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

Optimal. Leaf size=398 \[ -\frac {i b c^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {i b c^2 \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {b c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}}+\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 398, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {4693, 4627, 266, 63, 208, 4709, 4183, 2531, 2282, 6589} \[ -\frac {i b c^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {i b c^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {b c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}}-\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 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 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4693

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

Rule 4709

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[d, 0] && IGtQ[n, 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 {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{x^3} \, dx &=-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{x^2} \, dx}{\sqrt {1-c^2 x^2}}-\frac {\left (c^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}-\frac {\left (c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}+\frac {\left (b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{x \sqrt {1-c^2 x^2}} \, dx}{\sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {\left (b c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {\left (b c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {\left (b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {i b c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {i b c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {\left (b^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}+\frac {\left (i b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {\left (i b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}-\frac {i b c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {i b c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {\left (b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {\left (b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}-\frac {i b c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {i b c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {b^2 c^2 \sqrt {d-c^2 d x^2} \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 5.28, size = 480, normalized size = 1.21 \[ \frac {1}{8} \left (-\frac {4 a^2 \sqrt {d-c^2 d x^2}}{x^2}+4 a^2 c^2 \sqrt {d} \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )-4 a^2 c^2 \sqrt {d} \log (x)+\frac {2 a b c^2 d \sqrt {1-c^2 x^2} \left (-4 i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )+4 i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-4 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+4 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )-2 \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )-2 \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin ^{-1}(c x) \csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+\sin ^{-1}(c x) \sec ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{\sqrt {d-c^2 d x^2}}+\frac {b^2 c^2 d \sqrt {1-c^2 x^2} \left (-8 i \sin ^{-1}(c x) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )+8 i \sin ^{-1}(c x) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )+8 \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )-8 \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )-4 \sin ^{-1}(c x)^2 \log \left (1-e^{i \sin ^{-1}(c x)}\right )+4 \sin ^{-1}(c x)^2 \log \left (1+e^{i \sin ^{-1}(c x)}\right )-4 \sin ^{-1}(c x) \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )-4 \sin ^{-1}(c x) \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\sin ^{-1}(c x)^2 \left (-\csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )+\sin ^{-1}(c x)^2 \sec ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+8 \log \left (\tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )}{\sqrt {d-c^2 d x^2}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-4*a^2*Sqrt[d - c^2*d*x^2])/x^2 - 4*a^2*c^2*Sqrt[d]*Log[x] + 4*a^2*c^2*Sqrt[d]*Log[d + Sqrt[d]*Sqrt[d - c^2*
d*x^2]] + (2*a*b*c^2*d*Sqrt[1 - c^2*x^2]*(-2*Cot[ArcSin[c*x]/2] - ArcSin[c*x]*Csc[ArcSin[c*x]/2]^2 - 4*ArcSin[
c*x]*Log[1 - E^(I*ArcSin[c*x])] + 4*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - (4*I)*PolyLog[2, -E^(I*ArcSin[c*x
])] + (4*I)*PolyLog[2, E^(I*ArcSin[c*x])] + ArcSin[c*x]*Sec[ArcSin[c*x]/2]^2 - 2*Tan[ArcSin[c*x]/2]))/Sqrt[d -
c^2*d*x^2] + (b^2*c^2*d*Sqrt[1 - c^2*x^2]*(-4*ArcSin[c*x]*Cot[ArcSin[c*x]/2] - ArcSin[c*x]^2*Csc[ArcSin[c*x]/
2]^2 - 4*ArcSin[c*x]^2*Log[1 - E^(I*ArcSin[c*x])] + 4*ArcSin[c*x]^2*Log[1 + E^(I*ArcSin[c*x])] + 8*Log[Tan[Arc
Sin[c*x]/2]] - (8*I)*ArcSin[c*x]*PolyLog[2, -E^(I*ArcSin[c*x])] + (8*I)*ArcSin[c*x]*PolyLog[2, E^(I*ArcSin[c*x
])] + 8*PolyLog[3, -E^(I*ArcSin[c*x])] - 8*PolyLog[3, E^(I*ArcSin[c*x])] + ArcSin[c*x]^2*Sec[ArcSin[c*x]/2]^2
- 4*ArcSin[c*x]*Tan[ArcSin[c*x]/2]))/Sqrt[d - c^2*d*x^2])/8

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2/x^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.55, size = 1082, normalized size = 2.72 \[ -\frac {a^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}+\frac {a^{2} \sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right ) c^{2}}{2}-\frac {a^{2} \sqrt {-c^{2} d \,x^{2}+d}\, c^{2}}{2}-\frac {b^{2} \arcsin \left (c x \right )^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2}}{2 \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \arcsin \left (c x \right ) \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{x \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \arcsin \left (c x \right )^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}}{2 x^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}+\frac {2 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 \left (c^{2} x^{2}-1\right )}+\frac {i b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right ) \arcsin \left (c x \right )}{c^{2} x^{2}-1}+\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \arctanh \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}-\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{2}}{c^{2} x^{2}-1}+\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{x \left (c^{2} x^{2}-1\right )}+\frac {a b \arcsin \left (c x \right ) \sqrt {-d \left (c^{2} x^{2}-1\right )}}{x^{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 )}\, c^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right ) \arcsin \left (c x \right )}{2 c^{2} x^{2}-2}-\frac {2 a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right ) \arcsin \left (c x \right )}{2 c^{2} x^{2}-2}-\frac {i b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right ) \arcsin \left (c x \right )}{c^{2} x^{2}-1}-\frac {2 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{2 c^{2} x^{2}-2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2/x^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2/x^3,x, algorithm="maxima")

[Out]

1/2*(c^2*sqrt(d)*log(2*sqrt(-c^2*d*x^2 + d)*sqrt(d)/abs(x) + 2*d/abs(x)) - sqrt(-c^2*d*x^2 + d)*c^2 - (-c^2*d*
x^2 + d)^(3/2)/(d*x^2))*a^2 + sqrt(d)*integrate((b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b*arct
an2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d-c^2\,d\,x^2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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