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

Optimal. Leaf size=403 \[ \frac {3 i b c^2 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}-\frac {3 i b c^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}-\frac {6 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d^3}-\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b^2 c^2 \text {Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}+\frac {b^2 c^2}{12 d^3 \left (1-c^2 x^2\right )}-\frac {7 b^2 c^2 \log \left (1-c^2 x^2\right )}{6 d^3}+\frac {b^2 c^2 \log (x)}{d^3} \]

[Out]

1/12*b^2*c^2/d^3/(-c^2*x^2+1)-b*c*(a+b*arcsin(c*x))/d^3/x/(-c^2*x^2+1)^(3/2)+5/6*b*c^3*x*(a+b*arcsin(c*x))/d^3
/(-c^2*x^2+1)^(3/2)+3/4*c^2*(a+b*arcsin(c*x))^2/d^3/(-c^2*x^2+1)^2-1/2*(a+b*arcsin(c*x))^2/d^3/x^2/(-c^2*x^2+1
)^2+3/2*c^2*(a+b*arcsin(c*x))^2/d^3/(-c^2*x^2+1)-6*c^2*(a+b*arcsin(c*x))^2*arctanh((I*c*x+(-c^2*x^2+1)^(1/2))^
2)/d^3+b^2*c^2*ln(x)/d^3-7/6*b^2*c^2*ln(-c^2*x^2+1)/d^3+3*I*b*c^2*(a+b*arcsin(c*x))*polylog(2,-(I*c*x+(-c^2*x^
2+1)^(1/2))^2)/d^3-3*I*b*c^2*(a+b*arcsin(c*x))*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3-3/2*b^2*c^2*polylog
(3,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3+3/2*b^2*c^2*polylog(3,(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3-4/3*b*c^3*x*(a+b
*arcsin(c*x))/d^3/(-c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.78, antiderivative size = 403, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 19, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.704, Rules used = {4701, 4705, 4679, 4419, 4183, 2531, 2282, 6589, 4651, 260, 4655, 261, 271, 192, 191, 4689, 12, 1251, 893} \[ \frac {3 i b c^2 \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}-\frac {3 i b c^2 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}-\frac {3 b^2 c^2 \text {PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}-\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}-\frac {6 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d^3}+\frac {b^2 c^2}{12 d^3 \left (1-c^2 x^2\right )}-\frac {7 b^2 c^2 \log \left (1-c^2 x^2\right )}{6 d^3}+\frac {b^2 c^2 \log (x)}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*c^2)/(12*d^3*(1 - c^2*x^2)) - (b*c*(a + b*ArcSin[c*x]))/(d^3*x*(1 - c^2*x^2)^(3/2)) + (5*b*c^3*x*(a + b*A
rcSin[c*x]))/(6*d^3*(1 - c^2*x^2)^(3/2)) - (4*b*c^3*x*(a + b*ArcSin[c*x]))/(3*d^3*Sqrt[1 - c^2*x^2]) + (3*c^2*
(a + b*ArcSin[c*x])^2)/(4*d^3*(1 - c^2*x^2)^2) - (a + b*ArcSin[c*x])^2/(2*d^3*x^2*(1 - c^2*x^2)^2) + (3*c^2*(a
+ b*ArcSin[c*x])^2)/(2*d^3*(1 - c^2*x^2)) - (6*c^2*(a + b*ArcSin[c*x])^2*ArcTanh[E^((2*I)*ArcSin[c*x])])/d^3
+ (b^2*c^2*Log[x])/d^3 - (7*b^2*c^2*Log[1 - c^2*x^2])/(6*d^3) + ((3*I)*b*c^2*(a + b*ArcSin[c*x])*PolyLog[2, -E
^((2*I)*ArcSin[c*x])])/d^3 - ((3*I)*b*c^2*(a + b*ArcSin[c*x])*PolyLog[2, E^((2*I)*ArcSin[c*x])])/d^3 - (3*b^2*
c^2*PolyLog[3, -E^((2*I)*ArcSin[c*x])])/(2*d^3) + (3*b^2*c^2*PolyLog[3, E^((2*I)*ArcSin[c*x])])/(2*d^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -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]

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]

Rule 271

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

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

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 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 4183

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

Rule 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rule 4651

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a + b*
ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 4679

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

Rule 4689

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(1 - c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSin[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 - c
^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2
, 0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rule 4701

