3.256 \(\int \frac{x^3 (a+b \sin ^{-1}(c x))^2}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=332 \[ \frac{5 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{5 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{10 i b \sqrt{1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}+\frac{b^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}} \]

[Out]

b^2/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) - (b*x*(a + b*ArcSin[c*x]))/(3*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^
2]) + (x^2*(a + b*ArcSin[c*x])^2)/(3*c^2*d*(d - c^2*d*x^2)^(3/2)) - (2*(a + b*ArcSin[c*x])^2)/(3*c^4*d^2*Sqrt[
d - c^2*d*x^2]) - (((10*I)/3)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/(c^4*d^2*Sqrt
[d - c^2*d*x^2]) + (((5*I)/3)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^4*d^2*Sqrt[d - c^2*
d*x^2]) - (((5*I)/3)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^4*d^2*Sqrt[d - c^2*d*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.489147, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {4703, 4677, 4657, 4181, 2279, 2391, 261} \[ \frac{5 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{5 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{10 i b \sqrt{1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}+\frac{b^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^(5/2),x]

[Out]

b^2/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) - (b*x*(a + b*ArcSin[c*x]))/(3*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^
2]) + (x^2*(a + b*ArcSin[c*x])^2)/(3*c^2*d*(d - c^2*d*x^2)^(3/2)) - (2*(a + b*ArcSin[c*x])^2)/(3*c^4*d^2*Sqrt[
d - c^2*d*x^2]) - (((10*I)/3)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/(c^4*d^2*Sqrt
[d - c^2*d*x^2]) + (((5*I)/3)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^4*d^2*Sqrt[d - c^2*
d*x^2]) - (((5*I)/3)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^4*d^2*Sqrt[d - c^2*d*x^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 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 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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -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 )^{5/2}} \, dx &=\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \int \frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{3 c^2 d}-\frac{\left (2 b \sqrt{1-c^2 x^2}\right ) \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{a+b \sin ^{-1}(c x)}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (4 b \sqrt{1-c^2 x^2}\right ) \int \frac{a+b \sin ^{-1}(c x)}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b^2 \sqrt{1-c^2 x^2}\right ) \int \frac{x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{3 c^2 d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{b^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (4 b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{b^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{10 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (4 b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (4 b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{b^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{10 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (i b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (i b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (4 i b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (4 i b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{b^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{10 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{5 i b^2 \sqrt{1-c^2 x^2} \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{5 i b^2 \sqrt{1-c^2 x^2} \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.833875, size = 511, normalized size = 1.54 \[ \frac{-20 i b^2 \left (1-c^2 x^2\right )^{3/2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+20 i b^2 \left (1-c^2 x^2\right )^{3/2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )-12 a^2 c^2 x^2+8 a^2+15 a b \sqrt{1-c^2 x^2} \log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )-15 a b \sqrt{1-c^2 x^2} \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )+4 a b \sin ^{-1}(c x)+2 a b \sin \left (2 \sin ^{-1}(c x)\right )+12 a b \sin ^{-1}(c x) \cos \left (2 \sin ^{-1}(c x)\right )+5 a b \cos \left (3 \sin ^{-1}(c x)\right ) \log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )-5 a b \cos \left (3 \sin ^{-1}(c x)\right ) \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )-15 b^2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+15 b^2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+2 b^2 \sin ^{-1}(c x)^2+2 b^2 \sin ^{-1}(c x) \sin \left (2 \sin ^{-1}(c x)\right )-2 b^2 \cos \left (2 \sin ^{-1}(c x)\right )+6 b^2 \sin ^{-1}(c x)^2 \cos \left (2 \sin ^{-1}(c x)\right )-5 b^2 \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right ) \cos \left (3 \sin ^{-1}(c x)\right )+5 b^2 \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right ) \cos \left (3 \sin ^{-1}(c x)\right )-2 b^2}{12 c^4 d^2 \left (c^2 x^2-1\right ) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^(5/2),x]

[Out]

(8*a^2 - 2*b^2 - 12*a^2*c^2*x^2 + 4*a*b*ArcSin[c*x] + 2*b^2*ArcSin[c*x]^2 - 2*b^2*Cos[2*ArcSin[c*x]] + 12*a*b*
ArcSin[c*x]*Cos[2*ArcSin[c*x]] + 6*b^2*ArcSin[c*x]^2*Cos[2*ArcSin[c*x]] - 15*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]
*Log[1 - I*E^(I*ArcSin[c*x])] - 5*b^2*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 - I*E^(I*ArcSin[c*x])] + 15*b^2*Sqr
t[1 - c^2*x^2]*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 5*b^2*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 + I*E^(I*
ArcSin[c*x])] + 15*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 5*a*b*Cos[3*ArcSin[c*x
]]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - 15*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin
[c*x]/2]] - 5*a*b*Cos[3*ArcSin[c*x]]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] - (20*I)*b^2*(1 - c^2*x^2)^(
3/2)*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (20*I)*b^2*(1 - c^2*x^2)^(3/2)*PolyLog[2, I*E^(I*ArcSin[c*x])] + 2*a
*b*Sin[2*ArcSin[c*x]] + 2*b^2*ArcSin[c*x]*Sin[2*ArcSin[c*x]])/(12*c^4*d^2*(-1 + c^2*x^2)*Sqrt[d - c^2*d*x^2])

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Maple [B]  time = 0.283, size = 829, normalized size = 2.5 \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)^(5/2),x)

[Out]

a^2*x^2/c^2/d/(-c^2*d*x^2+d)^(3/2)-2/3*a^2/d/c^4/(-c^2*d*x^2+d)^(3/2)+b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-
1)^2/c^2*arcsin(c*x)^2*x^2-1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)
*x-1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^2*x^2-2/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c
^4*arcsin(c*x)^2+1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^4-5/3*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^
2-1))^(1/2)/c^4/d^3/(c^2*x^2-1)*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+5/3*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^
2-1))^(1/2)/c^4/d^3/(c^2*x^2-1)*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+5/3*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-
1))^(1/2)/c^4/d^3/(c^2*x^2-1)*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-5/3*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c
^2*x^2-1))^(1/2)/c^4/d^3/(c^2*x^2-1)*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*a*b*(-d*(c^2*x^2-1))^(1/
2)/d^3/(c^2*x^2-1)^2/c^2*arcsin(c*x)*x^2-1/3*a*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^3*(-c^2*x^2+1)^(1/
2)*x-4/3*a*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^4*arcsin(c*x)-5/3*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-
1))^(1/2)/c^4/d^3/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)+5/3*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)
/c^4/d^3/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)

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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)^(5/2),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{{\left (b^{2} x^{3} \arcsin \left (c x\right )^{2} + 2 \, a b x^{3} \arcsin \left (c x\right ) + a^{2} x^{3}\right )} \sqrt{-c^{2} d x^{2} + d}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, 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)^(5/2),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)*sqrt(-c^2*d*x^2 + d)/(c^6*d^3*x^6 - 3*c^4*
d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)**(5/2),x)

[Out]

Exception raised: TypeError

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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 )}^{\frac{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)^2*x^3/(-c^2*d*x^2 + d)^(5/2), x)