3.255 \(\int \frac {x^4 (a+b \sin ^{-1}(c x))^2}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=421 \[ \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {8 b \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {4 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {b^2 x}{3 c^4 d^2 \sqrt {d-c^2 d x^2}} \]

[Out]

1/3*x^3*(a+b*arcsin(c*x))^2/c^2/d/(-c^2*d*x^2+d)^(3/2)+1/3*b^2*x/c^4/d^2/(-c^2*d*x^2+d)^(1/2)-x*(a+b*arcsin(c*
x))^2/c^4/d^2/(-c^2*d*x^2+d)^(1/2)-1/3*b*x^2*(a+b*arcsin(c*x))/c^3/d^2/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)
-1/3*b^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/2)+4/3*I*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1
/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/2)+1/3*(a+b*arcsin(c*x))^3*(-c^2*x^2+1)^(1/2)/b/c^5/d^2/(-c^2*d*x^2+d)^(1/2)-8/3
*b*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/2)+4/3*I*
b^2*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.73, antiderivative size = 421, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {4703, 4643, 4641, 4675, 3719, 2190, 2279, 2391, 288, 216} \[ \frac {4 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {8 b \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {b^2 x}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{3 c^5 d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^(5/2),x]

[Out]

(b^2*x)/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) - (b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(3*c^5*d^2*Sqrt[d - c^2*d*x^2])
- (b*x^2*(a + b*ArcSin[c*x]))/(3*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcSin[c*x])^2)/
(3*c^2*d*(d - c^2*d*x^2)^(3/2)) - (x*(a + b*ArcSin[c*x])^2)/(c^4*d^2*Sqrt[d - c^2*d*x^2]) + (((4*I)/3)*Sqrt[1
- c^2*x^2]*(a + b*ArcSin[c*x])^2)/(c^5*d^2*Sqrt[d - c^2*d*x^2]) + (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(3
*b*c^5*d^2*Sqrt[d - c^2*d*x^2]) - (8*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(
3*c^5*d^2*Sqrt[d - c^2*d*x^2]) + (((4*I)/3)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/(c^5*d^2
*Sqrt[d - c^2*d*x^2])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 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 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 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 4643

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

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 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]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {\int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{c^2 d}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {x^3 \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^2 \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^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d-c^2 d x^2}} \, dx}{c^4 d^2}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{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^2}{\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 x}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x^2 \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^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}} \, dx}{c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b^2 x}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x^2 \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^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (4 i b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (4 i b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b^2 x}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x^2 \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^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {8 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b^2 x}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x^2 \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^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {8 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{3 c^5 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+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{3 c^5 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+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{c^5 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b^2 x}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x^2 \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^3 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {8 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.63, size = 374, normalized size = 0.89 \[ \frac {a^2 c \sqrt {d} x \left (4 c^2 x^2-3\right )+3 a^2 \left (c^2 x^2-1\right ) \sqrt {d-c^2 d x^2} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )-a b \sqrt {d} \left (\sqrt {1-c^2 x^2}+\left (1-c^2 x^2\right )^{3/2} \left (4 \log \left (1-c^2 x^2\right )-3 \sin ^{-1}(c x)^2\right )+2 \sin ^{-1}(c x) \sin \left (3 \sin ^{-1}(c x)\right )\right )+b^2 \sqrt {d} \left (-c^3 x^3+4 c^3 x^3 \sin ^{-1}(c x)^2+4 i \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )+\left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x)^3+4 i \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x)^2-\sqrt {1-c^2 x^2} \sin ^{-1}(c x)-8 \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )+c x-3 c x \sin ^{-1}(c x)^2\right )}{3 c^5 d^{5/2} \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^(5/2),x]

[Out]

