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

Optimal. Leaf size=429 \[ \frac{15 i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3}-\frac{15 i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3}+\frac{2 i b^2 c \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac{2 i b^2 c \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac{15 b^2 c \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}+\frac{15 b^2 c \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}-\frac{15 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3}-\frac{4 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}+\frac{b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}+\frac{11 b^2 c \tanh ^{-1}(c x)}{6 d^3} \]

[Out]

(b^2*c^2*x)/(12*d^3*(1 - c^2*x^2)) - (b*c*(a + b*ArcSin[c*x]))/(6*d^3*(1 - c^2*x^2)^(3/2)) - (7*b*c*(a + b*Arc
Sin[c*x]))/(4*d^3*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])^2/(d^3*x*(1 - c^2*x^2)^2) + (5*c^2*x*(a + b*ArcSin[
c*x])^2)/(4*d^3*(1 - c^2*x^2)^2) + (15*c^2*x*(a + b*ArcSin[c*x])^2)/(8*d^3*(1 - c^2*x^2)) - (((15*I)/4)*c*(a +
b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/d^3 - (4*b*c*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/d^3
+ (11*b^2*c*ArcTanh[c*x])/(6*d^3) + ((2*I)*b^2*c*PolyLog[2, -E^(I*ArcSin[c*x])])/d^3 + (((15*I)/4)*b*c*(a + b*
ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d^3 - (((15*I)/4)*b*c*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*A
rcSin[c*x])])/d^3 - ((2*I)*b^2*c*PolyLog[2, E^(I*ArcSin[c*x])])/d^3 - (15*b^2*c*PolyLog[3, (-I)*E^(I*ArcSin[c*
x])])/(4*d^3) + (15*b^2*c*PolyLog[3, I*E^(I*ArcSin[c*x])])/(4*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.759348, antiderivative size = 429, normalized size of antiderivative = 1., number of steps used = 27, number of rules used = 15, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {4701, 4655, 4657, 4181, 2531, 2282, 6589, 4677, 206, 199, 4705, 4709, 4183, 2279, 2391} \[ \frac{15 i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3}-\frac{15 i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3}+\frac{2 i b^2 c \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac{2 i b^2 c \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac{15 b^2 c \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}+\frac{15 b^2 c \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}-\frac{15 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3}-\frac{4 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^3}+\frac{b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}+\frac{11 b^2 c \tanh ^{-1}(c x)}{6 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*c^2*x)/(12*d^3*(1 - c^2*x^2)) - (b*c*(a + b*ArcSin[c*x]))/(6*d^3*(1 - c^2*x^2)^(3/2)) - (7*b*c*(a + b*Arc
Sin[c*x]))/(4*d^3*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])^2/(d^3*x*(1 - c^2*x^2)^2) + (5*c^2*x*(a + b*ArcSin[
c*x])^2)/(4*d^3*(1 - c^2*x^2)^2) + (15*c^2*x*(a + b*ArcSin[c*x])^2)/(8*d^3*(1 - c^2*x^2)) - (((15*I)/4)*c*(a +
b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/d^3 - (4*b*c*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/d^3
+ (11*b^2*c*ArcTanh[c*x])/(6*d^3) + ((2*I)*b^2*c*PolyLog[2, -E^(I*ArcSin[c*x])])/d^3 + (((15*I)/4)*b*c*(a + b*
ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d^3 - (((15*I)/4)*b*c*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*A
rcSin[c*x])])/d^3 - ((2*I)*b^2*c*PolyLog[2, E^(I*ArcSin[c*x])])/d^3 - (15*b^2*c*PolyLog[3, (-I)*E^(I*ArcSin[c*
x])])/(4*d^3) + (15*b^2*c*PolyLog[3, I*E^(I*ArcSin[c*x])])/(4*d^3)

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 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 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 206

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

Rule 199

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}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 4709

