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

Optimal. Leaf size=300 \[ -\frac {3 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac {3 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^5 d^2}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt {1-c^2 x^2}}+\frac {3 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}-\frac {3 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c^5 d^2}-\frac {2 b^2 x}{c^4 d^2} \]

[Out]

-2*b^2*x/c^4/d^2+3/2*x*(a+b*arcsin(c*x))^2/c^4/d^2+1/2*x^3*(a+b*arcsin(c*x))^2/c^2/d^2/(-c^2*x^2+1)+3*I*(a+b*a
rcsin(c*x))^2*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c^5/d^2+b^2*arctanh(c*x)/c^5/d^2-3*I*b*(a+b*arcsin(c*x))*polylo
g(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/d^2+3*I*b*(a+b*arcsin(c*x))*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5
/d^2+3*b^2*polylog(3,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/d^2-3*b^2*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/
d^2-b*(a+b*arcsin(c*x))/c^5/d^2/(-c^2*x^2+1)^(1/2)+2*b*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c^5/d^2

________________________________________________________________________________________

Rubi [A]  time = 0.53, antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 14, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.518, Rules used = {4703, 4715, 4657, 4181, 2531, 2282, 6589, 4677, 8, 266, 43, 4689, 388, 208} \[ -\frac {3 i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac {3 i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac {3 b^2 \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}-\frac {3 b^2 \text {PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt {1-c^2 x^2}}+\frac {3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac {3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^5 d^2}-\frac {2 b^2 x}{c^4 d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c^5 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 208

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

Rule 266

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

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 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 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 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 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 4703

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

Rule 4715

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

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

________________________________________________________________________________________

Mathematica [B]  time = 3.14, size = 614, normalized size = 2.05 \[ \frac {-\frac {2 a^2 c x}{c^2 x^2-1}+4 a^2 c x+3 a^2 \log (1-c x)-3 a^2 \log (c x+1)+8 a b \sqrt {1-c^2 x^2}+\frac {2 a b \sqrt {1-c^2 x^2}}{c x-1}-\frac {2 a b \sqrt {1-c^2 x^2}}{c x+1}-12 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+12 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+8 a b c x \sin ^{-1}(c x)-\frac {2 a b \sin ^{-1}(c x)}{c x-1}-\frac {2 a b \sin ^{-1}(c x)}{c x+1}+6 i \pi a b \sin ^{-1}(c x)-12 a b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-6 \pi a b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+12 a b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-6 \pi a b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+6 \pi a b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+6 \pi a b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+\frac {8 b^2 c x}{c^2 x^2-1}-\frac {6 b^2 c^2 x^2 \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+\frac {6 b^2 c x \sin ^{-1}(c x)^2}{1-c^2 x^2}+2 b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)+\frac {2 b^2 \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+\frac {8 b^2 c^3 x^3}{1-c^2 x^2}+\frac {4 b^2 c^3 x^3 \sin ^{-1}(c x)^2}{c^2 x^2-1}+12 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )-12 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )+4 b^2 \tanh ^{-1}(c x)+12 i b^2 \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^5 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*a^2*c*x + (8*b^2*c^3*x^3)/(1 - c^2*x^2) + 8*a*b*Sqrt[1 - c^2*x^2] + (2*a*b*Sqrt[1 - c^2*x^2])/(-1 + c*x) -
(2*a*b*Sqrt[1 - c^2*x^2])/(1 + c*x) - (2*a^2*c*x)/(-1 + c^2*x^2) + (8*b^2*c*x)/(-1 + c^2*x^2) + (6*I)*a*b*Pi*A
rcSin[c*x] + 8*a*b*c*x*ArcSin[c*x] - (2*a*b*ArcSin[c*x])/(-1 + c*x) - (2*a*b*ArcSin[c*x])/(1 + c*x) + (2*b^2*A
rcSin[c*x])/Sqrt[1 - c^2*x^2] - (6*b^2*c^2*x^2*ArcSin[c*x])/Sqrt[1 - c^2*x^2] + 2*b^2*Sqrt[1 - c^2*x^2]*ArcSin
[c*x] + (6*b^2*c*x*ArcSin[c*x]^2)/(1 - c^2*x^2) + (4*b^2*c^3*x^3*ArcSin[c*x]^2)/(-1 + c^2*x^2) + (12*I)*b^2*Ar
cSin[c*x]^2*ArcTan[E^(I*ArcSin[c*x])] + 4*b^2*ArcTanh[c*x] - 6*a*b*Pi*Log[1 - I*E^(I*ArcSin[c*x])] - 12*a*b*Ar
cSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 6*a*b*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 12*a*b*ArcSin[c*x]*Log[1 + I*
E^(I*ArcSin[c*x])] + 3*a^2*Log[1 - c*x] - 3*a^2*Log[1 + c*x] + 6*a*b*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 6*
a*b*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (12*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (1
2*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x])] + 12*b^2*PolyLog[3, (-I)*E^(I*ArcSin[c*x])] - 12*b^
2*PolyLog[3, I*E^(I*ArcSin[c*x])])/(4*c^5*d^2)

