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

Optimal. Leaf size=270 \[ \frac{2 i b c^2 \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{2 i b c^2 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{b^2 c^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\frac{b^2 c^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d^2 \left (1-c^2 x^2\right )}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 x \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac{4 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d^2}-\frac{b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d^2}+\frac{b^2 c^2 \log (x)}{d^2} \]

[Out]

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

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Rubi [A]  time = 0.550665, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 15, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {4701, 4705, 4679, 4419, 4183, 2531, 2282, 6589, 4651, 260, 271, 191, 4689, 446, 72} \[ \frac{2 i b c^2 \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{2 i b c^2 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac{b^2 c^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\frac{b^2 c^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\frac{c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d^2 \left (1-c^2 x^2\right )}-\frac{b c \left (a+b \sin ^{-1}(c x)\right )}{d^2 x \sqrt{1-c^2 x^2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac{4 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d^2}-\frac{b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d^2}+\frac{b^2 c^2 \log (x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 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 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 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 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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

Mathematica [A]  time = 1.55839, size = 430, normalized size = 1.59 \[ \frac{2 a b \left (2 c^2 \left (i \left (\text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )-\text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )\right )+2 \sin ^{-1}(c x) \left (\log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )\right )-\frac{c^3 x}{\sqrt{1-c^2 x^2}}-\frac{c \sqrt{1-c^2 x^2}}{x}+\frac{c^2 \sin ^{-1}(c x)}{1-c^2 x^2}-\frac{\sin ^{-1}(c x)}{x^2}\right )+b^2 c^2 \left (4 i \sin ^{-1}(c x) \left (\text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )-\text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )\right )+2 \left (\text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )-\text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )\right )+2 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+\frac{\sin ^{-1}(c x)^2}{1-c^2 x^2}-\frac{\sin ^{-1}(c x)^2}{c^2 x^2}-\frac{2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{c x}-\frac{2 c x \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}}+4 \sin ^{-1}(c x)^2 \left (\log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )\right )+\frac{a^2 c^2}{1-c^2 x^2}-2 a^2 c^2 \log \left (1-c^2 x^2\right )+4 a^2 c^2 \log (x)-\frac{a^2}{x^2}}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.246, size = 903, normalized size = 3.3 \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^3/(-c^2*d*x^2+d)^2,x)

[Out]

-1/2*a^2/d^2/x^2-4*I*c^2*a*b/d^2*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+2*I*c^2*a*b/d^2*polylog(2,-(I*c*x+(-c^2*x
^2+1)^(1/2))^2)-4*I*c^2*b^2/d^2*arcsin(c*x)*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-4*I*c^2*b^2/d^2*arcsin(c*x)*p
olylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+2*I*c^2*b^2/d^2*arcsin(c*x)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-4*I*c^
2*a*b/d^2*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))+4*c^2*a*b/d^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-2*c^2*
a*b/d^2*arcsin(c*x)/(c^2*x^2-1)-4*c^2*a*b/d^2*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+4*c^2*a*b/d^2*arc
sin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+a*b/d^2/(c^2*x^2-1)/x^2*arcsin(c*x)-1/4*c^2*a^2/d^2/(c*x-1)+4*c^2*b^2/
d^2*polylog(3,I*c*x+(-c^2*x^2+1)^(1/2))+2*c^2*a^2/d^2*ln(c*x)+c^2*b^2/d^2*ln(I*c*x+(-c^2*x^2+1)^(1/2)-1)+c^2*b
^2/d^2*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-c^2*b^2/d^2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-c^2*a^2/d^2*ln(c*x-1)-c^2
*a^2/d^2*ln(c*x+1)+4*c^2*b^2/d^2*polylog(3,-I*c*x-(-c^2*x^2+1)^(1/2))+1/4*c^2*a^2/d^2/(c*x+1)+1/2*b^2/d^2/(c^2
*x^2-1)/x^2*arcsin(c*x)^2-b^2*c^2*polylog(3,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d^2+2*c^2*b^2/d^2*arcsin(c*x)^2*ln(
1+I*c*x+(-c^2*x^2+1)^(1/2))-c^2*b^2/d^2*arcsin(c*x)^2/(c^2*x^2-1)-2*c^2*b^2/d^2*arcsin(c*x)^2*ln(1+(I*c*x+(-c^
2*x^2+1)^(1/2))^2)+2*c^2*b^2/d^2*arcsin(c*x)^2*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+c*b^2/d^2/(c^2*x^2-1)/x*arcsin(c
*x)*(-c^2*x^2+1)^(1/2)+c*a*b/d^2/x/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a^{2}{\left (\frac{2 \, c^{2} \log \left (c x + 1\right )}{d^{2}} + \frac{2 \, c^{2} \log \left (c x - 1\right )}{d^{2}} - \frac{4 \, c^{2} \log \left (x\right )}{d^{2}} + \frac{2 \, c^{2} x^{2} - 1}{c^{2} d^{2} x^{4} - d^{2} x^{2}}\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^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}\,{d x} \end{align*}

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)^2,x, algorithm="maxima")

[Out]

-1/2*a^2*(2*c^2*log(c*x + 1)/d^2 + 2*c^2*log(c*x - 1)/d^2 - 4*c^2*log(x)/d^2 + (2*c^2*x^2 - 1)/(c^2*d^2*x^4 -
d^2*x^2)) + integrate((b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b*arctan2(c*x, sqrt(c*x + 1)*sqr
t(-c*x + 1)))/(c^4*d^2*x^7 - 2*c^2*d^2*x^5 + d^2*x^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^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}, x\right ) \end{align*}

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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{4} x^{7} - 2 c^{2} x^{5} + x^{3}}\, dx + \int \frac{b^{2} \operatorname{asin}^{2}{\left (c x \right )}}{c^{4} x^{7} - 2 c^{2} x^{5} + x^{3}}\, dx + \int \frac{2 a b \operatorname{asin}{\left (c x \right )}}{c^{4} x^{7} - 2 c^{2} x^{5} + x^{3}}\, dx}{d^{2}} \end{align*}

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)**2,x)

[Out]

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

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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}}{{\left (c^{2} d x^{2} - d\right )}^{2} x^{3}}\,{d x} \end{align*}

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)^2,x, algorithm="giac")

[Out]

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