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

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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x^2 \left (d-c^2 d x^2\right )^3} \, dx &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\left (5 c^2\right ) \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^3} \, dx+\frac{(2 b c) \int \frac{a+b \sin ^{-1}(c x)}{x \left (1-c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=\frac{2 b c \left (a+b \sin ^{-1}(c x)\right )}{3 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{(2 b c) \int \frac{a+b \sin ^{-1}(c x)}{x \left (1-c^2 x^2\right )^{3/2}} \, dx}{d^3}-\frac{\left (2 b^2 c^2\right ) \int \frac{1}{\left (1-c^2 x^2\right )^2} \, dx}{3 d^3}-\frac{\left (5 b c^3\right ) \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{2 d^3}+\frac{\left (15 c^2\right ) \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac{b^2 c^2 x}{3 d^3 \left (1-c^2 x^2\right )}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{2 b c \left (a+b \sin ^{-1}(c x)\right )}{d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{(2 b c) \int \frac{a+b \sin ^{-1}(c x)}{x \sqrt{1-c^2 x^2}} \, dx}{d^3}-\frac{\left (b^2 c^2\right ) \int \frac{1}{1-c^2 x^2} \, dx}{3 d^3}+\frac{\left (5 b^2 c^2\right ) \int \frac{1}{\left (1-c^2 x^2\right )^2} \, dx}{6 d^3}-\frac{\left (2 b^2 c^2\right ) \int \frac{1}{1-c^2 x^2} \, dx}{d^3}-\frac{\left (15 b c^3\right ) \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{4 d^3}+\frac{\left (15 c^2\right ) \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{7 b^2 c \tanh ^{-1}(c x)}{3 d^3}+\frac{(15 c) \operatorname{Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 d^3}+\frac{(2 b c) \operatorname{Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}+\frac{\left (5 b^2 c^2\right ) \int \frac{1}{1-c^2 x^2} \, dx}{12 d^3}+\frac{\left (15 b^2 c^2\right ) \int \frac{1}{1-c^2 x^2} \, dx}{4 d^3}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{15 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac{11 b^2 c \tanh ^{-1}(c x)}{6 d^3}-\frac{(15 b c) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d^3}+\frac{(15 b c) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d^3}-\frac{\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}+\frac{\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^3}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{15 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac{11 b^2 c \tanh ^{-1}(c x)}{6 d^3}+\frac{15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}+\frac{\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac{\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac{\left (15 i b^2 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d^3}+\frac{\left (15 i b^2 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d^3}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{15 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac{11 b^2 c \tanh ^{-1}(c x)}{6 d^3}+\frac{2 i b^2 c \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac{15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{2 i b^2 c \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac{\left (15 b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{4 d^3}+\frac{\left (15 b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{4 d^3}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d^3 x \left (1-c^2 x^2\right )^2}+\frac{5 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{15 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{15 i c \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac{11 b^2 c \tanh ^{-1}(c x)}{6 d^3}+\frac{2 i b^2 c \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d^3}+\frac{15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{15 i b c \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac{2 i b^2 c \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d^3}-\frac{15 b^2 c \text{Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}+\frac{15 b^2 c \text{Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{4 d^3}\\ \end{align*}

