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

Optimal. Leaf size=287 \[ 2 i b c^2 d^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {2 i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-\frac {1}{4} c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2-2 c^2 d^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{2} b c^3 d^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{4} b^2 c^4 d^2 x^2-b^2 c^2 d^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d^2 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4695, 4699, 4625, 3717, 2190, 2531, 2282, 6589, 4647, 4641, 30, 14} \[ 2 i b c^2 d^2 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-b^2 c^2 d^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{2} b c^3 d^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {2 i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-\frac {1}{4} c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2-2 c^2 d^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{4} b^2 c^4 d^2 x^2+b^2 c^2 d^2 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 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 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(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]

Rule 4695

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*(a + b*ArcSin[c*x])^n)/(f*(m + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^
(m + 2)*(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])/
(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}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rule 4699

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*(a + b*ArcSin[c*x])^n)/(f*(m + 2*p + 1)), x] + (Dist[(2*d*p)/(m + 2*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)^FracPart[p])/(
f*(m + 2*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] && GtQ[p, 0] &&  !LtQ[m, -1]
&& (RationalQ[m] || 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 (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{x^3} \, dx &=-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}-\left (2 c^2 d\right ) \int \frac {\left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{x} \, dx+\left (b c d^2\right ) \int \frac {\left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x^2} \, dx\\ &=-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}-\left (2 c^2 d^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x} \, dx+\left (b^2 c^2 d^2\right ) \int \frac {1-c^2 x^2}{x} \, dx+\left (2 b c^3 d^2\right ) \int \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx-\left (3 b c^3 d^2\right ) \int \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx\\ &=-\frac {1}{2} b c^3 d^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}-\left (2 c^2 d^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c x)\right )+\left (b^2 c^2 d^2\right ) \int \left (\frac {1}{x}-c^2 x\right ) \, dx+\left (b c^3 d^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx-\frac {1}{2} \left (3 b c^3 d^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx-\left (b^2 c^4 d^2\right ) \int x \, dx+\frac {1}{2} \left (3 b^2 c^4 d^2\right ) \int x \, dx\\ &=-\frac {1}{4} b^2 c^4 d^2 x^2-\frac {1}{2} b c^3 d^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{4} c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {2 i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+b^2 c^2 d^2 \log (x)+\left (4 i c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {1}{4} b^2 c^4 d^2 x^2-\frac {1}{2} b c^3 d^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{4} c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {2 i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d^2 \log (x)+\left (4 b c^2 d^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {1}{4} b^2 c^4 d^2 x^2-\frac {1}{2} b c^3 d^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{4} c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {2 i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d^2 \log (x)+2 i b c^2 d^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-\left (2 i b^2 c^2 d^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {1}{4} b^2 c^4 d^2 x^2-\frac {1}{2} b c^3 d^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{4} c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {2 i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d^2 \log (x)+2 i b c^2 d^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-\left (b^2 c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=-\frac {1}{4} b^2 c^4 d^2 x^2-\frac {1}{2} b c^3 d^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{4} c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {2 i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d^2 \log (x)+2 i b c^2 d^2 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-b^2 c^2 d^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.86, size = 343, normalized size = 1.20 \[ \frac {1}{2} d^2 \left (a^2 c^4 x^2-4 a^2 c^2 \log (x)-\frac {a^2}{x^2}+2 a b c^4 x^2 \sin ^{-1}(c x)+4 i a b c^2 \left (\sin ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\right )+a b c^2 \left (c x \sqrt {1-c^2 x^2}-\sin ^{-1}(c x)\right )-\frac {2 a b \left (c x \sqrt {1-c^2 x^2}+\sin ^{-1}(c x)\right )}{x^2}-8 a b c^2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+\frac {1}{6} i b^2 c^2 \left (-24 \sin ^{-1}(c x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c x)}\right )+12 i \text {Li}_3\left (e^{-2 i \sin ^{-1}(c x)}\right )-8 \sin ^{-1}(c x)^3+24 i \sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )+\pi ^3\right )-\frac {b^2 \left (-2 c^2 x^2 \log (c x)+2 c x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)+\sin ^{-1}(c x)^2\right )}{x^2}+\frac {1}{2} b^2 c^2 \sin ^{-1}(c x) \sin \left (2 \sin ^{-1}(c x)\right )-\frac {1}{4} b^2 c^2 \left (2 \sin ^{-1}(c x)^2-1\right ) \cos \left (2 \sin ^{-1}(c x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^2*(-(a^2/x^2) + a^2*c^4*x^2 + a*b*c^2*(c*x*Sqrt[1 - c^2*x^2] - ArcSin[c*x]) + 2*a*b*c^4*x^2*ArcSin[c*x] - (
2*a*b*(c*x*Sqrt[1 - c^2*x^2] + ArcSin[c*x]))/x^2 - (b^2*c^2*(-1 + 2*ArcSin[c*x]^2)*Cos[2*ArcSin[c*x]])/4 - 8*a
*b*c^2*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] - 4*a^2*c^2*Log[x] - (b^2*(2*c*x*Sqrt[1 - c^2*x^2]*ArcSin[c*
x] + ArcSin[c*x]^2 - 2*c^2*x^2*Log[c*x]))/x^2 + (4*I)*a*b*c^2*(ArcSin[c*x]^2 + PolyLog[2, E^((2*I)*ArcSin[c*x]
)]) + (I/6)*b^2*c^2*(Pi^3 - 8*ArcSin[c*x]^3 + (24*I)*ArcSin[c*x]^2*Log[1 - E^((-2*I)*ArcSin[c*x])] - 24*ArcSin
[c*x]*PolyLog[2, E^((-2*I)*ArcSin[c*x])] + (12*I)*PolyLog[3, E^((-2*I)*ArcSin[c*x])]) + (b^2*c^2*ArcSin[c*x]*S
in[2*ArcSin[c*x]])/2))/2

