3.3.46 \(\int \frac {x^3 (a+b \text {ArcSin}(c x))^2}{(d-c^2 d x^2)^{3/2}} \, dx\) [246]

Optimal. Leaf size=412 \[ -\frac {4 a b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {2 b^2 \left (1-c^2 x^2\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {4 b^2 x \sqrt {1-c^2 x^2} \text {ArcSin}(c x)}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b x \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 (a+b \text {ArcSin}(c x))^2}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^2}{c^4 d^2}+\frac {4 i b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.32, antiderivative size = 412, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {4791, 4767, 4715, 267, 4795, 4749, 4266, 2317, 2438} \begin {gather*} \frac {4 i b \sqrt {1-c^2 x^2} \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {x^2 (a+b \text {ArcSin}(c x))^2}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^2}{c^4 d^2}+\frac {2 b x \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {4 a b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {4 b^2 x \sqrt {1-c^2 x^2} \text {ArcSin}(c x)}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {2 b^2 \left (1-c^2 x^2\right )}{c^4 d \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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 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 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 4715

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSin[c*x])^n, x] - Dist[b*c*n, Int[
x*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 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 4767

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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 4791

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

Rule 4795

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/(c*(m + 2*p + 1)))*S
imp[(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]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {2 \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d-c^2 d x^2}} \, dx}{c^2 d}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {2 b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{1-c^2 x^2} \, dx}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\left (4 b \sqrt {1-c^2 x^2}\right ) \int \left (a+b \sin ^{-1}(c x)\right ) \, dx}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{c^2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {4 a b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b^2 \left (1-c^2 x^2\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {2 b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {\left (4 b^2 \sqrt {1-c^2 x^2}\right ) \int \sin ^{-1}(c x) \, dx}{c^3 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {4 a b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b^2 \left (1-c^2 x^2\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {4 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2}+\frac {4 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {\left (4 b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{c^2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {4 a b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {2 b^2 \left (1-c^2 x^2\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {4 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2}+\frac {4 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {\left (2 i b^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {\left (2 i b^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {4 a b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {2 b^2 \left (1-c^2 x^2\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {4 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^4 d^2}+\frac {4 i b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 369, normalized size = 0.90 \begin {gather*} \frac {4 a^2-2 b^2-2 a^2 c^2 x^2+6 a b \text {ArcSin}(c x)+3 b^2 \text {ArcSin}(c x)^2-2 b^2 \cos (2 \text {ArcSin}(c x))+2 a b \text {ArcSin}(c x) \cos (2 \text {ArcSin}(c x))+b^2 \text {ArcSin}(c x)^2 \cos (2 \text {ArcSin}(c x))-4 b^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )+4 b^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+4 a b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-4 a b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-4 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+4 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )-2 a b \sin (2 \text {ArcSin}(c x))-2 b^2 \text {ArcSin}(c x) \sin (2 \text {ArcSin}(c x))}{2 c^4 d \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2 - 2*b^2 - 2*a^2*c^2*x^2 + 6*a*b*ArcSin[c*x] + 3*b^2*ArcSin[c*x]^2 - 2*b^2*Cos[2*ArcSin[c*x]] + 2*a*b*Ar
cSin[c*x]*Cos[2*ArcSin[c*x]] + b^2*ArcSin[c*x]^2*Cos[2*ArcSin[c*x]] - 4*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[
1 - I*E^(I*ArcSin[c*x])] + 4*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 4*a*b*Sqrt[1 - c
^2*x^2]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - 4*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] + Sin[Ar
cSin[c*x]/2]] - (4*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (4*I)*b^2*Sqrt[1 - c^2*x^2]*P
olyLog[2, I*E^(I*ArcSin[c*x])] - 2*a*b*Sin[2*ArcSin[c*x]] - 2*b^2*ArcSin[c*x]*Sin[2*ArcSin[c*x]])/(2*c^4*d*Sqr
t[d - c^2*d*x^2])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 828 vs. \(2 (401 ) = 802\).
time = 0.40, size = 829, normalized size = 2.01

method	result	size
default	\(a^{2} \left (-\frac {x^{2}}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {-c^{2} d \,x^{2}+d}}\right )+\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, x}{c^{3} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2}}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x}{c^{3} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {4 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}\)	\(829\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(-x^2/c^2/d/(-c^2*d*x^2+d)^(1/2)+2/d/c^4/(-c^2*d*x^2+d)^(1/2))+2*b^2*(-d*(c^2*x^2-1))^(1/2)/c^3/d^2/(c^2*x
^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x+b^2*(-d*(c^2*x^2-1))^(1/2)/c^2/d^2/(c^2*x^2-1)*arcsin(c*x)^2*x^2-2*b^2*
(-d*(c^2*x^2-1))^(1/2)/c^2/d^2/(c^2*x^2-1)*x^2-2*b^2*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2*x^2-1)*arcsin(c*x)^2+
2*b^2*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2*x^2-1)+2*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^
2*x^2-1)*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2*
x^2-1)*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2*x^2-
1)*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2
*x^2-1)*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*a*b*(-d*(c^2*x^2-1))^(1/2)/c^3/d^2/(c^2*x^2-1)*(-c^2*
x^2+1)^(1/2)*x+2*a*b*(-d*(c^2*x^2-1))^(1/2)/c^2/d^2/(c^2*x^2-1)*arcsin(c*x)*x^2-4*a*b*(-d*(c^2*x^2-1))^(1/2)/c
^4/d^2/(c^2*x^2-1)*arcsin(c*x)-2*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2*x^2-1)*ln(I*c*x+(-
c^2*x^2+1)^(1/2)-I)+2*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^
(1/2)+I)

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

[Out]

-a*b*c*(2*x/(c^4*d^(3/2)) + log(c*x + 1)/(c^5*d^(3/2)) - log(c*x - 1)/(c^5*d^(3/2))) - 2*a*b*(x^2/(sqrt(-c^2*d
*x^2 + d)*c^2*d) - 2/(sqrt(-c^2*d*x^2 + d)*c^4*d))*arcsin(c*x) - a^2*(x^2/(sqrt(-c^2*d*x^2 + d)*c^2*d) - 2/(sq
rt(-c^2*d*x^2 + d)*c^4*d)) + ((c^2*x^2 - 2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*sqrt(d)*arctan2(c*x, sqrt(c*x + 1)*sq
rt(-c*x + 1))^2 - (c^6*d^2*x^2 - c^4*d^2)*sqrt(d)*integrate(2*(c^2*x^4 - 2*x^2)*arctan2(c*x, sqrt(c*x + 1)*sqr
t(-c*x + 1))/(c^3*d^2*x^2 - c*d^2), x))*b^2/(c^6*d^2*x^2 - c^4*d^2)

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

[Out]

integral((b^2*x^3*arcsin(c*x)^2 + 2*a*b*x^3*arcsin(c*x) + a^2*x^3)*sqrt(-c^2*d*x^2 + d)/(c^4*d^2*x^4 - 2*c^2*d
^2*x^2 + d^2), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real zoo

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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