3.3.5 \(\int \frac {(a+b \text {ArcSin}(c x))^2}{(d-c^2 d x^2)^3} \, dx\) [205]

Optimal. Leaf size=332 \[ \frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac {b (a+b \text {ArcSin}(c x))}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b (a+b \text {ArcSin}(c x))}{4 c d^3 \sqrt {1-c^2 x^2}}+\frac {x (a+b \text {ArcSin}(c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 x (a+b \text {ArcSin}(c x))^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac {3 i (a+b \text {ArcSin}(c x))^2 \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{4 c d^3}+\frac {5 b^2 \tanh ^{-1}(c x)}{6 c d^3}+\frac {3 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{4 c d^3}-\frac {3 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{4 c d^3}-\frac {3 b^2 \text {PolyLog}\left (3,-i e^{i \text {ArcSin}(c x)}\right )}{4 c d^3}+\frac {3 b^2 \text {PolyLog}\left (3,i e^{i \text {ArcSin}(c x)}\right )}{4 c d^3} \]

[Out]

1/12*b^2*x/d^3/(-c^2*x^2+1)-1/6*b*(a+b*arcsin(c*x))/c/d^3/(-c^2*x^2+1)^(3/2)+1/4*x*(a+b*arcsin(c*x))^2/d^3/(-c
^2*x^2+1)^2+3/8*x*(a+b*arcsin(c*x))^2/d^3/(-c^2*x^2+1)-3/4*I*(a+b*arcsin(c*x))^2*arctan(I*c*x+(-c^2*x^2+1)^(1/
2))/c/d^3+5/6*b^2*arctanh(c*x)/c/d^3+3/4*I*b*(a+b*arcsin(c*x))*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d^3-
3/4*I*b*(a+b*arcsin(c*x))*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d^3-3/4*b^2*polylog(3,-I*(I*c*x+(-c^2*x^2+
1)^(1/2)))/c/d^3+3/4*b^2*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d^3-3/4*b*(a+b*arcsin(c*x))/c/d^3/(-c^2*x^2
+1)^(1/2)

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Rubi [A]
time = 0.25, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4747, 4749, 4266, 2611, 2320, 6724, 4767, 212, 205} \begin {gather*} -\frac {3 i \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{4 c d^3}-\frac {3 b (a+b \text {ArcSin}(c x))}{4 c d^3 \sqrt {1-c^2 x^2}}-\frac {b (a+b \text {ArcSin}(c x))}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {3 x (a+b \text {ArcSin}(c x))^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {x (a+b \text {ArcSin}(c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {3 i b \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{4 c d^3}-\frac {3 i b \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{4 c d^3}-\frac {3 b^2 \text {Li}_3\left (-i e^{i \text {ArcSin}(c x)}\right )}{4 c d^3}+\frac {3 b^2 \text {Li}_3\left (i e^{i \text {ArcSin}(c x)}\right )}{4 c d^3}+\frac {b^2 x}{12 d^3 \left (1-c^2 x^2\right )}+\frac {5 b^2 \tanh ^{-1}(c x)}{6 c d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x)/(12*d^3*(1 - c^2*x^2)) - (b*(a + b*ArcSin[c*x]))/(6*c*d^3*(1 - c^2*x^2)^(3/2)) - (3*b*(a + b*ArcSin[c*
x]))/(4*c*d^3*Sqrt[1 - c^2*x^2]) + (x*(a + b*ArcSin[c*x])^2)/(4*d^3*(1 - c^2*x^2)^2) + (3*x*(a + b*ArcSin[c*x]
)^2)/(8*d^3*(1 - c^2*x^2)) - (((3*I)/4)*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/(c*d^3) + (5*b^2*ArcT
anh[c*x])/(6*c*d^3) + (((3*I)/4)*b*(a + b*ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c*d^3) - (((3*I)/4
)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c*d^3) - (3*b^2*PolyLog[3, (-I)*E^(I*ArcSin[c*x])])/
(4*c*d^3) + (3*b^2*PolyLog[3, I*E^(I*ArcSin[c*x])])/(4*c*d^3)

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

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

Rule 2320

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 2611

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 4266

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 4747

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/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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 4749

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 4767

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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 6724

