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

Optimal. Leaf size=210 \[ \frac {i b c^2 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {i b c^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac {2 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}-\frac {b^2 c^2 \text {Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac {b^2 c^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac {b^2 c^2 \log (x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4701, 4679, 4419, 4183, 2531, 2282, 6589, 4681, 29} \[ \frac {i b c^2 \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {i b c^2 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {b^2 c^2 \text {PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac {b^2 c^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac {2 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}+\frac {b^2 c^2 \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rule 4679

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

Rule 4681

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[(b*c*n*d^IntPart[p]*(d + e*x
^2)^FracPart[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*ArcSi
n[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

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

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Mathematica [A]  time = 1.19, size = 353, normalized size = 1.68 \[ -\frac {a^2 c^2 \log \left (1-c^2 x^2\right )-2 a^2 c^2 \log (x)+\frac {a^2}{x^2}+2 a b c^2 \left (\frac {\sqrt {1-c^2 x^2}}{c x}+\frac {\sin ^{-1}(c x)}{c^2 x^2}-i \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )+i \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+2 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )+2 b^2 c^2 \left (\frac {\sin ^{-1}(c x)^2}{2 c^2 x^2}+\frac {\sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c x}-i \sin ^{-1}(c x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c x)}\right )-i \sin ^{-1}(c x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{-2 i \sin ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )-\log (c x)-\frac {2}{3} i \sin ^{-1}(c x)^3-\sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x)^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )+\frac {i \pi ^3}{24}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(a^2/x^2 - 2*a^2*c^2*Log[x] + a^2*c^2*Log[1 - c^2*x^2] + 2*a*b*c^2*(Sqrt[1 - c^2*x^2]/(c*x) + ArcSin[c*x]
/(c^2*x^2) - 2*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] + 2*ArcSin[c*x]*Log[1 + E^((2*I)*ArcSin[c*x])] - I*P
olyLog[2, -E^((2*I)*ArcSin[c*x])] + I*PolyLog[2, E^((2*I)*ArcSin[c*x])]) + 2*b^2*c^2*((I/24)*Pi^3 + (Sqrt[1 -
c^2*x^2]*ArcSin[c*x])/(c*x) + ArcSin[c*x]^2/(2*c^2*x^2) - ((2*I)/3)*ArcSin[c*x]^3 - ArcSin[c*x]^2*Log[1 - E^((
-2*I)*ArcSin[c*x])] + ArcSin[c*x]^2*Log[1 + E^((2*I)*ArcSin[c*x])] - Log[c*x] - I*ArcSin[c*x]*PolyLog[2, E^((-
2*I)*ArcSin[c*x])] - I*ArcSin[c*x]*PolyLog[2, -E^((2*I)*ArcSin[c*x])] - PolyLog[3, E^((-2*I)*ArcSin[c*x])]/2 +
PolyLog[3, -E^((2*I)*ArcSin[c*x])]/2))/d

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{2} d x^{5} - d x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.53, size = 793, normalized size = 3.78 \[ \frac {i c^{2} a b}{d}+\frac {c^{2} b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {c^{2} b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i c^{2} b^{2} \arcsin \left (c x \right )}{d}-\frac {c^{2} b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {2 c^{2} a b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {2 c^{2} a b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {2 i c^{2} b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {2 i c^{2} b^{2} \arcsin \left (c x \right ) \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i c^{2} a b \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i c^{2} b^{2} \arcsin \left (c x \right ) \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {2 i c^{2} a b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {2 i c^{2} a b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {2 c^{2} a b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {a b \arcsin \left (c x \right )}{d \,x^{2}}-\frac {a^{2}}{2 d \,x^{2}}+\frac {c^{2} a^{2} \ln \left (c x \right )}{d}-\frac {b^{2} \arcsin \left (c x \right )^{2}}{2 d \,x^{2}}-\frac {c^{2} a^{2} \ln \left (c x +1\right )}{2 d}-\frac {c^{2} a^{2} \ln \left (c x -1\right )}{2 d}+\frac {c^{2} b^{2} \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d}+\frac {c^{2} b^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {2 c^{2} b^{2} \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {b^{2} c^{2} \polylog \left (3, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d}+\frac {2 c^{2} b^{2} \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {2 c^{2} b^{2} \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {c \,b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{d x}-\frac {c a b \sqrt {-c^{2} x^{2}+1}}{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (\frac {c^{2} \log \left (c x + 1\right )}{d} + \frac {c^{2} \log \left (c x - 1\right )}{d} - \frac {2 \, c^{2} \log \relax (x)}{d} + \frac {1}{d x^{2}}\right )} a^{2} - \int \frac {b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{2} d x^{5} - d x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a^{2}}{c^{2} x^{5} - x^{3}}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c x \right )}}{c^{2} x^{5} - x^{3}}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c x \right )}}{c^{2} x^{5} - x^{3}}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a**2/(c**2*x**5 - x**3), x) + Integral(b**2*asin(c*x)**2/(c**2*x**5 - x**3), x) + Integral(2*a*b*as
in(c*x)/(c**2*x**5 - x**3), x))/d

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