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

Optimal. Leaf size=227 \[ -\frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2}+\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^4 d^2}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d^2}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{\log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2} \]

[Out]

-((b*x*(a + b*ArcSin[c*x]))/(c^3*d^2*Sqrt[1 - c^2*x^2])) + (a + b*ArcSin[c*x])^2/(2*c^4*d^2) + (x^2*(a + b*Arc
Sin[c*x])^2)/(2*c^2*d^2*(1 - c^2*x^2)) - ((I/3)*(a + b*ArcSin[c*x])^3)/(b*c^4*d^2) + ((a + b*ArcSin[c*x])^2*Lo
g[1 + E^((2*I)*ArcSin[c*x])])/(c^4*d^2) - (b^2*Log[1 - c^2*x^2])/(2*c^4*d^2) - (I*b*(a + b*ArcSin[c*x])*PolyLo
g[2, -E^((2*I)*ArcSin[c*x])])/(c^4*d^2) + (b^2*PolyLog[3, -E^((2*I)*ArcSin[c*x])])/(2*c^4*d^2)

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Rubi [A]  time = 0.394962, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4703, 4675, 3719, 2190, 2531, 2282, 6589, 4641, 260} \[ -\frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2}+\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^4 d^2}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d^2}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{\log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*x*(a + b*ArcSin[c*x]))/(c^3*d^2*Sqrt[1 - c^2*x^2])) + (a + b*ArcSin[c*x])^2/(2*c^4*d^2) + (x^2*(a + b*Arc
Sin[c*x])^2)/(2*c^2*d^2*(1 - c^2*x^2)) - ((I/3)*(a + b*ArcSin[c*x])^3)/(b*c^4*d^2) + ((a + b*ArcSin[c*x])^2*Lo
g[1 + E^((2*I)*ArcSin[c*x])])/(c^4*d^2) - (b^2*Log[1 - c^2*x^2])/(2*c^4*d^2) - (I*b*(a + b*ArcSin[c*x])*PolyLo
g[2, -E^((2*I)*ArcSin[c*x])])/(c^4*d^2) + (b^2*PolyLog[3, -E^((2*I)*ArcSin[c*x])])/(2*c^4*d^2)

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 4675

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

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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]

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 )^2} \, dx &=\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{b \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{c d^2}-\frac{\int \frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{c^2 d}\\ &=-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{\operatorname{Subst}\left (\int (a+b x)^2 \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d^2}+\frac{b \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{c^3 d^2}+\frac{b^2 \int \frac{x}{1-c^2 x^2} \, dx}{c^2 d^2}\\ &=-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^2}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d^2}\\ &=-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d^2}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d^2}\\ &=-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d^2}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d^2}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d^2}\\ &=-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d^2}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 c^4 d^2}\\ &=-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d^2}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d^2}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d^2}+\frac{b^2 \text{Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^4 d^2}\\ \end{align*}

