3.2.89 \(\int \frac {(a+b \text {ArcSin}(c x))^2}{x^2 (d-c^2 d x^2)} \, dx\) [189]

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

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {4789, 4749, 4266, 2611, 2320, 6724, 4803, 4268, 2317, 2438} \begin {gather*} -\frac {2 i c \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{d}+\frac {2 i b c \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d}-\frac {2 i b c \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d}-\frac {(a+b \text {ArcSin}(c x))^2}{d x}-\frac {4 b c \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d}+\frac {2 i b^2 c \text {Li}_2\left (-e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {2 i b^2 c \text {Li}_2\left (e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {2 b^2 c \text {Li}_3\left (-i e^{i \text {ArcSin}(c x)}\right )}{d}+\frac {2 b^2 c \text {Li}_3\left (i e^{i \text {ArcSin}(c x)}\right )}{d} \end {gather*}

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 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 4789

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/(f*(m + 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]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -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^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 \text {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {(2 b c) \text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.44, size = 391, normalized size = 1.64 \begin {gather*} -\frac {\frac {2 a^2}{x}+a^2 c \log (1-c x)-a^2 c \log (1+c x)+4 a b c \left (\frac {\text {ArcSin}(c x)}{c x}-\text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )+\text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+\log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-\log \left (\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-i \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+i \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )\right )+2 b^2 c \left (\frac {\text {ArcSin}(c x)^2}{c x}-2 \text {ArcSin}(c x) \log \left (1-e^{i \text {ArcSin}(c x)}\right )-\text {ArcSin}(c x)^2 \log \left (1-i e^{i \text {ArcSin}(c x)}\right )+\text {ArcSin}(c x)^2 \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+2 \text {ArcSin}(c x) \log \left (1+e^{i \text {ArcSin}(c x)}\right )-2 i \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )-2 i \text {ArcSin}(c x) \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+2 i \text {ArcSin}(c x) \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )+2 i \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )+2 \text {PolyLog}\left (3,-i e^{i \text {ArcSin}(c x)}\right )-2 \text {PolyLog}\left (3,i e^{i \text {ArcSin}(c x)}\right )\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((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
)*PolyLog[2, E^(I*ArcSin[c*x])] + 2*PolyLog[3, (-I)*E^(I*ArcSin[c*x])] - 2*PolyLog[3, I*E^(I*ArcSin[c*x])]))/d

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Maple [A]
time = 0.27, size = 569, normalized size = 2.39

method	result	size
derivativedivides	\(c \left (-\frac {a^{2}}{d c x}+\frac {a^{2} \ln \left (c x +1\right )}{2 d}-\frac {a^{2} \ln \left (c x -1\right )}{2 d}-\frac {b^{2} \arcsin \left (c x \right )^{2}}{d c x}+\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 b^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {2 i b^{2} \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i b^{2} \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {2 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 a b \arcsin \left (c x \right )}{d c x}+\frac {2 a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 a b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d}-\frac {2 a b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {2 i a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}\right )\)	\(569\)
default	\(c \left (-\frac {a^{2}}{d c x}+\frac {a^{2} \ln \left (c x +1\right )}{2 d}-\frac {a^{2} \ln \left (c x -1\right )}{2 d}-\frac {b^{2} \arcsin \left (c x \right )^{2}}{d c x}+\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 b^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {2 i b^{2} \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i b^{2} \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {2 b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 a b \arcsin \left (c x \right )}{d c x}+\frac {2 a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 a b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d}-\frac {2 a b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {2 i a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}\right )\)	\(569\)

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,method=_RETURNVERBOSE)

[Out]

c*(-a^2/d/c/x+1/2*a^2/d*ln(c*x+1)-1/2*a^2/d*ln(c*x-1)-b^2/d/c/x*arcsin(c*x)^2-2*I*a*b/d*dilog(1-I*(I*c*x+(-c^2
*x^2+1)^(1/2)))-2*b^2/d*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+2*I*b^2/d*dilog(1+I*c*x+(-c^2*x^2+1)^(1/2))
+1/d*b^2*arcsin(c*x)^2*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*I*b^2/d*dilog(I*c*x+(-c^2*x^2+1)^(1/2))+2/d*b^2*po
lylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2)))-1/d*b^2*arcsin(c*x)^2*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*I/d*b^2*arcsi
n(c*x)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2/d*b^2*polylog(3,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*a*b/d*arcsi
n(c*x)/c/x+2*a*b/d*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*a*b/d*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+
1)^(1/2)))+2*a*b/d*ln(I*c*x+(-c^2*x^2+1)^(1/2)-1)-2*a*b/d*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+2*I*a*b/d*dilog(1+I*(
I*c*x+(-c^2*x^2+1)^(1/2)))-2*I/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 = 0.00, 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^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 = 0.00, 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^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {gather*}

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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