0.186361, size = 192, normalized size = 1.12 \[ \frac{6 a^2 c^2 x^2-3 a^2-8 a b c^3 x^3 \sqrt{1-c^2 x^2}+6 a b c x \sqrt{1-c^2 x^2}+2 b \sin ^{-1}(c x) \left (a \left (6 c^2 x^2-3\right )+b c x \sqrt{1-c^2 x^2} \left (3-4 c^2 x^2\right )\right )-b^2 c^2 x^2+4 b^2 \left (c^2 x^2-1\right )^2 \log \left (1-c^2 x^2\right )+3 b^2 \left (2 c^2 x^2-1\right ) \sin ^{-1}(c x)^2+b^2}{12 c^4 d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2 + b^2 + 6*a^2*c^2*x^2 - b^2*c^2*x^2 + 6*a*b*c*x*Sqrt[1 - c^2*x^2] - 8*a*b*c^3*x^3*Sqrt[1 - c^2*x^2] +
2*b*(b*c*x*(3 - 4*c^2*x^2)*Sqrt[1 - c^2*x^2] + a*(-3 + 6*c^2*x^2))*ArcSin[c*x] + 3*b^2*(-1 + 2*c^2*x^2)*ArcSin
[c*x]^2 + 4*b^2*(-1 + c^2*x^2)^2*Log[1 - c^2*x^2])/(12*c^4*d^3*(-1 + c^2*x^2)^2)

________________________________________________________________________________________

Maple [B]  time = 0.353, size = 472, normalized size = 2.7 \begin{align*}{\frac{{a}^{2}}{16\,{c}^{4}{d}^{3} \left ( cx-1 \right ) ^{2}}}+{\frac{3\,{a}^{2}}{16\,{c}^{4}{d}^{3} \left ( cx-1 \right ) }}+{\frac{{a}^{2}}{16\,{c}^{4}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{3\,{a}^{2}}{16\,{c}^{4}{d}^{3} \left ( cx+1 \right ) }}+{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{4\,{c}^{4}{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) ^{2}}}-{\frac{{b}^{2}\arcsin \left ( cx \right ) x}{6\,{c}^{3}{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) ^{2}}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{{b}^{2}}{12\,{c}^{4}{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{2\,{b}^{2}\arcsin \left ( cx \right ) x}{3\,{c}^{3}{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{{b}^{2}\ln \left ( -{c}^{2}{x}^{2}+1 \right ) }{3\,{c}^{4}{d}^{3}}}+{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{2\,{c}^{4}{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{ab\arcsin \left ( cx \right ) }{8\,{c}^{4}{d}^{3} \left ( cx-1 \right ) ^{2}}}+{\frac{3\,ab\arcsin \left ( cx \right ) }{8\,{c}^{4}{d}^{3} \left ( cx-1 \right ) }}+{\frac{ab\arcsin \left ( cx \right ) }{8\,{c}^{4}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{3\,ab\arcsin \left ( cx \right ) }{8\,{c}^{4}{d}^{3} \left ( cx+1 \right ) }}-{\frac{ab}{3\,{c}^{4}{d}^{3} \left ( cx-1 \right ) }\sqrt{- \left ( cx-1 \right ) ^{2}-2\,cx+2}}-{\frac{ab}{3\,{c}^{4}{d}^{3} \left ( cx+1 \right ) }\sqrt{- \left ( cx+1 \right ) ^{2}+2\,cx+2}}-{\frac{ab}{24\,{c}^{4}{d}^{3} \left ( cx-1 \right ) ^{2}}\sqrt{- \left ( cx-1 \right ) ^{2}-2\,cx+2}}+{\frac{ab}{24\,{c}^{4}{d}^{3} \left ( cx+1 \right ) ^{2}}\sqrt{- \left ( cx+1 \right ) ^{2}+2\,cx+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/c^4*a^2/d^3/(c*x-1)^2+3/16/c^4*a^2/d^3/(c*x-1)+1/16/c^4*a^2/d^3/(c*x+1)^2-3/16/c^4*a^2/d^3/(c*x+1)+1/4/c^
4*b^2/d^3*arcsin(c*x)^2/(c^2*x^2-1)^2-1/6/c^3*b^2/d^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)^2*x-1/12/c^4*
b^2/d^3/(c^2*x^2-1)-2/3/c^3*b^2/d^3*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)*x+1/3*b^2*ln(-c^2*x^2+1)/c^4/d^
3+1/2/c^4*b^2/d^3*arcsin(c*x)^2/(c^2*x^2-1)+1/8/c^4*a*b/d^3*arcsin(c*x)/(c*x-1)^2+3/8/c^4*a*b/d^3*arcsin(c*x)/
(c*x-1)+1/8/c^4*a*b/d^3*arcsin(c*x)/(c*x+1)^2-3/8/c^4*a*b/d^3*arcsin(c*x)/(c*x+1)-1/3/c^4*a*b/d^3/(c*x-1)*(-(c
*x-1)^2-2*c*x+2)^(1/2)-1/3/c^4*a*b/d^3/(c*x+1)*(-(c*x+1)^2+2*c*x+2)^(1/2)-1/24/c^4*a*b/d^3/(c*x-1)^2*(-(c*x-1)
^2-2*c*x+2)^(1/2)+1/24/c^4*a*b/d^3/(c*x+1)^2*(-(c*x+1)^2+2*c*x+2)^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (2 \, c^{2} x^{2} - 1\right )} a^{2}}{4 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} + \frac{{\left (2 \, b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} - 2 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )} \int \frac{4 \, a b c^{3} x^{3} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (2 \, b^{2} c^{2} x^{2} - b^{2}\right )} \sqrt{c x + 1} \sqrt{-c x + 1} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )}{c^{9} d^{3} x^{6} - 3 \, c^{7} d^{3} x^{4} + 3 \, c^{5} d^{3} x^{2} - c^{3} d^{3}}\,{d x}}{4 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*c^2*x^2 - 1)*a^2/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3) + 1/4*((2*b^2*c^2*x^2 - b^2)*arctan2(c*x, sqrt
(c*x + 1)*sqrt(-c*x + 1))^2 + 4*(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3)*integrate(-1/2*(4*a*b*c^3*x^3*arctan2(
c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) - (2*b^2*c^2*x^2 - b^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x
+ 1)*sqrt(-c*x + 1)))/(c^9*d^3*x^6 - 3*c^7*d^3*x^4 + 3*c^5*d^3*x^2 - c^3*d^3), x))/(c^8*d^3*x^4 - 2*c^6*d^3*x^
2 + c^4*d^3)

