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

Optimal. Leaf size=218 \[ \frac {2 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {2 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {2 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {2 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {2 b^2 x}{c^2 d} \]

[Out]

2*b^2*x/c^2/d-x*(a+b*arcsin(c*x))^2/c^2/d-2*I*(a+b*arcsin(c*x))^2*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/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*arcsin(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^2*x^2+1)^(1/2)/c^3/d

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Rubi [A]  time = 0.29, antiderivative size = 218, 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 = {4715, 4657, 4181, 2531, 2282, 6589, 4677, 8} \[ \frac {2 i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {2 i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {2 b^2 \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {2 b^2 \text {PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d}+\frac {2 b^2 x}{c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*x)/(c^2*d) - (2*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(c^3*d) - (x*(a + b*ArcSin[c*x])^2)/(c^2*d) -
((2*I)*(a + b*ArcSin[c*x])^2*ArcTan[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*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)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 4715

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)/(e*(m + 2*p + 1)), x] + (Dist[(f^2*(m - 1))/(c^2*(m
+ 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(c*(m + 2*p + 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] && GtQ[m,
1] && NeQ[m + 2*p + 1, 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 {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx &=-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d}+\frac {\int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{c^2}+\frac {(2 b) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{c d}\\ &=-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d}+\frac {\operatorname {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d}+\frac {\left (2 b^2\right ) \int 1 \, dx}{c^2 d}\\ &=\frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d}-\frac {(2 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}+\frac {(2 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}\\ &=\frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {2 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}-\frac {2 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}-\frac {\left (2 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}+\frac {\left (2 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}\\ &=\frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {2 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}-\frac {2 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}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^3 d}\\ &=\frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {2 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}-\frac {2 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}-\frac {2 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}+\frac {2 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 317, normalized size = 1.45 \[ -\frac {2 a^2 c x+a^2 \log (1-c x)-a^2 \log (c x+1)+4 a b \sqrt {1-c^2 x^2}-4 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+4 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+4 a b c x \sin ^{-1}(c x)-4 a b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+4 a b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+4 b^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)+4 b^2 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )-4 b^2 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )-4 b^2 c x+2 b^2 c x \sin ^{-1}(c x)^2-2 b^2 \sin ^{-1}(c x)^2 \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+2 b^2 \sin ^{-1}(c x)^2 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} x^{2} \arcsin \left (c x\right )^{2} + 2 \, a b x^{2} \arcsin \left (c x\right ) + a^{2} x^{2}}{c^{2} d x^{2} - d}, x\right ) \]

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),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^2*d*x^2 - d), x)

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

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),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a +b \arcsin \left (c x \right )\right )^{2}}{-c^{2} d \,x^{2}+d}\, dx \]

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),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a^{2} {\left (\frac {2 \, x}{c^{2} d} - \frac {\log \left (c x + 1\right )}{c^{3} d} + \frac {\log \left (c x - 1\right )}{c^{3} d}\right )} - \frac {2 \, b^{2} c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, c^{3} d \int \frac {2 \, a b c^{2} x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + {\left (2 \, b^{2} c x \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) + b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{4} d x^{2} - c^{2} d}\,{d x} - b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (c x + 1\right ) + b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left (-c x + 1\right )}{2 \, c^{3} d} \]

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),x, algorithm="maxima")

[Out]

-1/2*a^2*(2*x/(c^2*d) - log(c*x + 1)/(c^3*d) + log(c*x - 1)/(c^3*d)) - 1/2*(2*b^2*c*x*arctan2(c*x, sqrt(c*x +
1)*sqrt(-c*x + 1))^2 - 2*c^3*d*integrate(-(2*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, sqrt(c*x + 1)*sqrt(-c*x + 1)) - b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) +
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^4*d*x^2 - c^2*d
), x) - b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(c*x + 1) + b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c
*x + 1))^2*log(-c*x + 1))/(c^3*d)

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

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),x)

[Out]

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

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

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),x)

[Out]

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

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