3.3.56 \(\int \frac {x^3 (a+b \text {ArcSin}(c x))^2}{(d-c^2 d x^2)^{5/2}} \, dx\) [256]

Optimal. Leaf size=332 \[ \frac {b^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x (a+b \text {ArcSin}(c x))}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^2 (a+b \text {ArcSin}(c x))^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 (a+b \text {ArcSin}(c x))^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {10 i b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {5 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {5 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}} \]

[Out]

1/3*x^2*(a+b*arcsin(c*x))^2/c^2/d/(-c^2*d*x^2+d)^(3/2)+1/3*b^2/c^4/d^2/(-c^2*d*x^2+d)^(1/2)-2/3*(a+b*arcsin(c*
x))^2/c^4/d^2/(-c^2*d*x^2+d)^(1/2)-1/3*b*x*(a+b*arcsin(c*x))/c^3/d^2/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)-1
0/3*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^2/(-c^2*d*x^2+d)^(1/2)+5/3
*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^2/(-c^2*d*x^2+d)^(1/2)-5/3*I*b^2*poly
log(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))*(-c^2*x^2+1)^(1/2)/c^4/d^2/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.37, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {4791, 4767, 4749, 4266, 2317, 2438, 267} \begin {gather*} -\frac {10 i b \sqrt {1-c^2 x^2} \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {x^2 (a+b \text {ArcSin}(c x))^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 (a+b \text {ArcSin}(c x))^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x (a+b \text {ArcSin}(c x))}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {5 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {5 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b^2}{3 c^4 d^2 \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)^(5/2),x]

[Out]

b^2/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) - (b*x*(a + b*ArcSin[c*x]))/(3*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^
2]) + (x^2*(a + b*ArcSin[c*x])^2)/(3*c^2*d*(d - c^2*d*x^2)^(3/2)) - (2*(a + b*ArcSin[c*x])^2)/(3*c^4*d^2*Sqrt[
d - c^2*d*x^2]) - (((10*I)/3)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/(c^4*d^2*Sqrt
[d - c^2*d*x^2]) + (((5*I)/3)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^4*d^2*Sqrt[d - c^2*
d*x^2]) - (((5*I)/3)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^4*d^2*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 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]

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 )^{5/2}} \, dx &=\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{3 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 )}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (4 b \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{3 c^2 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (4 b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {10 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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (4 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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (4 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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {10 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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (4 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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (4 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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{3 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {10 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 )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {5 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {5 i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 511, normalized size = 1.54 \begin {gather*} \frac {8 a^2-2 b^2-12 a^2 c^2 x^2+4 a b \text {ArcSin}(c x)+2 b^2 \text {ArcSin}(c x)^2-2 b^2 \cos (2 \text {ArcSin}(c x))+12 a b \text {ArcSin}(c x) \cos (2 \text {ArcSin}(c x))+6 b^2 \text {ArcSin}(c x)^2 \cos (2 \text {ArcSin}(c x))-15 b^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-5 b^2 \text {ArcSin}(c x) \cos (3 \text {ArcSin}(c x)) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )+15 b^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+5 b^2 \text {ArcSin}(c x) \cos (3 \text {ArcSin}(c x)) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+15 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 )+5 a b \cos (3 \text {ArcSin}(c x)) \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-15 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 )-5 a b \cos (3 \text {ArcSin}(c x)) \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-20 i b^2 \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+20 i b^2 \left (1-c^2 x^2\right )^{3/2} \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))}{12 c^4 d^2 \left (-1+c^2 x^2\right ) \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)^(5/2),x]

[Out]

(8*a^2 - 2*b^2 - 12*a^2*c^2*x^2 + 4*a*b*ArcSin[c*x] + 2*b^2*ArcSin[c*x]^2 - 2*b^2*Cos[2*ArcSin[c*x]] + 12*a*b*
ArcSin[c*x]*Cos[2*ArcSin[c*x]] + 6*b^2*ArcSin[c*x]^2*Cos[2*ArcSin[c*x]] - 15*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]
*Log[1 - I*E^(I*ArcSin[c*x])] - 5*b^2*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 - I*E^(I*ArcSin[c*x])] + 15*b^2*Sqr
t[1 - c^2*x^2]*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 5*b^2*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 + I*E^(I*
ArcSin[c*x])] + 15*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 5*a*b*Cos[3*ArcSin[c*x
]]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - 15*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin
[c*x]/2]] - 5*a*b*Cos[3*ArcSin[c*x]]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] - (20*I)*b^2*(1 - c^2*x^2)^(
3/2)*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (20*I)*b^2*(1 - c^2*x^2)^(3/2)*PolyLog[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]])/(12*c^4*d^2*(-1 + c^2*x^2)*Sqrt[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. 827 vs. \(2 (315 ) = 630\).
time = 0.41, size = 828, normalized size = 2.49

method	result	size
default	\(a^{2} \left (\frac {x^{2}}{c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} x^{2}}{d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{2}}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, x}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{3}}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{2}}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{2}}-\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2}}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{4}}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{4}}+\frac {5 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 )}{3 c^{4} d^{3} \left (c^{2} x^{2}-1\right )}-\frac {5 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 )}{3 c^{4} d^{3} \left (c^{2} x^{2}-1\right )}-\frac {5 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 )}{3 c^{4} d^{3} \left (c^{2} x^{2}-1\right )}+\frac {5 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 )}{3 c^{4} d^{3} \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}}{d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{2}}-\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{3}}-\frac {4 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{4}}+\frac {5 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 )}{3 c^{4} d^{3} \left (c^{2} x^{2}-1\right )}-\frac {5 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 )}{3 c^{4} d^{3} \left (c^{2} x^{2}-1\right )}\)	\(828\)

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)^(5/2),x,method=_RETURNVERBOSE)

[Out]

a^2*(x^2/c^2/d/(-c^2*d*x^2+d)^(3/2)-2/3/d/c^4/(-c^2*d*x^2+d)^(3/2))+b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)
^2/c^2*arcsin(c*x)^2*x^2-1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x
-1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^2*x^2-2/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^4
*arcsin(c*x)^2+1/3*b^2*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^4+5/3*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1)
)^(1/2)/c^4/d^3/(c^2*x^2-1)*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-5/3*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2
*x^2-1))^(1/2)/c^4/d^3/(c^2*x^2-1)*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-5/3*I*b^2*(-c^2*x^2+1)^(1/2)
*(-d*(c^2*x^2-1))^(1/2)/c^4/d^3/(c^2*x^2-1)*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+5/3*I*b^2*(-c^2*x^2+1)^(1/2)
*(-d*(c^2*x^2-1))^(1/2)/c^4/d^3/(c^2*x^2-1)*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*a*b*(-d*(c^2*x^2-1))^(1/2)
/d^3/(c^2*x^2-1)^2/c^2*arcsin(c*x)*x^2-1/3*a*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^3*(-c^2*x^2+1)^(1/2)
*x-4/3*a*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^4*arcsin(c*x)+5/3*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1)
)^(1/2)/c^4/d^3/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)-5/3*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c
^4/d^3/(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)^(5/2),x, algorithm="maxima")

[Out]

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

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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)^(5/2),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^6*d^3*x^6 - 3*c^4*
d^3*x^4 + 3*c^2*d^3*x^2 - d^3), 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)**(5/2),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)^(5/2),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 )}^{5/2}} \,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)^(5/2),x)

[Out]

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

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