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

Optimal. Leaf size=210 \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^4 d}-\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{b x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d}+\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d}-\frac{\log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}+\frac{b^2 x^2}{4 c^2 d} \]

[Out]

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

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Rubi [A]  time = 0.37787, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {4715, 4675, 3719, 2190, 2531, 2282, 6589, 4707, 4641, 30} \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^4 d}-\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{b x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d}+\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d}-\frac{\log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}+\frac{b^2 x^2}{4 c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4715

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)/(e*(m + 2*p + 1)), x] + (Dist[(f^2*(m - 1))/(c^2*(m
+ 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(c*(m + 2*p + 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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[m,
1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

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 4707

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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [B]  time = 0.397319, size = 441, normalized size = 2.1 \[ -\frac{-48 i a b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )-48 i a b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )-24 i b^2 \sin ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )+12 b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )+12 a^2 c^2 x^2+12 a^2 \log \left (1-c^2 x^2\right )+12 a b c x \sqrt{1-c^2 x^2}+24 a b c^2 x^2 \sin ^{-1}(c x)-24 i a b \sin ^{-1}(c x)^2-12 a b \sin ^{-1}(c x)+48 i \pi a b \sin ^{-1}(c x)+48 a b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+48 a b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+96 \pi a b \log \left (1+e^{-i \sin ^{-1}(c x)}\right )+24 \pi a b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-24 \pi a b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-24 \pi a b \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-96 \pi a b \log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )+24 \pi a b \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-8 i b^2 \sin ^{-1}(c x)^3+6 b^2 \sin \left (2 \sin ^{-1}(c x)\right ) \sin ^{-1}(c x)+24 b^2 \sin ^{-1}(c x)^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )-6 b^2 \sin ^{-1}(c x)^2 \cos \left (2 \sin ^{-1}(c x)\right )+3 b^2 \cos \left (2 \sin ^{-1}(c x)\right )}{24 c^4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(12*a^2*c^2*x^2 + 12*a*b*c*x*Sqrt[1 - c^2*x^2] - 12*a*b*ArcSin[c*x] + (48*I)*a*b*Pi*ArcSin[c*x] + 24*a*b*c^2*
x^2*ArcSin[c*x] - (24*I)*a*b*ArcSin[c*x]^2 - (8*I)*b^2*ArcSin[c*x]^3 + 3*b^2*Cos[2*ArcSin[c*x]] - 6*b^2*ArcSin
[c*x]^2*Cos[2*ArcSin[c*x]] + 96*a*b*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + 24*a*b*Pi*Log[1 - I*E^(I*ArcSin[c*x])]
+ 48*a*b*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 24*a*b*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 48*a*b*ArcSin[c*x
]*Log[1 + I*E^(I*ArcSin[c*x])] + 24*b^2*ArcSin[c*x]^2*Log[1 + E^((2*I)*ArcSin[c*x])] + 12*a^2*Log[1 - c^2*x^2]
- 96*a*b*Pi*Log[Cos[ArcSin[c*x]/2]] + 24*a*b*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] - 24*a*b*Pi*Log[Sin[(Pi + 2
*ArcSin[c*x])/4]] - (48*I)*a*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - (48*I)*a*b*PolyLog[2, I*E^(I*ArcSin[c*x])]
- (24*I)*b^2*ArcSin[c*x]*PolyLog[2, -E^((2*I)*ArcSin[c*x])] + 12*b^2*PolyLog[3, -E^((2*I)*ArcSin[c*x])] + 6*b
^2*ArcSin[c*x]*Sin[2*ArcSin[c*x]])/(24*c^4*d)

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Maple [A]  time = 0.243, size = 416, normalized size = 2. \begin{align*} -{\frac{{a}^{2}{x}^{2}}{2\,{c}^{2}d}}-{\frac{{a}^{2}\ln \left ( cx-1 \right ) }{2\,d{c}^{4}}}-{\frac{{a}^{2}\ln \left ( cx+1 \right ) }{2\,d{c}^{4}}}+{\frac{{\frac{i}{3}}{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{3}}{d{c}^{4}}}-{\frac{{b}^{2}\arcsin \left ( cx \right ) x}{2\,d{c}^{3}}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}{x}^{2}}{2\,{c}^{2}d}}+{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{4\,d{c}^{4}}}+{\frac{{b}^{2}{x}^{2}}{4\,{c}^{2}d}}-{\frac{{b}^{2}}{8\,d{c}^{4}}}-{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{d{c}^{4}}\ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{iab}{d{c}^{4}}{\it polylog} \left ( 2,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{{b}^{2}}{2\,d{c}^{4}}{\it polylog} \left ( 3,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{iab \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{d{c}^{4}}}-{\frac{abx}{2\,d{c}^{3}}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{ab\arcsin \left ( cx \right ){x}^{2}}{{c}^{2}d}}+{\frac{ab\arcsin \left ( cx \right ) }{2\,d{c}^{4}}}-2\,{\frac{ab\arcsin \left ( cx \right ) \ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }{d{c}^{4}}}+{\frac{i{b}^{2}\arcsin \left ( cx \right ) }{d{c}^{4}}{\it polylog} \left ( 2,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) } \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),x)

[Out]

-1/2/c^2*a^2/d*x^2-1/2/c^4*a^2/d*ln(c*x-1)-1/2/c^4*a^2/d*ln(c*x+1)+1/3*I/c^4*b^2/d*arcsin(c*x)^3-1/2/c^3*b^2/d
*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x-1/2/c^2*b^2/d*arcsin(c*x)^2*x^2+1/4/c^4*b^2/d*arcsin(c*x)^2+1/4*b^2*x^2/c^2/
d-1/8/c^4*b^2/d-1/c^4*b^2/d*arcsin(c*x)^2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+I/c^4*a*b/d*polylog(2,-(I*c*x+(-c
^2*x^2+1)^(1/2))^2)-1/2*b^2*polylog(3,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c^4/d+I/c^4*a*b/d*arcsin(c*x)^2-1/2/c^3*a
*b/d*(-c^2*x^2+1)^(1/2)*x-1/c^2*a*b/d*arcsin(c*x)*x^2+1/2/c^4*a*b/d*arcsin(c*x)-2/c^4*a*b/d*arcsin(c*x)*ln(1+(
I*c*x+(-c^2*x^2+1)^(1/2))^2)+I/c^4*b^2/d*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),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^{2} d x^{2} - d}, 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),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^2*d*x^2 - d), x)

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

[Out]

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

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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}}{c^{2} d x^{2} - d}\,{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),x, algorithm="giac")

[Out]

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