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

Optimal. Leaf size=378 \[ \frac {2 i b \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {2 i b \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {2 a b c x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}+\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}}-\frac {2 b^2 \sqrt {d-c^2 d x^2} \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {2 b^2 \sqrt {d-c^2 d x^2} \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-2 b^2 \sqrt {d-c^2 d x^2}-\frac {2 b^2 c x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \]

[Out]

-2*b^2*(-c^2*d*x^2+d)^(1/2)+(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2-2*a*b*c*x*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+
1)^(1/2)-2*b^2*c*x*arcsin(c*x)*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-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)+2*I*b*(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)-2*I*b*(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)-2*b^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)+2*b^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 = 0.35, antiderivative size = 378, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {4697, 4709, 4183, 2531, 2282, 6589, 4619, 261} \[ \frac {2 i b \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {2 i b \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {2 b^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {2 b^2 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {2 a b c x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}+\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {2 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}}-2 b^2 \sqrt {d-c^2 d x^2}-\frac {2 b^2 c x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 261

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

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 4619

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

Rule 4697

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 + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1 -
c^2*x^2]), Int[((f*x)^m*(a + b*ArcSin[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*(m
+ 2)*Sqrt[1 - c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}
, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 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} \, dx &=\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\sqrt {d-c^2 d x^2} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt {1-c^2 x^2}} \, dx}{\sqrt {1-c^2 x^2}}-\frac {\left (2 b c \sqrt {d-c^2 d x^2}\right ) \int \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=-\frac {2 a b c x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}+\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\sqrt {d-c^2 d x^2} \operatorname {Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {\left (2 b^2 c \sqrt {d-c^2 d x^2}\right ) \int \sin ^{-1}(c x) \, dx}{\sqrt {1-c^2 x^2}}\\ &=-\frac {2 a b c x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}-\frac {2 b^2 c x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {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 (2 b \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 (2 b \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 (2 b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {1-c^2 x^2}}\\ &=-2 b^2 \sqrt {d-c^2 d x^2}-\frac {2 a b c x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}-\frac {2 b^2 c x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {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 {2 i b \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 {2 i b \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 (2 i b^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 (2 i b^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}}\\ &=-2 b^2 \sqrt {d-c^2 d x^2}-\frac {2 a b c x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}-\frac {2 b^2 c x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {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 {2 i b \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 {2 i b \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 (2 b^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 (2 b^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}}\\ &=-2 b^2 \sqrt {d-c^2 d x^2}-\frac {2 a b c x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}-\frac {2 b^2 c x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}}+\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {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 {2 i b \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 {2 i b \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 {2 b^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 {2 b^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 = 1.21, size = 391, normalized size = 1.03 \[ a^2 \sqrt {d-c^2 d x^2}-a^2 \sqrt {d} \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )+a^2 \sqrt {d} \log (c x)+\frac {2 a b \sqrt {d-c^2 d x^2} \left (\sqrt {1-c^2 x^2} \sin ^{-1}(c x)+i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-c x+\sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt {1-c^2 x^2}}+\frac {b^2 \sqrt {d-c^2 d x^2} \left (-2 \sqrt {1-c^2 x^2}+\sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2+2 i \sin ^{-1}(c x) \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-2 i \sin ^{-1}(c x) \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )-2 \text {Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )+2 \text {Li}_3\left (e^{i \sin ^{-1}(c x)}\right )-2 c x \sin ^{-1}(c x)+\sin ^{-1}(c x)^2 \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x)^2 \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt {1-c^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{x}, x\right ) \]

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,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, 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,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 = 0.38, size = 1017, normalized size = 2.69 \[ -\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right ) a^{2}+\sqrt {-c^{2} d \,x^{2}+d}\, a^{2}+\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x c}{c^{2} x^{2}-1}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} x^{2} c^{2}}{c^{2} x^{2}-1}-\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} x^{2}}{c^{2} x^{2}-1}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2}}{c^{2} x^{2}-1}+\frac {2 b^{2} \sqrt {-d \left (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}\, \arcsin \left (c x \right )^{2} \ln \left (1-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}\, \arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}-\frac {2 i b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}+\frac {2 i b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\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}\, \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\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}\, \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}+\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x c}{c^{2} x^{2}-1}+\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x^{2} c^{2}}{c^{2} x^{2}-1}-\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{c^{2} x^{2}-1}-\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 )}{c^{2} x^{2}-1}+\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 )}{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 )}\, \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\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 )}\, \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1} \]

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,x)

[Out]

-d^(1/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)*a^2+(-c^2*d*x^2+d)^(1/2)*a^2+2*b^2*(-d*(c^2*x^2-1))^(1/2)/
(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*x*c+b^2*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)*arcsin(c*x)^2*x^2*c^2-2*
b^2*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)*c^2*x^2-b^2*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)*arcsin(c*x)^2+2*b^2*(-d*
(c^2*x^2-1))^(1/2)/(c^2*x^2-1)-b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)^2*ln(1-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*x^2-1)*arcsin(c*x)^2*ln(1+I*c*x+(-c
^2*x^2+1)^(1/2))+2*I*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)*polylog(2,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*x^2-1)*polylog(2,I*c*x+(-c^2*x^2+1)^(1
/2))-2*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*polylog(3,I*c*x+(-c^2*x^2+1)^(1/2))+2*b^2*(-d
*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*polylog(3,-I*c*x-(-c^2*x^2+1)^(1/2))+2*a*b*(-d*(c^2*x^2-1))
^(1/2)/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x*c+2*a*b*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)*arcsin(c*x)*x^2*c^2-2*a*b*(
-d*(c^2*x^2-1))^(1/2)/(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)/(c^2*x^2-1)*arcs
in(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+2*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)*arcsin(c*x)
*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-2*I*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)*poly
log(2,-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*x^2-1)*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 \[ -{\left (\sqrt {d} \log \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {d}}{{\left | x \right |}} + \frac {2 \, d}{{\left | x \right |}}\right ) - \sqrt {-c^{2} d x^{2} + d}\right )} a^{2} + \sqrt {d} \int \frac {{\left (b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{x}\,{d x} \]

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,x, algorithm="maxima")

[Out]

-(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))*a^2 + sqrt(d)*integr
ate((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)/x, x)

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

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,x)

[Out]

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

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

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,x)

[Out]

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

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