________________________________________________________________________________________

fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{4} \arcsin \left (c x\right )^{2} + 2 \, a b x^{4} \arcsin \left (c x\right ) + a^{2} x^{4}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.62, size = 705, normalized size = 2.35 \[ \frac {3 i a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d^{2}}+\frac {a b \sqrt {-c^{2} x^{2}+1}}{c^{5} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{c^{5} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 a b \arcsin \left (c x \right ) x}{c^{4} d^{2}}-\frac {b^{2} \arcsin \left (c x \right )^{2} x}{2 c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {3 a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d^{2}}-\frac {3 a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d^{2}}+\frac {3 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d^{2}}-\frac {3 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d^{2}}-\frac {3 i a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d^{2}}-\frac {a b \arcsin \left (c x \right ) x}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 a b \sqrt {-c^{2} x^{2}+1}}{c^{5} d^{2}}+\frac {b^{2} \arcsin \left (c x \right )^{2} x}{c^{4} d^{2}}+\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )}{c^{5} d^{2}}-\frac {3 b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c^{5} d^{2}}-\frac {3 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d^{2}}-\frac {2 b^{2} x}{c^{4} d^{2}}-\frac {a^{2}}{4 c^{5} d^{2} \left (c x +1\right )}-\frac {a^{2}}{4 c^{5} d^{2} \left (c x -1\right )}+\frac {a^{2} x}{c^{4} d^{2}}+\frac {3 a^{2} \ln \left (c x -1\right )}{4 c^{5} d^{2}}-\frac {3 a^{2} \ln \left (c x +1\right )}{4 c^{5} d^{2}}+\frac {3 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d^{2}}+\frac {3 b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c^{5} d^{2}}-\frac {2 i b^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{5} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*I/c^5*b^2/d^2*arcsin(c*x)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3*I/c^5*b^2/d^2*arcsin(c*x)*polylog(2,-I*(
I*c*x+(-c^2*x^2+1)^(1/2)))-3*I/c^5*a*b/d^2*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3*I/c^5*a*b/d^2*dilog(1-I*(I*
c*x+(-c^2*x^2+1)^(1/2)))+1/c^5*a*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+3/c^5*a*b/d^2*arcsin(c*x)*ln(1+I*(I*c*x+
(-c^2*x^2+1)^(1/2)))-3/c^5*a*b/d^2*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+1/c^5*b^2/d^2/(c^2*x^2-1)*ar
csin(c*x)*(-c^2*x^2+1)^(1/2)+2/c^4*a*b/d^2*arcsin(c*x)*x-1/2/c^4*b^2/d^2/(c^2*x^2-1)*arcsin(c*x)^2*x-1/c^4*a*b
/d^2/(c^2*x^2-1)*arcsin(c*x)*x+2/c^5*a*b/d^2*(-c^2*x^2+1)^(1/2)+1/c^4*b^2/d^2*arcsin(c*x)^2*x+2/c^5*b^2/d^2*(-
c^2*x^2+1)^(1/2)*arcsin(c*x)-3/2/c^5*b^2/d^2*arcsin(c*x)^2*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/2/c^5*b^2/d^2*
arcsin(c*x)^2*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*I/c^5*b^2/d^2*arctan(I*c*x+(-c^2*x^2+1)^(1/2))+3*b^2*polylo
g(3,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/d^2-3*b^2*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/d^2-2*b^2*x/c^4/d
^2-1/4/c^5*a^2/d^2/(c*x+1)-1/4/c^5*a^2/d^2/(c*x-1)+1/c^4*a^2/d^2*x+3/4/c^5*a^2/d^2*ln(c*x-1)-3/4/c^5*a^2/d^2*l
n(c*x+1)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, a^{2} {\left (\frac {2 \, x}{c^{6} d^{2} x^{2} - c^{4} d^{2}} - \frac {4 \, x}{c^{4} d^{2}} + \frac {3 \, \log \left (c x + 1\right )}{c^{5} d^{2}} - \frac {3 \, \log \left (c x - 1\right )}{c^{5} d^{2}}\right )} - \frac {3 \, {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (c x + 1\right ) - 3 \, {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \, {\left (2 \, b^{2} c^{3} x^{3} - 3 \, b^{2} c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} - 2 \, {\left (c^{7} d^{2} x^{2} - c^{5} d^{2}\right )} \int \frac {4 \, a b c^{4} x^{4} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (3 \, {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - 3 \, {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \, {\left (2 \, b^{2} c^{3} x^{3} - 3 \, b^{2} c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{8} d^{2} x^{4} - 2 \, c^{6} d^{2} x^{2} + c^{4} d^{2}}\,{d x}}{4 \, {\left (c^{7} d^{2} x^{2} - c^{5} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

-1/4*a^2*(2*x/(c^6*d^2*x^2 - c^4*d^2) - 4*x/(c^4*d^2) + 3*log(c*x + 1)/(c^5*d^2) - 3*log(c*x - 1)/(c^5*d^2)) -
1/4*(3*(b^2*c^2*x^2 - b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(c*x + 1) - 3*(b^2*c^2*x^2 - b^2)*
arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(-c*x + 1) - 2*(2*b^2*c^3*x^3 - 3*b^2*c*x)*arctan2(c*x, sqrt(c
*x + 1)*sqrt(-c*x + 1))^2 + 4*(c^7*d^2*x^2 - c^5*d^2)*integrate(-1/2*(4*a*b*c^4*x^4*arctan2(c*x, sqrt(c*x + 1)
*sqrt(-c*x + 1)) - (3*(b^2*c^2*x^2 - b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 3*(b^2*c^2
*x^2 - b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(2*b^2*c^3*x^3 - 3*b^2*c*x)*arctan2(c
*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^8*d^2*x^4 - 2*c^6*d^2*x^2 + c^4*d^2), x))/
(c^7*d^2*x^2 - c^5*d^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________