(a^2*c*Sqrt[d]*x*(-3 + 4*c^2*x^2) + 3*a^2*(-1 + c^2*x^2)*Sqrt[d - c^2*d*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/
(Sqrt[d]*(-1 + c^2*x^2))] + b^2*Sqrt[d]*(c*x - c^3*x^3 - Sqrt[1 - c^2*x^2]*ArcSin[c*x] - 3*c*x*ArcSin[c*x]^2 +
4*c^3*x^3*ArcSin[c*x]^2 + (4*I)*(1 - c^2*x^2)^(3/2)*ArcSin[c*x]^2 + (1 - c^2*x^2)^(3/2)*ArcSin[c*x]^3 - 8*(1
- c^2*x^2)^(3/2)*ArcSin[c*x]*Log[1 + E^((2*I)*ArcSin[c*x])] + (4*I)*(1 - c^2*x^2)^(3/2)*PolyLog[2, -E^((2*I)*A
rcSin[c*x])]) - a*b*Sqrt[d]*(Sqrt[1 - c^2*x^2] + (1 - c^2*x^2)^(3/2)*(-3*ArcSin[c*x]^2 + 4*Log[1 - c^2*x^2]) +
2*ArcSin[c*x]*Sin[3*ArcSin[c*x]]))/(3*c^5*d^(5/2)*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2])

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b^{2} x^{4} \arcsin \left (c x\right )^{2} + 2 \, a b x^{4} \arcsin \left (c x\right ) + a^{2} x^{4}\right )} \sqrt {-c^{2} d x^{2} + d}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \]

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)^(5/2),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)*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{4}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)^2*x^4/(-c^2*d*x^2 + d)^(5/2), x)

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maple [B]  time = 1.00, size = 1304, normalized size = 3.10 \[ \text {result too large to display} \]

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)^(5/2),x)

[Out]

1/3*a^2*x^3/c^2/d/(-c^2*d*x^2+d)^(3/2)-a^2/c^4/d^2*x/(-c^2*d*x^2+d)^(1/2)+a^2/c^4/d^2/(c^2*d)^(1/2)*arctan((c^
2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^5*arcsin(c*x)*(-
c^2*x^2+1)^(1/2)+8/3*I*a*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)
*x^2-8/3*I*a*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^5*arcsin(c*x)*(-c^2*x^2+1)^(1/2)+4/3*I*b^2*(
-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^3*arcsin(c*x)^2*(-c^2*x^2+1)^(1/2)*x^2-4/3*I*b^2*(-c^2*x^2+1
)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^5/d^3/(c^2*x^2-1)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)+4/3*b^2*(-d*(c^2*x
^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^2*arcsin(c*x)^2*x^3-b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2
+1)/c^4*arcsin(c*x)^2*x-1/3*I*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^3*x^2*(-c^2*x^2+1)^(1/2)+
1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^4*x-1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c
^2*x^2+1)/c^2*x^3-4/3*I*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^5*arcsin(c*x)^2*(-c^2*x^2+1)^(1
/2)+8/3*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^5/d^3/(c^2*x^2-1)*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)
^(1/2))^2)-1/3*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^3/(c^2*x^2-1)*arcsin(c*x)^3-a*b*(-d*(c^2*x^
2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^3/(c^2*x^2-1)*arcsin(c*x)^2+1/3*I*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4
-2*c^2*x^2+1)/c^5*(-c^2*x^2+1)^(1/2)-1/3*a*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^5*(-c^2*x^2+1)
^(1/2)-16/3*I*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^5/d^3/(c^2*x^2-1)*arcsin(c*x)-2*a*b*(-d*(c^2*x^2
-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^4*arcsin(c*x)*x+8/3*a*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d
^3/(c^2*x^2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-8/3*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^5/d^3/
(c^2*x^2-1)*arcsin(c*x)^2+8/3*a*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^2*arcsin(c*x)*x^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (x {\left (\frac {3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} - \frac {2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} - \frac {x}{\sqrt {-c^{2} d x^{2} + d} c^{4} d^{2}} + \frac {3 \, \arcsin \left (c x\right )}{c^{5} d^{\frac {5}{2}}}\right )} a^{2} - \sqrt {d} \int \frac {{\left (b^{2} x^{4} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, a b x^{4} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}\,{d x} \]

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)^(5/2),x, algorithm="maxima")

[Out]

1/3*(x*(3*x^2/((-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d)) - x/(sqrt(-c^2*d*x^2 + d)*c^4
*d^2) + 3*arcsin(c*x)/(c^5*d^(5/2)))*a^2 - sqrt(d)*integrate((b^2*x^4*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1
))^2 + 2*a*b*x^4*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*asin(c*x))^2)/(d - c^2*d*x^2)^(5/2),x)

[Out]

int((x^4*(a + b*asin(c*x))^2)/(d - c^2*d*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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)**(5/2),x)

[Out]

Integral(x**4*(a + b*asin(c*x))**2/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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