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 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] && LtQ[m, -1] && Inte
gerQ[m]

Rule 4705

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p +
1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(2*f*(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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x^3 \left (d-c^2 d x^2\right )^3} \, dx &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\left (3 c^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x \left (d-c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {a+b \sin ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {8 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}-\frac {\left (b^2 c^2\right ) \int \frac {-3+12 c^2 x^2-8 c^4 x^4}{3 x \left (1-c^2 x^2\right )^2} \, dx}{d^3}-\frac {\left (3 b c^3\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{2 d^3}+\frac {\left (3 c^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x \left (d-c^2 d x^2\right )^2} \, dx}{d}\\ &=-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {8 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}-\frac {\left (b^2 c^2\right ) \int \frac {-3+12 c^2 x^2-8 c^4 x^4}{x \left (1-c^2 x^2\right )^2} \, dx}{3 d^3}-\frac {\left (b c^3\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{d^3}-\frac {\left (3 b c^3\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{d^3}+\frac {\left (b^2 c^4\right ) \int \frac {x}{\left (1-c^2 x^2\right )^2} \, dx}{2 d^3}+\frac {\left (3 c^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x \left (d-c^2 d x^2\right )} \, dx}{d^2}\\ &=\frac {b^2 c^2}{4 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}+\frac {\left (3 c^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \csc (x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}-\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {-3+12 c^2 x-8 c^4 x^2}{x \left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^3}+\frac {\left (b^2 c^4\right ) \int \frac {x}{1-c^2 x^2} \, dx}{d^3}+\frac {\left (3 b^2 c^4\right ) \int \frac {x}{1-c^2 x^2} \, dx}{d^3}\\ &=\frac {b^2 c^2}{4 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}-\frac {2 b^2 c^2 \log \left (1-c^2 x^2\right )}{d^3}+\frac {\left (6 c^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \csc (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}-\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \left (-\frac {3}{x}+\frac {c^2}{\left (-1+c^2 x\right )^2}-\frac {5 c^2}{-1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}-\frac {6 c^2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1-c^2 x^2\right )}{6 d^3}-\frac {\left (6 b c^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}+\frac {\left (6 b c^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}-\frac {6 c^2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1-c^2 x^2\right )}{6 d^3}+\frac {3 i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d^3}-\frac {3 i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}-\frac {\left (3 i b^2 c^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}+\frac {\left (3 i b^2 c^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}-\frac {6 c^2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1-c^2 x^2\right )}{6 d^3}+\frac {3 i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d^3}-\frac {3 i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}-\frac {\left (3 