________________________________________________________________________________________

Fricas [A]  time = 2.65076, size = 421, normalized size = 2.45 \begin{align*} \frac{{\left (6 \, a^{2} - b^{2}\right )} c^{2} x^{2} + 3 \,{\left (2 \, b^{2} c^{2} x^{2} - b^{2}\right )} \arcsin \left (c x\right )^{2} - 3 \, a^{2} + b^{2} + 6 \,{\left (2 \, a b c^{2} x^{2} - a b\right )} \arcsin \left (c x\right ) + 4 \,{\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c^{2} x^{2} - 1\right ) - 2 \,{\left (4 \, a b c^{3} x^{3} - 3 \, a b c x +{\left (4 \, b^{2} c^{3} x^{3} - 3 \, b^{2} c x\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} x^{2} + 1}}{12 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((6*a^2 - b^2)*c^2*x^2 + 3*(2*b^2*c^2*x^2 - b^2)*arcsin(c*x)^2 - 3*a^2 + b^2 + 6*(2*a*b*c^2*x^2 - a*b)*ar
csin(c*x) + 4*(b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*log(c^2*x^2 - 1) - 2*(4*a*b*c^3*x^3 - 3*a*b*c*x + (4*b^2*c^3
*x^3 - 3*b^2*c*x)*arcsin(c*x))*sqrt(-c^2*x^2 + 1))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a^{2} x^{3}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac{b^{2} x^{3} \operatorname{asin}^{2}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac{2 a b x^{3} \operatorname{asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a**2*x**3/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(b**2*x**3*asin(c*x)**2/(c**6*x
**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(2*a*b*x**3*asin(c*x)/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**
2 - 1), x))/d**3

________________________________________________________________________________________

Giac [B]  time = 2.13606, size = 429, normalized size = 2.49 \begin{align*} \frac{b^{2} x^{4} \arcsin \left (c x\right )^{2}}{4 \,{\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac{a b x^{4} \arcsin \left (c x\right )}{2 \,{\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac{a^{2} x^{4}}{4 \,{\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac{b^{2} x^{3} \arcsin \left (c x\right )}{6 \,{\left (c^{2} x^{2} - 1\right )} \sqrt{-c^{2} x^{2} + 1} c d^{3}} + \frac{a b x^{3}}{6 \,{\left (c^{2} x^{2} - 1\right )} \sqrt{-c^{2} x^{2} + 1} c d^{3}} - \frac{b^{2} x^{2}}{12 \,{\left (c^{2} x^{2} - 1\right )} c^{2} d^{3}} + \frac{b^{2} x \arcsin \left (c x\right )}{2 \, \sqrt{-c^{2} x^{2} + 1} c^{3} d^{3}} - \frac{b^{2} \arcsin \left (c x\right )^{2}}{4 \, c^{4} d^{3}} + \frac{a b x}{2 \, \sqrt{-c^{2} x^{2} + 1} c^{3} d^{3}} - \frac{a b \arcsin \left (c x\right )}{2 \, c^{4} d^{3}} + \frac{2 \, b^{2} \log \left (2\right )}{3 \, c^{4} d^{3}} + \frac{b^{2} \log \left ({\left | -c^{2} x^{2} + 1 \right |}\right )}{3 \, c^{4} d^{3}} - \frac{a^{2}}{4 \, c^{4} d^{3}} + \frac{b^{2}}{12 \, c^{4} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4*arcsin(c*x)^2/((c^2*x^2 - 1)^2*d^3) + 1/2*a*b*x^4*arcsin(c*x)/((c^2*x^2 - 1)^2*d^3) + 1/4*a^2*x^4/
((c^2*x^2 - 1)^2*d^3) + 1/6*b^2*x^3*arcsin(c*x)/((c^2*x^2 - 1)*sqrt(-c^2*x^2 + 1)*c*d^3) + 1/6*a*b*x^3/((c^2*x
^2 - 1)*sqrt(-c^2*x^2 + 1)*c*d^3) - 1/12*b^2*x^2/((c^2*x^2 - 1)*c^2*d^3) + 1/2*b^2*x*arcsin(c*x)/(sqrt(-c^2*x^
2 + 1)*c^3*d^3) - 1/4*b^2*arcsin(c*x)^2/(c^4*d^3) + 1/2*a*b*x/(sqrt(-c^2*x^2 + 1)*c^3*d^3) - 1/2*a*b*arcsin(c*
x)/(c^4*d^3) + 2/3*b^2*log(2)/(c^4*d^3) + 1/3*b^2*log(abs(-c^2*x^2 + 1))/(c^4*d^3) - 1/4*a^2/(c^4*d^3) + 1/12*
b^2/(c^4*d^3)