Mathematica [B]  time = 11.5663, size = 1351, normalized size = 3.15 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(a^2/(d^3*x)) + (a^2*c^2*x)/(4*d^3*(-1 + c^2*x^2)^2) - (7*a^2*c^2*x)/(8*d^3*(-1 + c^2*x^2)) - (15*a^2*c*Log[1
- c*x])/(16*d^3) + (15*a^2*c*Log[1 + c*x])/(16*d^3) - (b^2*c*((-2*I)*PolyLog[2, -E^(I*ArcSin[c*x])] + (44*Arc
Sin[c*x] + 15*ArcSin[c*x]^3 - 45*ArcSin[c*x]^2*Log[1 - I*E^(I*ArcSin[c*x])] - 45*Pi*ArcSin[c*x]*Log[((-1)^(1/4
)*(1 - I*E^(I*ArcSin[c*x])))/(2*E^((I/2)*ArcSin[c*x]))] + 45*ArcSin[c*x]^2*Log[1 + I*E^(I*ArcSin[c*x])] + 45*A
rcSin[c*x]^2*Log[((1/2 + I/2)*(-I + E^(I*ArcSin[c*x])))/E^((I/2)*ArcSin[c*x])] - 45*Pi*ArcSin[c*x]*Log[-((-1)^
(1/4)*(-I + E^(I*ArcSin[c*x])))/(2*E^((I/2)*ArcSin[c*x]))] - 45*ArcSin[c*x]^2*Log[((1 + I) + (1 - I)*E^(I*ArcS
in[c*x]))/(2*E^((I/2)*ArcSin[c*x]))] + 45*Pi*ArcSin[c*x]*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 44*Log[Cos[ArcSin
[c*x]/2] - Sin[ArcSin[c*x]/2]] - 45*ArcSin[c*x]^2*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - 44*Log[Cos[Ar
cSin[c*x]/2] + Sin[ArcSin[c*x]/2]] + 45*ArcSin[c*x]^2*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] + 45*Pi*Arc
Sin[c*x]*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (90*I)*ArcSin[c*x]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (90*I)*Arc
Sin[c*x]*PolyLog[2, I*E^(I*ArcSin[c*x])] + 90*PolyLog[3, (-I)*E^(I*ArcSin[c*x])] - 90*PolyLog[3, I*E^(I*ArcSin
[c*x])])/24 - (4 + 88*c*x*ArcSin[c*x] - 54*ArcSin[c*x]^2 + 30*c*x*ArcSin[c*x]^3 - 240*ArcSin[c*x]^2*Cos[2*ArcS
in[c*x]] - 4*Cos[4*ArcSin[c*x]] - 90*ArcSin[c*x]^2*Cos[4*ArcSin[c*x]] + 96*c*x*ArcSin[c*x]*Log[1 - E^(I*ArcSin
[c*x])] - 96*c*x*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - (768*I)*c*x*(1 - c^2*x^2)^2*PolyLog[2, E^(I*ArcSin[c
*x])] - 200*ArcSin[c*x]*Sin[2*ArcSin[c*x]] + 132*ArcSin[c*x]*Sin[3*ArcSin[c*x]] + 45*ArcSin[c*x]^3*Sin[3*ArcSi
n[c*x]] + 144*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])]*Sin[3*ArcSin[c*x]] - 144*ArcSin[c*x]*Log[1 + E^(I*ArcSin[
c*x])]*Sin[3*ArcSin[c*x]] - 84*ArcSin[c*x]*Sin[4*ArcSin[c*x]] + 44*ArcSin[c*x]*Sin[5*ArcSin[c*x]] + 15*ArcSin[
c*x]^3*Sin[5*ArcSin[c*x]] + 48*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])]*Sin[5*ArcSin[c*x]] - 48*ArcSin[c*x]*Log[
1 + E^(I*ArcSin[c*x])]*Sin[5*ArcSin[c*x]])/(384*c*x*(1 - c^2*x^2)^2)))/d^3 - (a*b*c*(24*ArcSin[c*x]*Cot[ArcSin
[c*x]/2] - 90*ArcSin[c*x]*(Log[1 - I*E^(I*ArcSin[c*x])] - Log[1 + I*E^(I*ArcSin[c*x])]) + 48*Log[Cos[ArcSin[c*
x]/2]] - 48*Log[Sin[ArcSin[c*x]/2]] - (90*I)*(PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - PolyLog[2, I*E^(I*ArcSin[c*
x])]) - (3*ArcSin[c*x])/(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^4 - (-1 + 21*ArcSin[c*x])/(Cos[ArcSin[c*x]/2
] - Sin[ArcSin[c*x]/2])^2 + (2*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^3 + (44*Sin[ArcSi
n[c*x]/2])/(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]) + (3*ArcSin[c*x])/(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2
])^4 - (2*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^3 + (1 + 21*ArcSin[c*x])/(Cos[ArcSin[c
*x]/2] + Sin[ArcSin[c*x]/2])^2 - (44*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + 24*ArcSin
[c*x]*Tan[ArcSin[c*x]/2]))/(24*d^3)