b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}+\frac {\left (3 b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 x \left (1-c^2 x^2\right )^{3/2}}+\frac {5 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {4 b c^3 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 d^3 \left (1-c^2 x^2\right )}-\frac {6 c^2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1-c^2 x^2\right )}{6 d^3}+\frac {3 i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d^3}-\frac {3 i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^3}-\frac {3 b^2 c^2 \text {Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 7.91, size = 569, normalized size = 1.41 \[ -\frac {\frac {12 a^2 c^2}{c^2 x^2-1}-\frac {3 a^2 c^2}{\left (c^2 x^2-1\right )^2}+18 a^2 c^2 \log \left (1-c^2 x^2\right )-36 a^2 c^2 \log (x)+\frac {6 a^2}{x^2}+2 a b c^2 \left (\frac {14 c x}{\sqrt {1-c^2 x^2}}+\frac {c x}{\left (1-c^2 x^2\right )^{3/2}}+\frac {6 \sqrt {1-c^2 x^2}}{c x}+\frac {12 \sin ^{-1}(c x)}{c^2 x^2-1}-\frac {3 \sin ^{-1}(c x)}{\left (c^2 x^2-1\right )^2}+\frac {6 \sin ^{-1}(c x)}{c^2 x^2}-18 i \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )+18 i \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-36 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+36 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )+12 b^2 c^2 \left (\frac {1}{24} \left (\frac {2}{c^2 x^2-1}+28 \log \left (1-c^2 x^2\right )+\frac {24 \sin ^{-1}(c x)^2}{c^2 x^2-1}+\frac {12 \sin ^{-1}(c x)^2}{c^2 x^2}-\frac {6 \sin ^{-1}(c x)^2}{\left (c^2 x^2-1\right )^2}+\frac {24 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c x}+\frac {56 c x \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+\frac {4 c x \sin ^{-1}(c x)}{\left (1-c^2 x^2\right )^{3/2}}-36 \text {Li}_3\left (e^{-2 i \sin ^{-1}(c x)}\right )+36 \text {Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )-24 \log (c x)-48 i \sin ^{-1}(c x)^3-72 \sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )+72 \sin ^{-1}(c x)^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )+3 i \pi ^3\right )-3 i \sin ^{-1}(c x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c x)}\right )-3 i \sin ^{-1}(c x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )\right )}{12 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/12*((6*a^2)/x^2 - (3*a^2*c^2)/(-1 + c^2*x^2)^2 + (12*a^2*c^2)/(-1 + c^2*x^2) - 36*a^2*c^2*Log[x] + 18*a^2*c
^2*Log[1 - c^2*x^2] + 2*a*b*c^2*((c*x)/(1 - c^2*x^2)^(3/2) + (14*c*x)/Sqrt[1 - c^2*x^2] + (6*Sqrt[1 - c^2*x^2]
)/(c*x) + (6*ArcSin[c*x])/(c^2*x^2) - (3*ArcSin[c*x])/(-1 + c^2*x^2)^2 + (12*ArcSin[c*x])/(-1 + c^2*x^2) - 36*
ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] + 36*ArcSin[c*x]*Log[1 + E^((2*I)*ArcSin[c*x])] - (18*I)*PolyLog[2,
-E^((2*I)*ArcSin[c*x])] + (18*I)*PolyLog[2, E^((2*I)*ArcSin[c*x])]) + 12*b^2*c^2*((-3*I)*ArcSin[c*x]*PolyLog[
2, E^((-2*I)*ArcSin[c*x])] - (3*I)*ArcSin[c*x]*PolyLog[2, -E^((2*I)*ArcSin[c*x])] + ((3*I)*Pi^3 + 2/(-1 + c^2*
x^2) + (4*c*x*ArcSin[c*x])/(1 - c^2*x^2)^(3/2) + (56*c*x*ArcSin[c*x])/Sqrt[1 - c^2*x^2] + (24*Sqrt[1 - c^2*x^2
]*ArcSin[c*x])/(c*x) + (12*ArcSin[c*x]^2)/(c^2*x^2) - (6*ArcSin[c*x]^2)/(-1 + c^2*x^2)^2 + (24*ArcSin[c*x]^2)/
(-1 + c^2*x^2) - (48*I)*ArcSin[c*x]^3 - 72*ArcSin[c*x]^2*Log[1 - E^((-2*I)*ArcSin[c*x])] + 72*ArcSin[c*x]^2*Lo
g[1 + E^((2*I)*ArcSin[c*x])] - 24*Log[c*x] + 28*Log[1 - c^2*x^2] - 36*PolyLog[3, E^((-2*I)*ArcSin[c*x])] + 36*
PolyLog[3, -E^((2*I)*ArcSin[c*x])])/24))/d^3

