3.183 \(\int \frac{x^4 (a+b \sin ^{-1}(c x))^2}{d-c^2 d x^2} \, dx\)

Optimal. Leaf size=297 \[ \frac{2 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d}-\frac{2 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d}-\frac{2 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{2 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac{22 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac{2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^5 d}+\frac{2 b^2 x^3}{27 c^2 d}+\frac{22 b^2 x}{9 c^4 d} \]

[Out]

(22*b^2*x)/(9*c^4*d) + (2*b^2*x^3)/(27*c^2*d) - (22*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c^5*d) - (2*b*
x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c^3*d) - (x*(a + b*ArcSin[c*x])^2)/(c^4*d) - (x^3*(a + b*ArcSin[
c*x])^2)/(3*c^2*d) - ((2*I)*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/(c^5*d) + ((2*I)*b*(a + b*ArcSin[
c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^5*d) - ((2*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x]
)])/(c^5*d) - (2*b^2*PolyLog[3, (-I)*E^(I*ArcSin[c*x])])/(c^5*d) + (2*b^2*PolyLog[3, I*E^(I*ArcSin[c*x])])/(c^
5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.549021, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {4715, 4657, 4181, 2531, 2282, 6589, 4677, 8, 4707, 30} \[ \frac{2 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d}-\frac{2 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d}-\frac{2 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{2 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac{22 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac{2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^5 d}+\frac{2 b^2 x^3}{27 c^2 d}+\frac{22 b^2 x}{9 c^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2),x]

[Out]

(22*b^2*x)/(9*c^4*d) + (2*b^2*x^3)/(27*c^2*d) - (22*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c^5*d) - (2*b*
x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c^3*d) - (x*(a + b*ArcSin[c*x])^2)/(c^4*d) - (x^3*(a + b*ArcSin[
c*x])^2)/(3*c^2*d) - ((2*I)*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/(c^5*d) + ((2*I)*b*(a + b*ArcSin[
c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^5*d) - ((2*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x]
)])/(c^5*d) - (2*b^2*PolyLog[3, (-I)*E^(I*ArcSin[c*x])])/(c^5*d) + (2*b^2*PolyLog[3, I*E^(I*ArcSin[c*x])])/(c^
5*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 4657

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

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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^4 \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx &=-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}+\frac{\int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{c^2}+\frac{(2 b) \int \frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{3 c d}\\ &=-\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}+\frac{\int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{c^4}+\frac{(4 b) \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{9 c^3 d}+\frac{(2 b) \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{c^3 d}+\frac{\left (2 b^2\right ) \int x^2 \, dx}{9 c^2 d}\\ &=\frac{2 b^2 x^3}{27 c^2 d}-\frac{22 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}+\frac{\operatorname{Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}+\frac{\left (4 b^2\right ) \int 1 \, dx}{9 c^4 d}+\frac{\left (2 b^2\right ) \int 1 \, dx}{c^4 d}\\ &=\frac{22 b^2 x}{9 c^4 d}+\frac{2 b^2 x^3}{27 c^2 d}-\frac{22 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}\\ &=\frac{22 b^2 x}{9 c^4 d}+\frac{2 b^2 x^3}{27 c^2 d}-\frac{22 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}+\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}\\ &=\frac{22 b^2 x}{9 c^4 d}+\frac{2 b^2 x^3}{27 c^2 d}-\frac{22 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^5 d}\\ &=\frac{22 b^2 x}{9 c^4 d}+\frac{2 b^2 x^3}{27 c^2 d}-\frac{22 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^5 d}-\frac{2 b x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3 d}-\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d}-\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d}-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac{2 b^2 \text{Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac{2 b^2 \text{Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}\\ \end{align*}

Mathematica [A]  time = 0.824935, size = 508, normalized size = 1.71 \[ -\frac{-216 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+216 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+216 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )-216 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )+36 a^2 c^3 x^3+108 a^2 c x+54 a^2 \log (1-c x)-54 a^2 \log (c x+1)+24 a b c^2 x^2 \sqrt{1-c^2 x^2}+264 a b \sqrt{1-c^2 x^2}+72 a b c^3 x^3 \sin ^{-1}(c x)+216 a b c x \sin ^{-1}(c x)+108 i \pi a b \sin ^{-1}(c x)-216 a b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-108 \pi a b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+216 a b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-108 \pi a b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+108 \pi a b \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+108 \pi a b \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+270 b^2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x)-270 b^2 c x+135 b^2 c x \sin ^{-1}(c x)^2+2 b^2 \sin \left (3 \sin ^{-1}(c x)\right )-9 b^2 \sin ^{-1}(c x)^2 \sin \left (3 \sin ^{-1}(c x)\right )-108 b^2 \sin ^{-1}(c x)^2 \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+108 b^2 \sin ^{-1}(c x)^2 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-6 b^2 \sin ^{-1}(c x) \cos \left (3 \sin ^{-1}(c x)\right )}{108 c^5 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2),x]

[Out]

-(108*a^2*c*x - 270*b^2*c*x + 36*a^2*c^3*x^3 + 264*a*b*Sqrt[1 - c^2*x^2] + 24*a*b*c^2*x^2*Sqrt[1 - c^2*x^2] +
(108*I)*a*b*Pi*ArcSin[c*x] + 216*a*b*c*x*ArcSin[c*x] + 72*a*b*c^3*x^3*ArcSin[c*x] + 270*b^2*Sqrt[1 - c^2*x^2]*
ArcSin[c*x] + 135*b^2*c*x*ArcSin[c*x]^2 - 6*b^2*ArcSin[c*x]*Cos[3*ArcSin[c*x]] - 108*a*b*Pi*Log[1 - I*E^(I*Arc
Sin[c*x])] - 216*a*b*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 108*b^2*ArcSin[c*x]^2*Log[1 - I*E^(I*ArcSin[c*
x])] - 108*a*b*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 216*a*b*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 108*b^2*Ar
cSin[c*x]^2*Log[1 + I*E^(I*ArcSin[c*x])] + 54*a^2*Log[1 - c*x] - 54*a^2*Log[1 + c*x] + 108*a*b*Pi*Log[-Cos[(Pi
+ 2*ArcSin[c*x])/4]] + 108*a*b*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (216*I)*b*(a + b*ArcSin[c*x])*PolyLog[2,
(-I)*E^(I*ArcSin[c*x])] + (216*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x])] + 216*b^2*PolyLog[3,
(-I)*E^(I*ArcSin[c*x])] - 216*b^2*PolyLog[3, I*E^(I*ArcSin[c*x])] + 2*b^2*Sin[3*ArcSin[c*x]] - 9*b^2*ArcSin[c*
x]^2*Sin[3*ArcSin[c*x]])/(108*c^5*d)

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Maple [F]  time = 0.418, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{4} \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}}{-{c}^{2}d{x}^{2}+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral(-(b^2*x^4*arcsin(c*x)^2 + 2*a*b*x^4*arcsin(c*x) + a^2*x^4)/(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^{4}}{c^{2} x^{2} - 1}\, dx + \int \frac{b^{2} x^{4} \operatorname{asin}^{2}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx + \int \frac{2 a b x^{4} \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**4*(a+b*asin(c*x))**2/(-c**2*d*x**2+d),x)

[Out]

-(Integral(a**2*x**4/(c**2*x**2 - 1), x) + Integral(b**2*x**4*asin(c*x)**2/(c**2*x**2 - 1), x) + Integral(2*a*
b*x**4*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^{4}}{c^{2} d x^{2} - d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d),x, algorithm="giac")

[Out]

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