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

Optimal. Leaf size=238 \[ \frac{2 i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{2 i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}+\frac{2 i b^2 c \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{2 i b^2 c \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{2 b^2 c \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{d}+\frac{2 b^2 c \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d x}-\frac{2 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac{4 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d} \]

[Out]

-((a + b*ArcSin[c*x])^2/(d*x)) - ((2*I)*c*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/d - (4*b*c*(a + b*A
rcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/d + ((2*I)*b^2*c*PolyLog[2, -E^(I*ArcSin[c*x])])/d + ((2*I)*b*c*(a + b
*ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d - ((2*I)*b*c*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[
c*x])])/d - ((2*I)*b^2*c*PolyLog[2, E^(I*ArcSin[c*x])])/d - (2*b^2*c*PolyLog[3, (-I)*E^(I*ArcSin[c*x])])/d + (
2*b^2*c*PolyLog[3, I*E^(I*ArcSin[c*x])])/d

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Rubi [A]  time = 0.347538, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {4701, 4657, 4181, 2531, 2282, 6589, 4709, 4183, 2279, 2391} \[ \frac{2 i b c \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{2 i b c \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}+\frac{2 i b^2 c \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{2 i b^2 c \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{2 b^2 c \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{d}+\frac{2 b^2 c \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d x}-\frac{2 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac{4 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSin[c*x])^2/(d*x)) - ((2*I)*c*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/d - (4*b*c*(a + b*A
rcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/d + ((2*I)*b^2*c*PolyLog[2, -E^(I*ArcSin[c*x])])/d + ((2*I)*b*c*(a + b
*ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d - ((2*I)*b*c*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[
c*x])])/d - ((2*I)*b^2*c*PolyLog[2, E^(I*ArcSin[c*x])])/d - (2*b^2*c*PolyLog[3, (-I)*E^(I*ArcSin[c*x])])/d + (
2*b^2*c*PolyLog[3, I*E^(I*ArcSin[c*x])])/d

Rule 4701

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)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 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] && LtQ[m, -1] && Inte
gerQ[m]

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 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 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 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 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]

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 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 2279

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 2391

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]

Rubi steps

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

Mathematica [A]  time = 0.714095, size = 391, normalized size = 1.64 \[ -\frac{4 a b c \left (-i \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+i \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+\frac{\sin ^{-1}(c x)}{c x}+\sin ^{-1}(c x) \left (-\log \left (1-i e^{i \sin ^{-1}(c x)}\right )\right )+\sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )+\log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )\right )+2 b^2 c \left (-2 i \sin ^{-1}(c x) \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+2 i \sin ^{-1}(c x) \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )-2 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )+2 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )-2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )+\frac{\sin ^{-1}(c x)^2}{c x}+\sin ^{-1}(c x)^2 \left (-\log \left (1-i e^{i \sin ^{-1}(c x)}\right )\right )+\sin ^{-1}(c x)^2 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-2 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+2 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )+a^2 c \log (1-c x)-a^2 c \log (c x+1)+\frac{2 a^2}{x}}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2*a^2)/x + a^2*c*Log[1 - c*x] - a^2*c*Log[1 + c*x] + 4*a*b*c*(ArcSin[c*x]/(c*x) - ArcSin[c*x]*Log[1 - I*E^(
I*ArcSin[c*x])] + ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + Log[Cos[ArcSin[c*x]/2]] - Log[Sin[ArcSin[c*x]/2]]
- I*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + I*PolyLog[2, I*E^(I*ArcSin[c*x])]) + 2*b^2*c*(ArcSin[c*x]^2/(c*x) -
2*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] - ArcSin[c*x]^2*Log[1 - I*E^(I*ArcSin[c*x])] + ArcSin[c*x]^2*Log[1 +
I*E^(I*ArcSin[c*x])] + 2*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - (2*I)*PolyLog[2, -E^(I*ArcSin[c*x])] - (2*I)
*ArcSin[c*x]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (2*I)*ArcSin[c*x]*PolyLog[2, I*E^(I*ArcSin[c*x])] + (2*I)*Po
lyLog[2, E^(I*ArcSin[c*x])] + 2*PolyLog[3, (-I)*E^(I*ArcSin[c*x])] - 2*PolyLog[3, I*E^(I*ArcSin[c*x])]))/(2*d)

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Maple [A]  time = 0.21, size = 575, normalized size = 2.4 \begin{align*} -{\frac{c{a}^{2}\ln \left ( cx-1 \right ) }{2\,d}}+{\frac{c{a}^{2}\ln \left ( cx+1 \right ) }{2\,d}}-{\frac{{a}^{2}}{dx}}-{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{dx}}-{\frac{2\,icab}{d}{\it dilog} \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-2\,{\frac{c{b}^{2}\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }{d}}+{\frac{2\,ic{b}^{2}}{d}{\it dilog} \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{c{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{d}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{2\,ic{b}^{2}\arcsin \left ( cx \right ) }{d}{\it polylog} \left ( 2,i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-2\,{\frac{c{b}^{2}{\it polylog} \left ( 3,-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }{d}}+{\frac{c{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{d}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{2\,ic{b}^{2}}{d}{\it dilog} \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+2\,{\frac{c{b}^{2}{\it polylog} \left ( 3,i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }{d}}-2\,{\frac{ab\arcsin \left ( cx \right ) }{dx}}+2\,{\frac{cab\arcsin \left ( cx \right ) \ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }{d}}-2\,{\frac{cab\arcsin \left ( cx \right ) \ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }{d}}+2\,{\frac{cab\ln \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1}-1 \right ) }{d}}-2\,{\frac{cab\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }{d}}+{\frac{2\,icab}{d}{\it dilog} \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{2\,ic{b}^{2}\arcsin \left ( cx \right ) }{d}{\it polylog} \left ( 2,-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{2}{\left (\frac{c \log \left (c x + 1\right )}{d} - \frac{c \log \left (c x - 1\right )}{d} - \frac{2}{d x}\right )} + \frac{b^{2} c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (c x + 1\right ) - b^{2} c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \, b^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} - 2 \, d x \int \frac{2 \, a b \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (b^{2} c^{2} x^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) - b^{2} c^{2} x^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \, b^{2} c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{2} d x^{4} - d x^{2}}\,{d x}}{2 \, d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*(c*log(c*x + 1)/d - c*log(c*x - 1)/d - 2/(d*x)) + 1/2*(b^2*c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x +
1))^2*log(c*x + 1) - b^2*c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(-c*x + 1) - 2*b^2*arctan2(c*x, s
qrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*d*x*integrate(-(2*a*b*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) - (b^2*c^2
*x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - b^2*c^2*x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*
x + 1))*log(-c*x + 1) - 2*b^2*c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sqrt(-c*x + 1))/(c
^2*d*x^4 - d*x^2), x))/(d*x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{2} d x^{4} - d x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a^{2}}{c^{2} x^{4} - x^{2}}\, dx + \int \frac{b^{2} \operatorname{asin}^{2}{\left (c x \right )}}{c^{2} x^{4} - x^{2}}\, dx + \int \frac{2 a b \operatorname{asin}{\left (c x \right )}}{c^{2} x^{4} - x^{2}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out