Mathematica [B]  time = 1.05886, size = 502, normalized size = 2.21 \[ \frac{-12 i a b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )-12 i a b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )-6 i b^2 \sin ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )+3 b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )-\frac{3 a^2}{c^2 x^2-1}+3 a^2 \log \left (1-c^2 x^2\right )+\frac{3 a b \sqrt{1-c^2 x^2}}{c x-1}+\frac{3 a b \sqrt{1-c^2 x^2}}{c x+1}-6 i a b \sin ^{-1}(c x)^2-\frac{3 a b \sin ^{-1}(c x)}{c x-1}+\frac{3 a b \sin ^{-1}(c x)}{c x+1}+12 i \pi a b \sin ^{-1}(c x)+12 a b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+12 a b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+24 \pi a b \log \left (1+e^{-i \sin ^{-1}(c x)}\right )+6 \pi a b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-6 \pi a b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-6 \pi a b \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-24 \pi a b \log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )+6 \pi a b \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-3 b^2 \log \left (1-c^2 x^2\right )+\frac{3 b^2 \sin ^{-1}(c x)^2}{1-c^2 x^2}-\frac{6 b^2 c x \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}}-2 i b^2 \sin ^{-1}(c x)^3+6 b^2 \sin ^{-1}(c x)^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{6 c^4 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((3*a*b*Sqrt[1 - c^2*x^2])/(-1 + c*x) + (3*a*b*Sqrt[1 - c^2*x^2])/(1 + c*x) - (3*a^2)/(-1 + c^2*x^2) + (12*I)*
a*b*Pi*ArcSin[c*x] - (3*a*b*ArcSin[c*x])/(-1 + c*x) + (3*a*b*ArcSin[c*x])/(1 + c*x) - (6*b^2*c*x*ArcSin[c*x])/
Sqrt[1 - c^2*x^2] - (6*I)*a*b*ArcSin[c*x]^2 + (3*b^2*ArcSin[c*x]^2)/(1 - c^2*x^2) - (2*I)*b^2*ArcSin[c*x]^3 +
24*a*b*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + 6*a*b*Pi*Log[1 - I*E^(I*ArcSin[c*x])] + 12*a*b*ArcSin[c*x]*Log[1 - I
*E^(I*ArcSin[c*x])] - 6*a*b*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 12*a*b*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])]
+ 6*b^2*ArcSin[c*x]^2*Log[1 + E^((2*I)*ArcSin[c*x])] + 3*a^2*Log[1 - c^2*x^2] - 3*b^2*Log[1 - c^2*x^2] - 24*a*
b*Pi*Log[Cos[ArcSin[c*x]/2]] + 6*a*b*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] - 6*a*b*Pi*Log[Sin[(Pi + 2*ArcSin[c*
x])/4]] - (12*I)*a*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - (12*I)*a*b*PolyLog[2, I*E^(I*ArcSin[c*x])] - (6*I)*b
^2*ArcSin[c*x]*PolyLog[2, -E^((2*I)*ArcSin[c*x])] + 3*b^2*PolyLog[3, -E^((2*I)*ArcSin[c*x])])/(6*c^4*d^2)

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Maple [B]  time = 0.327, size = 585, normalized size = 2.6 \begin{align*} \text{result too large to display} \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)^2,x)

[Out]

-1/4/c^4*a^2/d^2/(c*x-1)+1/2/c^4*a^2/d^2*ln(c*x-1)+1/4/c^4*a^2/d^2/(c*x+1)+1/2/c^4*a^2/d^2*ln(c*x+1)-I/c^4*a*b
/d^2*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-I/c^2*a*b/d^2/(c^2*x^2-1)*x^2+1/c^3*b^2/d^2*arcsin(c*x)/(c^2*x^2
-1)*(-c^2*x^2+1)^(1/2)*x-1/2/c^4*b^2/d^2*arcsin(c*x)^2/(c^2*x^2-1)-I/c^4*a*b/d^2*arcsin(c*x)^2+1/c^4*b^2/d^2*a
rcsin(c*x)^2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+I/c^4*b^2/d^2*arcsin(c*x)/(c^2*x^2-1)+1/2*b^2*polylog(3,-(I*c*
x+(-c^2*x^2+1)^(1/2))^2)/c^4/d^2-1/c^4*b^2/d^2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+2/c^4*b^2/d^2*ln(I*c*x+(-c^2
*x^2+1)^(1/2))-1/3*I/c^4*b^2/d^2*arcsin(c*x)^3+I/c^4*a*b/d^2/(c^2*x^2-1)+1/c^3*a*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1
)^(1/2)*x-1/c^4*a*b/d^2*arcsin(c*x)/(c^2*x^2-1)-I/c^2*b^2/d^2*arcsin(c*x)/(c^2*x^2-1)*x^2+2/c^4*a*b/d^2*arcsin
(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-I/c^4*b^2/d^2*arcsin(c*x)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} x^{3} \arcsin \left (c x\right )^{2} + 2 \, a b x^{3} \arcsin \left (c x\right ) + a^{2} x^{3}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\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)^2,x, algorithm="fricas")

[Out]

integral((b^2*x^3*arcsin(c*x)^2 + 2*a*b*x^3*arcsin(c*x) + a^2*x^3)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)

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

[Out]

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

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

[Out]

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