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

Optimal. Leaf size=233 \[ -\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}+\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}+\frac {i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d^2}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac {b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c^3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {4703, 4657, 4181, 2531, 2282, 6589, 4677, 206} \[ -\frac {i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}+\frac {i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}+\frac {b^2 \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac {b^2 \text {PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac {b \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d^2}+\frac {b^2 \tanh ^{-1}(c x)}{c^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 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 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 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 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4703

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

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

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Mathematica [A]  time = 2.64, size = 383, normalized size = 1.64 \[ -\frac {\frac {2 a^2 c x}{c^2 x^2-1}+a^2 (-\log (1-c x))+a^2 \log (c x+1)+\frac {2 a b \left (-2 \sqrt {1-c^2 x^2}+\cos \left (2 \sin ^{-1}(c x)\right )+\sin ^{-1}(c x) \left (2 c x-\log \left (1-i e^{i \sin ^{-1}(c x)}\right )+\log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\left (\log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\log \left (1-i e^{i \sin ^{-1}(c x)}\right )\right ) \cos \left (2 \sin ^{-1}(c x)\right )\right )+1\right )}{c^2 x^2-1}+4 i a b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )-4 i a b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+\frac {2 b^2 \sin ^{-1}(c x) \left (c x \sin ^{-1}(c x)-2 \sqrt {1-c^2 x^2}\right )}{c^2 x^2-1}-4 b^2 \left (-i \sin ^{-1}(c x) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+i \sin ^{-1}(c x) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+\text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )-\text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )+\tanh ^{-1}(c x)+i \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )\right )}{4 c^3 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 1.20, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{2} \arcsin \left (c x\right )^{2} + 2 \, a b x^{2} \arcsin \left (c x\right ) + a^{2} x^{2}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.40, size = 599, normalized size = 2.57 \[ -\frac {a^{2}}{4 c^{3} d^{2} \left (c x +1\right )}-\frac {a^{2} \ln \left (c x +1\right )}{4 c^{3} d^{2}}-\frac {a^{2}}{4 c^{3} d^{2} \left (c x -1\right )}+\frac {a^{2} \ln \left (c x -1\right )}{4 c^{3} d^{2}}-\frac {b^{2} \arcsin \left (c x \right )^{2} x}{2 c^{2} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{c^{3} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c^{3} d^{2}}+\frac {i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d^{2}}-\frac {b^{2} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d^{2}}+\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 c^{3} d^{2}}-\frac {i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d^{2}}+\frac {b^{2} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d^{2}}-\frac {2 i b^{2} \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{3} d^{2}}-\frac {a b \arcsin \left (c x \right ) x}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {a b \sqrt {-c^{2} x^{2}+1}}{c^{3} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d^{2}}-\frac {a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d^{2}}-\frac {i a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d^{2}}+\frac {i a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/c^3*a^2/d^2/(c*x+1)-1/4/c^3*a^2/d^2*ln(c*x+1)-1/4/c^3*a^2/d^2/(c*x-1)+1/4/c^3*a^2/d^2*ln(c*x-1)-1/2/c^2*b
^2/d^2/(c^2*x^2-1)*arcsin(c*x)^2*x+1/c^3*b^2/d^2/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-1/2/c^3*b^2/d^2*ar
csin(c*x)^2*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*I/c^3*b^2/d^2*arctan(I*c*x+(-c^2*x^2+1)^(1/2))-b^2*polylog(3,
I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^3/d^2+1/2/c^3*b^2/d^2*arcsin(c*x)^2*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-I/c^3*a
*b/d^2*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+b^2*polylog(3,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^3/d^2+I/c^3*a*b/d^
2*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-1/c^2*a*b/d^2/(c^2*x^2-1)*arcsin(c*x)*x+1/c^3*a*b/d^2/(c^2*x^2-1)*(-c^
2*x^2+1)^(1/2)+1/c^3*a*b/d^2*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-1/c^3*a*b/d^2*arcsin(c*x)*ln(1-I*(
I*c*x+(-c^2*x^2+1)^(1/2)))+I/c^3*b^2/d^2*arcsin(c*x)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))-I/c^3*b^2/d^2*arc
sin(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 \[ -\frac {1}{4} \, a^{2} {\left (\frac {2 \, x}{c^{4} d^{2} x^{2} - c^{2} d^{2}} + \frac {\log \left (c x + 1\right )}{c^{3} d^{2}} - \frac {\log \left (c x - 1\right )}{c^{3} d^{2}}\right )} - \frac {2 \, b^{2} c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (c x + 1\right ) - {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \, {\left (c^{5} d^{2} x^{2} - c^{3} d^{2}\right )} \int \frac {4 \, a b c^{2} x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (2 \, b^{2} c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{6} d^{2} x^{4} - 2 \, c^{4} d^{2} x^{2} + c^{2} d^{2}}\,{d x}}{4 \, {\left (c^{5} d^{2} x^{2} - c^{3} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*(2*x/(c^4*d^2*x^2 - c^2*d^2) + log(c*x + 1)/(c^3*d^2) - log(c*x - 1)/(c^3*d^2)) - 1/4*(2*b^2*c*x*arct
an2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + (b^2*c^2*x^2 - b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*lo
g(c*x + 1) - (b^2*c^2*x^2 - b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(-c*x + 1) + 4*(c^5*d^2*x^2 -
c^3*d^2)*integrate(-1/2*(4*a*b*c^2*x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) - (2*b^2*c*x*arctan2(c*x, s
qrt(c*x + 1)*sqrt(-c*x + 1)) + (b^2*c^2*x^2 - b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - (
b^2*c^2*x^2 - b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^
6*d^2*x^4 - 2*c^4*d^2*x^2 + c^2*d^2), x))/(c^5*d^2*x^2 - c^3*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x^{2}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{2} \operatorname {asin}^{2}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{2} \operatorname {asin}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*x**2/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b**2*x**2*asin(c*x)**2/(c**4*x**4 - 2*c**2*x*
*2 + 1), x) + Integral(2*a*b*x**2*asin(c*x)/(c**4*x**4 - 2*c**2*x**2 + 1), x))/d**2

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