________________________________________________________________________________________

fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{6} d^{3} x^{9} - 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} - d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/(c^6*d^3*x^9 - 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 - d^3*x^3
), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c^{2} d x^{2} - d\right )}^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.59, size = 1547, normalized size = 3.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/2*c^2*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)-3/2*c^4*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)^2*x^2+6*c^
2*a*b/d^3*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-6*c^2*a*b/d^3*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))
^2)+6*c^2*a*b/d^3*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-a*b/d^3/(c^4*x^4-2*c^2*x^2+1)/x^2*arcsin(c*x)-4/3
*I*c^2*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)-6*I*c^2*a*b/d^3*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+3*I*c^2*a*b/d^3*polyl
og(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-6*I*c^2*a*b/d^3*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-4/3*I*c^2*b^2/d^3/(c^
4*x^4-2*c^2*x^2+1)*arcsin(c*x)+3*I*c^2*b^2/d^3*arcsin(c*x)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-6*I*c^2*b^
2/d^3*arcsin(c*x)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))-6*I*c^2*b^2/d^3*arcsin(c*x)*polylog(2,-I*c*x-(-c^2*x^2+1
)^(1/2))-1/2*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)/x^2*arcsin(c*x)^2+9/4*c^2*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)
^2-3*c^2*b^2/d^3*arcsin(c*x)^2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+3*c^2*b^2/d^3*arcsin(c*x)^2*ln(1-I*c*x-(-c^2
*x^2+1)^(1/2))+3*c^2*b^2/d^3*arcsin(c*x)^2*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+6*c^2*b^2/d^3*polylog(3,-I*c*x-(-c^2
*x^2+1)^(1/2))+c^2*b^2/d^3*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+c^2*b^2/d^3*ln(I*c*x+(-c^2*x^2+1)^(1/2)-1)-3/2*c^2*a
^2/d^3*ln(c*x+1)-3/2*c^2*a^2/d^3*ln(c*x-1)+3*c^2*a^2/d^3*ln(c*x)+8/3*c^2*b^2/d^3*ln(I*c*x+(-c^2*x^2+1)^(1/2))-
7/3*c^2*b^2/d^3*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+6*c^2*b^2/d^3*polylog(3,I*c*x+(-c^2*x^2+1)^(1/2))+1/12*c^2*
b^2/d^3/(c^4*x^4-2*c^2*x^2+1)-9/16*c^2*a^2/d^3/(c*x-1)+1/16*c^2*a^2/d^3/(c*x+1)^2+9/16*c^2*a^2/d^3/(c*x+1)+1/1
6*c^2*a^2/d^3/(c*x-1)^2-3/2*b^2*c^2*polylog(3,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^3-c*b^2/d^3/(c^4*x^4-2*c^2*x^2+
1)/x*(-c^2*x^2+1)^(1/2)*arcsin(c*x)-c*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)/x*(-c^2*x^2+1)^(1/2)+4/3*c^5*a*b/d^3/(c^4*
x^4-2*c^2*x^2+1)*x^3*(-c^2*x^2+1)^(1/2)-3*c^4*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x^2-1/2*c^3*a*b/d^3/(c
^4*x^4-2*c^2*x^2+1)*x*(-c^2*x^2+1)^(1/2)+4/3*c^5*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*
x^3-1/2*c^3*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*x-4/3*I*c^6*a*b/d^3/(c^4*x^4-2*c^2*x^
2+1)*x^4+8/3*I*c^4*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*x^2-4/3*I*c^6*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x^4+8
/3*I*c^4*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x^2-1/2*a^2/d^3/x^2-1/12*c^4*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*
x^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, a^{2} {\left (\frac {6 \, c^{4} x^{4} - 9 \, c^{2} x^{2} + 2}{c^{4} d^{3} x^{6} - 2 \, c^{2} d^{3} x^{4} + d^{3} x^{2}} + \frac {6 \, c^{2} \log \left (c x + 1\right )}{d^{3}} + \frac {6 \, c^{2} \log \left (c x - 1\right )}{d^{3}} - \frac {12 \, c^{2} \log \relax (x)}{d^{3}}\right )} - \int \frac {b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{6} d^{3} x^{9} - 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} - d^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*((6*c^4*x^4 - 9*c^2*x^2 + 2)/(c^4*d^3*x^6 - 2*c^2*d^3*x^4 + d^3*x^2) + 6*c^2*log(c*x + 1)/d^3 + 6*c^2
*log(c*x - 1)/d^3 - 12*c^2*log(x)/d^3) - integrate((b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b*a
rctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^6*d^3*x^9 - 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 - d^3*x^3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))^2/(x^3*(d - c^2*d*x^2)^3),x)

[Out]

int((a + b*asin(c*x))^2/(x^3*(d - c^2*d*x^2)^3), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a^{2}}{c^{6} x^{9} - 3 c^{4} x^{7} + 3 c^{2} x^{5} - x^{3}}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c x \right )}}{c^{6} x^{9} - 3 c^{4} x^{7} + 3 c^{2} x^{5} - x^{3}}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c x \right )}}{c^{6} x^{9} - 3 c^{4} x^{7} + 3 c^{2} x^{5} - x^{3}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a**2/(c**6*x**9 - 3*c**4*x**7 + 3*c**2*x**5 - x**3), x) + Integral(b**2*asin(c*x)**2/(c**6*x**9 - 3
*c**4*x**7 + 3*c**2*x**5 - x**3), x) + Integral(2*a*b*asin(c*x)/(c**6*x**9 - 3*c**4*x**7 + 3*c**2*x**5 - x**3)
, x))/d**3

________________________________________________________________________________________