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Maple [B]  time = 0.392, size = 1093, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*I*c*b^2/d^3*arcsin(c*x)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))+15/4*I*c*a*b/d^3*dilog(1+I*(I*c*x+(-c^2*
x^2+1)^(1/2)))-15/4*I*c*a*b/d^3*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+25/8*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsi
n(c*x)^2*c^2*x-15/8*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)^2*c^4*x^3-23/12*c*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*
arcsin(c*x)*(-c^2*x^2+1)^(1/2)-23/12*c*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*(-c^2*x^2+1)^(1/2)-15/4*c*a*b/d^3*arcsin(
c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+15/4*c*a*b/d^3*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*a*b/d^
3/(c^4*x^4-2*c^2*x^2+1)/x*arcsin(c*x)+15/4*I*c*b^2/d^3*arcsin(c*x)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+1/
12*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*c^2*x-1/12*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*c^4*x^3+2*c*a*b/d^3*ln(I*c*x+(-c^2*x
^2+1)^(1/2)-1)-b^2/d^3/(c^4*x^4-2*c^2*x^2+1)/x*arcsin(c*x)^2-2*c*a*b/d^3*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-2*c*b^
2/d^3*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-15/8*c*b^2/d^3*arcsin(c*x)^2*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)
))+15/8*c*b^2/d^3*arcsin(c*x)^2*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*I*c*b^2/d^3*dilog(I*c*x+(-c^2*x^2+1)^(1/2
))+2*I*c*b^2/d^3*dilog(1+I*c*x+(-c^2*x^2+1)^(1/2))-11/3*I*c*b^2/d^3*arctan(I*c*x+(-c^2*x^2+1)^(1/2))+1/16*c*a^
2/d^3/(c*x-1)^2-7/16*c*a^2/d^3/(c*x-1)-1/16*c*a^2/d^3/(c*x+1)^2-7/16*c*a^2/d^3/(c*x+1)-15/16*c*a^2/d^3*ln(c*x-
1)+15/16*c*a^2/d^3*ln(c*x+1)+7/4*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^3*x^2-15/4*a*b
/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c^4*x^3+7/4*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*c^3*x^2*(-c^2*x^2+1)^(1/2)+25
/4*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c^2*x-a^2/d^3/x-15/4*b^2*c*polylog(3,-I*(I*c*x+(-c^2*x^2+1)^(1/2)
))/d^3+15/4*b^2*c*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/d^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{16} \, a^{2}{\left (\frac{2 \,{\left (15 \, c^{4} x^{4} - 25 \, c^{2} x^{2} + 8\right )}}{c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x} - \frac{15 \, c \log \left (c x + 1\right )}{d^{3}} + \frac{15 \, c \log \left (c x - 1\right )}{d^{3}}\right )} + \frac{15 \,{\left (b^{2} c^{5} x^{5} - 2 \, b^{2} c^{3} x^{3} + b^{2} c x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (c x + 1\right ) - 15 \,{\left (b^{2} c^{5} x^{5} - 2 \, b^{2} c^{3} x^{3} + b^{2} c x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \,{\left (15 \, b^{2} c^{4} x^{4} - 25 \, b^{2} c^{2} x^{2} + 8 \, b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} - 2 \,{\left (c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x\right )} \int \frac{16 \, a b \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (15 \,{\left (b^{2} c^{6} x^{6} - 2 \, b^{2} c^{4} x^{4} + b^{2} c^{2} x^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) - 15 \,{\left (b^{2} c^{6} x^{6} - 2 \, b^{2} c^{4} x^{4} + b^{2} c^{2} x^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \,{\left (15 \, b^{2} c^{5} x^{5} - 25 \, b^{2} c^{3} x^{3} + 8 \, 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^{3} x^{8} - 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} - d^{3} x^{2}}\,{d x}}{16 \,{\left (c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*a^2*(2*(15*c^4*x^4 - 25*c^2*x^2 + 8)/(c^4*d^3*x^5 - 2*c^2*d^3*x^3 + d^3*x) - 15*c*log(c*x + 1)/d^3 + 15*
c*log(c*x - 1)/d^3) + 1/16*(15*(b^2*c^5*x^5 - 2*b^2*c^3*x^3 + b^2*c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x +
1))^2*log(c*x + 1) - 15*(b^2*c^5*x^5 - 2*b^2*c^3*x^3 + b^2*c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*l
og(-c*x + 1) - 2*(15*b^2*c^4*x^4 - 25*b^2*c^2*x^2 + 8*b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 16*(
c^4*d^3*x^5 - 2*c^2*d^3*x^3 + d^3*x)*integrate(-1/8*(16*a*b*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) - (15*(
b^2*c^6*x^6 - 2*b^2*c^4*x^4 + b^2*c^2*x^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 15*(b^2*c
^6*x^6 - 2*b^2*c^4*x^4 + b^2*c^2*x^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(15*b^2*c^5
*x^5 - 25*b^2*c^3*x^3 + 8*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^3*x^8 - 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 - d^3*x^2), x))/(c^4*d^3*x^5 - 2*c^2*d^3*x^3 + d^3*x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{8} - 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} - d^{3} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/x^2/(-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^8 - 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 - d^3*x^2
), x)

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

[Out]

Timed out

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

[Out]

Timed out