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

Rubi steps

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

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Mathematica [A]
time = 3.28, size = 556, normalized size = 1.67 \begin {gather*} \frac {\frac {24 a^2 x}{\left (-1+c^2 x^2\right )^2}-\frac {36 a^2 x}{-1+c^2 x^2}-\frac {18 a^2 \log (1-c x)}{c}+\frac {18 a^2 \log (1+c x)}{c}+\frac {72 i a b \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{c}-\frac {4 b^2 \left (\frac {2 c x}{-1+c^2 x^2}+\frac {4 \text {ArcSin}(c x)}{\left (1-c^2 x^2\right )^{3/2}}+\frac {18 \text {ArcSin}(c x)}{\sqrt {1-c^2 x^2}}-\frac {6 c x \text {ArcSin}(c x)^2}{\left (-1+c^2 x^2\right )^2}+\frac {9 c x \text {ArcSin}(c x)^2}{-1+c^2 x^2}+18 i \text {ArcSin}(c x)^2 \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )-20 \tanh ^{-1}(c x)-18 i \text {ArcSin}(c x) \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+18 i \text {ArcSin}(c x) \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )+18 \text {PolyLog}\left (3,-i e^{i \text {ArcSin}(c x)}\right )-18 \text {PolyLog}\left (3,i e^{i \text {ArcSin}(c x)}\right )\right )}{c}+\frac {a b \left (30-70 \sqrt {1-c^2 x^2}+40 \cos (2 \text {ArcSin}(c x))-18 \cos (3 \text {ArcSin}(c x))+10 \cos (4 \text {ArcSin}(c x))-72 i \left (-1+c^2 x^2\right )^2 \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )+3 \text {ArcSin}(c x) \left (22 c x+9 \log \left (1-i e^{i \text {ArcSin}(c x)}\right )+12 \cos (2 \text {ArcSin}(c x)) \left (\log \left (1-i e^{i \text {ArcSin}(c x)}\right )-\log \left (1+i e^{i \text {ArcSin}(c x)}\right )\right )+3 \cos (4 \text {ArcSin}(c x)) \left (\log \left (1-i e^{i \text {ArcSin}(c x)}\right )-\log \left (1+i e^{i \text {ArcSin}(c x)}\right )\right )-9 \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+6 \sin (3 \text {ArcSin}(c x))\right )\right )}{c \left (-1+c^2 x^2\right )^2}}{96 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((24*a^2*x)/(-1 + c^2*x^2)^2 - (36*a^2*x)/(-1 + c^2*x^2) - (18*a^2*Log[1 - c*x])/c + (18*a^2*Log[1 + c*x])/c +
((72*I)*a*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/c - (4*b^2*((2*c*x)/(-1 + c^2*x^2) + (4*ArcSin[c*x])/(1 - c^2
*x^2)^(3/2) + (18*ArcSin[c*x])/Sqrt[1 - c^2*x^2] - (6*c*x*ArcSin[c*x]^2)/(-1 + c^2*x^2)^2 + (9*c*x*ArcSin[c*x]
^2)/(-1 + c^2*x^2) + (18*I)*ArcSin[c*x]^2*ArcTan[E^(I*ArcSin[c*x])] - 20*ArcTanh[c*x] - (18*I)*ArcSin[c*x]*Pol
yLog[2, (-I)*E^(I*ArcSin[c*x])] + (18*I)*ArcSin[c*x]*PolyLog[2, I*E^(I*ArcSin[c*x])] + 18*PolyLog[3, (-I)*E^(I
*ArcSin[c*x])] - 18*PolyLog[3, I*E^(I*ArcSin[c*x])]))/c + (a*b*(30 - 70*Sqrt[1 - c^2*x^2] + 40*Cos[2*ArcSin[c*
x]] - 18*Cos[3*ArcSin[c*x]] + 10*Cos[4*ArcSin[c*x]] - (72*I)*(-1 + c^2*x^2)^2*PolyLog[2, I*E^(I*ArcSin[c*x])]
+ 3*ArcSin[c*x]*(22*c*x + 9*Log[1 - I*E^(I*ArcSin[c*x])] + 12*Cos[2*ArcSin[c*x]]*(Log[1 - I*E^(I*ArcSin[c*x])]
- Log[1 + I*E^(I*ArcSin[c*x])]) + 3*Cos[4*ArcSin[c*x]]*(Log[1 - I*E^(I*ArcSin[c*x])] - Log[1 + I*E^(I*ArcSin[
c*x])]) - 9*Log[1 + I*E^(I*ArcSin[c*x])] + 6*Sin[3*ArcSin[c*x]])))/(c*(-1 + c^2*x^2)^2))/(96*d^3)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 843 vs. \(2 (339 ) = 678\).
time = 0.22, size = 844, normalized size = 2.54

