3.3.50 \(\int \frac {(a+b \text {ArcSin}(c x))^2}{x (d-c^2 d x^2)^{3/2}} \, dx\) [250]

Optimal. Leaf size=467 \[ \frac {(a+b \text {ArcSin}(c x))^2}{d \sqrt {d-c^2 d x^2}}+\frac {4 i b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2 \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {2 i b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {2 i b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,-e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}} \]

[Out]

(a+b*arcsin(c*x))^2/d/(-c^2*d*x^2+d)^(1/2)+4*I*b*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+
1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)-2*(a+b*arcsin(c*x))^2*arctanh(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/d/(
-c^2*d*x^2+d)^(1/2)+2*I*b*(a+b*arcsin(c*x))*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*
x^2+d)^(1/2)-2*I*b^2*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)+2*I*b^
2*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)-2*I*b*(a+b*arcsin(c*x))*po
lylog(2,I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)-2*b^2*polylog(3,-I*c*x-(-c^2*x^2+1
)^(1/2))*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)+2*b^2*polylog(3,I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2
)/d/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.34, antiderivative size = 467, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {4793, 4803, 4268, 2611, 2320, 6724, 4749, 4266, 2317, 2438} \begin {gather*} \frac {4 i b \sqrt {1-c^2 x^2} \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d \sqrt {d-c^2 d x^2}}+\frac {2 i b \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d \sqrt {d-c^2 d x^2}}-\frac {2 i b \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d \sqrt {d-c^2 d x^2}}+\frac {(a+b \text {ArcSin}(c x))^2}{d \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{d \sqrt {d-c^2 d x^2}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (-e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \text {Li}_3\left (e^{i \text {ArcSin}(c x)}\right )}{d \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2317

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 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 2438

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]

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 4268

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 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 4793

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/(2*f*(p + 1)))*Simp[(d + e*
x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; Fre
eQ[{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 4803

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

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

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Mathematica [A]
time = 1.11, size = 667, normalized size = 1.43 \begin {gather*} \frac {a^2 d+a^2 \sqrt {d} \sqrt {d-c^2 d x^2} \log (c x)-a^2 \sqrt {d} \sqrt {d-c^2 d x^2} \log \left (d+\sqrt {d} \sqrt {d-c^2 d x^2}\right )+2 a b d \left (\text {ArcSin}(c x)+\sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1-e^{i \text {ArcSin}(c x)}\right )-\sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1+e^{i \text {ArcSin}(c x)}\right )+\sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-\sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )+i \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )-i \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )\right )+b^2 d \left (\text {ArcSin}(c x)^2+\sqrt {1-c^2 x^2} \text {ArcSin}(c x)^2 \log \left (1-e^{i \text {ArcSin}(c x)}\right )-2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )+2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )-\sqrt {1-c^2 x^2} \text {ArcSin}(c x)^2 \log \left (1+e^{i \text {ArcSin}(c x)}\right )+2 i \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )-2 i \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+2 i \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )-2 i \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )-2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,-e^{i \text {ArcSin}(c x)}\right )+2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (3,e^{i \text {ArcSin}(c x)}\right )\right )}{d^2 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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. 1095 vs. \(2 (490 ) = 980\).
time = 0.28, size = 1096, normalized size = 2.35

method	result	size
default	\(\frac {a^{2}}{d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {a^{2} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{d^{\frac {3}{2}}}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2}}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {4 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2} \left (c^{2} x^{2}-1\right )}\)	\(1096\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2/d/(-c^2*d*x^2+d)^(1/2)-a^2/d^(3/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)-b^2*(-d*(c^2*x^2-1))^(1/2)/d
^2/(c^2*x^2-1)*arcsin(c*x)^2+b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)^2*ln(1+
I*c*x+(-c^2*x^2+1)^(1/2))-b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)^2*ln(1-I*c
*x-(-c^2*x^2+1)^(1/2))-2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*ln(1+I*(I*c
*x+(-c^2*x^2+1)^(1/2)))+2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*ln(1-I*(I*
c*x+(-c^2*x^2+1)^(1/2)))+2*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*dilog(1+I*(I*c*x+(-
c^2*x^2+1)^(1/2)))-2*I*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*dilog(1+I*c*x+(-c^2*x^2+1
)^(1/2))+2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*polylog(3,-I*c*x-(-c^2*x^2+1)^(1/2))-
2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*polylog(3,I*c*x+(-c^2*x^2+1)^(1/2))-2*I*b^2*(-
c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-2*I*b
^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*a*b*(-d*(
c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)+2*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*
arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-4*I*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*a
rctan(I*c*x+(-c^2*x^2+1)^(1/2))-2*I*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*dilog(I*c*x+
(-c^2*x^2+1)^(1/2))+2*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*polylog(2,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/x/(-c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

-a^2*(log(2*sqrt(-c^2*d*x^2 + d)*sqrt(d)/abs(x) + 2*d/abs(x))/d^(3/2) - 1/(sqrt(-c^2*d*x^2 + d)*d)) + sqrt(d)*
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)*sqrt(-c*x + 1))
)*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2*x), x)

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{x \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))**2/(x*(-d*(c*x - 1)*(c*x + 1))**(3/2)), x)

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

[Out]

integrate((b*arcsin(c*x) + a)^2/((-c^2*d*x^2 + d)^(3/2)*x), 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}{x\,{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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