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} c^{4} d^{2} x^{4} - 2 \, a^{2} c^{2} d^{2} x^{2} + a^{2} d^{2} + {\left (b^{2} c^{4} d^{2} x^{4} - 2 \, b^{2} c^{2} d^{2} x^{2} + b^{2} d^{2}\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (a b c^{4} d^{2} x^{4} - 2 \, a b c^{2} d^{2} x^{2} + a b d^{2}\right )} \arcsin \left (c x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.84, size = 767, normalized size = 2.67 \[ 4 i c^{2} d^{2} b^{2} \arcsin \left (c x \right ) \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+4 i c^{2} d^{2} b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+4 i c^{2} d^{2} a b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+4 i c^{2} d^{2} a b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 i c^{2} d^{2} a b \arcsin \left (c x \right )^{2}-4 c^{2} d^{2} a b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-4 c^{2} d^{2} a b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\frac {c^{3} d^{2} a b \sqrt {-c^{2} x^{2}+1}\, x}{2}+c^{4} d^{2} a b \arcsin \left (c x \right ) x^{2}-\frac {c \,d^{2} a b \sqrt {-c^{2} x^{2}+1}}{x}-\frac {c \,d^{2} b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{x}+\frac {c^{3} d^{2} b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, x}{2}-\frac {c^{2} d^{2} a b \arcsin \left (c x \right )}{2}+\frac {c^{4} d^{2} b^{2} \arcsin \left (c x \right )^{2} x^{2}}{2}+i c^{2} d^{2} b^{2} \arcsin \left (c x \right )-2 c^{2} d^{2} b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 c^{2} d^{2} b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\frac {2 i c^{2} d^{2} b^{2} \arcsin \left (c x \right )^{3}}{3}+i c^{2} d^{2} a b -\frac {d^{2} a^{2}}{2 x^{2}}+\frac {d^{2} b^{2} c^{2}}{8}-\frac {b^{2} c^{4} d^{2} x^{2}}{4}-\frac {d^{2} a b \arcsin \left (c x \right )}{x^{2}}+\frac {c^{4} d^{2} a^{2} x^{2}}{2}-2 c^{2} d^{2} a^{2} \ln \left (c x \right )-\frac {c^{2} d^{2} b^{2} \arcsin \left (c x \right )^{2}}{4}-\frac {d^{2} b^{2} \arcsin \left (c x \right )^{2}}{2 x^{2}}+c^{2} d^{2} b^{2} \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )+c^{2} d^{2} b^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 c^{2} d^{2} b^{2} \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )-4 c^{2} d^{2} b^{2} \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-4 c^{2} d^{2} b^{2} \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*c^4*d^2*x^2 - 2*a^2*c^2*d^2*log(x) - a*b*d^2*(sqrt(-c^2*x^2 + 1)*c/x + arcsin(c*x)/x^2) - 1/2*a^2*d^2/
x^2 + integrate(((b^2*c^4*d^2*x^4 - 2*b^2*c^2*d^2*x^2 + b^2*d^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2
+ 2*(a*b*c^4*d^2*x^4 - 2*a*b*c^2*d^2*x^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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