method	result	size
derivativedivides	\(\frac {-\frac {3 b^{2} \arcsin \left (c x \right )^{2} c^{3} x^{3}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 d^{3}}-\frac {3 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 d^{3}}+\frac {3 a^{2} \ln \left (c x +1\right )}{16 d^{3}}-\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 )}{4 d^{3}}+\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 )}{4 d^{3}}+\frac {3 i a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 d^{3}}-\frac {3 i a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 d^{3}}-\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 )}{4 d^{3}}+\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 )}{4 d^{3}}+\frac {5 b^{2} \arcsin \left (c x \right )^{2} c x}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\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 )}{8 d^{3}}-\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 )}{8 d^{3}}-\frac {5 i b^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3 d^{3}}+\frac {3 b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 a b \arcsin \left (c x \right ) c^{3} x^{3}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 a b \,c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {5 a b \arcsin \left (c x \right ) c x}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {11 b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {11 a b \sqrt {-c^{2} x^{2}+1}}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {b^{2} c^{3} x^{3}}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {b^{2} c x}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 a^{2} \ln \left (c x -1\right )}{16 d^{3}}-\frac {a^{2}}{16 d^{3} \left (c x +1\right )^{2}}-\frac {3 a^{2}}{16 d^{3} \left (c x +1\right )}+\frac {a^{2}}{16 d^{3} \left (c x -1\right )^{2}}-\frac {3 a^{2}}{16 d^{3} \left (c x -1\right )}}{c}\)	\(844\)
default	\(\frac {-\frac {3 b^{2} \arcsin \left (c x \right )^{2} c^{3} x^{3}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 d^{3}}-\frac {3 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 d^{3}}+\frac {3 a^{2} \ln \left (c x +1\right )}{16 d^{3}}-\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 )}{4 d^{3}}+\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 )}{4 d^{3}}+\frac {3 i a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 d^{3}}-\frac {3 i a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{4 d^{3}}-\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 )}{4 d^{3}}+\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 )}{4 d^{3}}+\frac {5 b^{2} \arcsin \left (c x \right )^{2} c x}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\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 )}{8 d^{3}}-\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 )}{8 d^{3}}-\frac {5 i b^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3 d^{3}}+\frac {3 b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 a b \arcsin \left (c x \right ) c^{3} x^{3}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {3 a b \,c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {5 a b \arcsin \left (c x \right ) c x}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {11 b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {11 a b \sqrt {-c^{2} x^{2}+1}}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {b^{2} c^{3} x^{3}}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {b^{2} c x}{12 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 a^{2} \ln \left (c x -1\right )}{16 d^{3}}-\frac {a^{2}}{16 d^{3} \left (c x +1\right )^{2}}-\frac {3 a^{2}}{16 d^{3} \left (c x +1\right )}+\frac {a^{2}}{16 d^{3} \left (c x -1\right )^{2}}-\frac {3 a^{2}}{16 d^{3} \left (c x -1\right )}}{c}\)	\(844\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-3/8*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)^2*c^3*x^3+3/16*a^2/d^3*ln(c*x+1)-5/3*I*b^2/d^3*arctan(I*c*
x+(-c^2*x^2+1)^(1/2))+5/8*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)^2*c*x+3/4*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*ar
csin(c*x)*(-c^2*x^2+1)^(1/2)*c^2*x^2-3/4*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c^3*x^3+3/4*a*b/d^3/(c^4*x^
4-2*c^2*x^2+1)*c^2*x^2*(-c^2*x^2+1)^(1/2)+5/4*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c*x-11/12*b^2/d^3/(c^4
*x^4-2*c^2*x^2+1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-3/4*I*b^2/d^3*arcsin(c*x)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/
2)))+3/4*I*b^2/d^3*arcsin(c*x)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-11/12*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*(-
c^2*x^2+1)^(1/2)-3/4*a*b/d^3*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/4*a*b/d^3*arcsin(c*x)*ln(1-I*(I*
c*x+(-c^2*x^2+1)^(1/2)))+3/4*I*a*b/d^3*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/4*I*a*b/d^3*dilog(1-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^3*x^3+1/12*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*c*x-3/16*a^2/
d^3*ln(c*x-1)-1/16*a^2/d^3/(c*x+1)^2-3/16*a^2/d^3/(c*x+1)+1/16*a^2/d^3/(c*x-1)^2-3/16*a^2/d^3/(c*x-1)+3/4*b^2/
d^3*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/4*b^2/d^3*polylog(3,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/8*b^2/d^3*a
rcsin(c*x)^2*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/8*b^2/d^3*arcsin(c*x)^2*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*a^2*(2*(3*c^2*x^3 - 5*x)/(c^4*d^3*x^4 - 2*c^2*d^3*x^2 + d^3) - 3*log(c*x + 1)/(c*d^3) + 3*log(c*x - 1)/(
c*d^3)) + 1/16*(3*(b^2*c^4*x^4 - 2*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^4*x^4 - 2*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*(3*b^
2*c^3*x^3 - 5*b^2*c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 16*(c^5*d^3*x^4 - 2*c^3*d^3*x^2 + c*d^3)
*integrate(-1/8*(16*a*b*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) - (3*(b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*ar
ctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 3*(b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*arctan2(c*x, sqr
t(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(3*b^2*c^3*x^3 - 5*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^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x))/(c^5*d^3*x^4 - 2
*c^3*d^3*x^2 + c*d^3)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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