3.3.21 \(\int (d-c^2 d x^2)^{3/2} (a+b \text {ArcSin}(c x))^2 \, dx\) [221]

Optimal. Leaf size=305 \[ -\frac {17}{64} b^2 d x \sqrt {d-c^2 d x^2}+\frac {1}{32} b^2 c^2 d x^3 \sqrt {d-c^2 d x^2}+\frac {17 b^2 d \sqrt {d-c^2 d x^2} \text {ArcSin}(c x)}{64 c \sqrt {1-c^2 x^2}}-\frac {5 b c d x^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{8 \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^4 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{8 \sqrt {1-c^2 x^2}}+\frac {3}{8} d x \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^2+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))^2+\frac {d \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^3}{8 b c \sqrt {1-c^2 x^2}} \]

[Out]

1/4*x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^2-17/64*b^2*d*x*(-c^2*d*x^2+d)^(1/2)+1/32*b^2*c^2*d*x^3*(-c^2*d*x
^2+d)^(1/2)+3/8*d*x*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)+17/64*b^2*d*arcsin(c*x)*(-c^2*d*x^2+d)^(1/2)/c/(-
c^2*x^2+1)^(1/2)-5/8*b*c*d*x^2*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/8*b*c^3*d*x^4*(a+b*
arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/8*d*(a+b*arcsin(c*x))^3*(-c^2*d*x^2+d)^(1/2)/b/c/(-c^2*
x^2+1)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 307, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4743, 4741, 4737, 4723, 327, 222, 4767, 201} \begin {gather*} \frac {d \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^3}{8 b c \sqrt {1-c^2 x^2}}+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))^2+\frac {3}{8} d x \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^2+\frac {b d \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{8 c}-\frac {3 b c d x^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{8 \sqrt {1-c^2 x^2}}+\frac {9 b^2 d \text {ArcSin}(c x) \sqrt {d-c^2 d x^2}}{64 c \sqrt {1-c^2 x^2}}-\frac {15}{64} b^2 d x \sqrt {d-c^2 d x^2}-\frac {1}{32} b^2 d x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*b^2*d*x*Sqrt[d - c^2*d*x^2])/64 - (b^2*d*x*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2])/32 + (9*b^2*d*Sqrt[d - c^2*
d*x^2]*ArcSin[c*x])/(64*c*Sqrt[1 - c^2*x^2]) - (3*b*c*d*x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*Sqrt[1
- c^2*x^2]) + (b*d*(1 - c^2*x^2)^(3/2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*c) + (3*d*x*Sqrt[d - c^2*d
*x^2]*(a + b*ArcSin[c*x])^2)/8 + (x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x])^2)/4 + (d*Sqrt[d - c^2*d*x^2]*(a
+ b*ArcSin[c*x])^3)/(8*b*c*Sqrt[1 - c^2*x^2])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 222

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

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSi
n[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 -
c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4737

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

Rule 4741

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

Rule 4743

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

Rubi steps

\begin {align*} \int \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{4} (3 d) \int \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx-\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=\frac {b d \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac {3}{8} d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\left (3 d \sqrt {d-c^2 d x^2}\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}} \, dx}{8 \sqrt {1-c^2 x^2}}-\frac {\left (b^2 d \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \, dx}{8 \sqrt {1-c^2 x^2}}-\frac {\left (3 b c d \sqrt {d-c^2 d x^2}\right ) \int x \left (a+b \sin ^{-1}(c x)\right ) \, dx}{4 \sqrt {1-c^2 x^2}}\\ &=-\frac {1}{32} b^2 d x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}-\frac {3 b c d x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt {1-c^2 x^2}}+\frac {b d \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac {3}{8} d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{8 b c \sqrt {1-c^2 x^2}}-\frac {\left (3 b^2 d \sqrt {d-c^2 d x^2}\right ) \int \sqrt {1-c^2 x^2} \, dx}{32 \sqrt {1-c^2 x^2}}+\frac {\left (3 b^2 c^2 d \sqrt {d-c^2 d x^2}\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{8 \sqrt {1-c^2 x^2}}\\ &=-\frac {15}{64} b^2 d x \sqrt {d-c^2 d x^2}-\frac {1}{32} b^2 d x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}-\frac {3 b c d x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt {1-c^2 x^2}}+\frac {b d \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac {3}{8} d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{8 b c \sqrt {1-c^2 x^2}}-\frac {\left (3 b^2 d \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{64 \sqrt {1-c^2 x^2}}+\frac {\left (3 b^2 d \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{16 \sqrt {1-c^2 x^2}}\\ &=-\frac {15}{64} b^2 d x \sqrt {d-c^2 d x^2}-\frac {1}{32} b^2 d x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}+\frac {9 b^2 d \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{64 c \sqrt {1-c^2 x^2}}-\frac {3 b c d x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 \sqrt {1-c^2 x^2}}+\frac {b d \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c}+\frac {3}{8} d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{8 b c \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.71, size = 329, normalized size = 1.08 \begin {gather*} \frac {32 b^2 d \sqrt {d-c^2 d x^2} \text {ArcSin}(c x)^3-96 a^2 d^{3/2} \sqrt {1-c^2 x^2} \text {ArcTan}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+8 b d \sqrt {d-c^2 d x^2} \text {ArcSin}(c x)^2 (12 a+8 b \sin (2 \text {ArcSin}(c x))+b \sin (4 \text {ArcSin}(c x)))+d \sqrt {d-c^2 d x^2} \left (160 a^2 c x \sqrt {1-c^2 x^2}-64 a^2 c^3 x^3 \sqrt {1-c^2 x^2}+64 a b \cos (2 \text {ArcSin}(c x))+4 a b \cos (4 \text {ArcSin}(c x))-32 b^2 \sin (2 \text {ArcSin}(c x))-b^2 \sin (4 \text {ArcSin}(c x))\right )+4 b d \sqrt {d-c^2 d x^2} \text {ArcSin}(c x) (16 b \cos (2 \text {ArcSin}(c x))+b \cos (4 \text {ArcSin}(c x))+4 a (8 \sin (2 \text {ArcSin}(c x))+\sin (4 \text {ArcSin}(c x))))}{256 c \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^2*d*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^3 - 96*a^2*d^(3/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2]
)/(Sqrt[d]*(-1 + c^2*x^2))] + 8*b*d*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^2*(12*a + 8*b*Sin[2*ArcSin[c*x]] + b*Sin[4
*ArcSin[c*x]]) + d*Sqrt[d - c^2*d*x^2]*(160*a^2*c*x*Sqrt[1 - c^2*x^2] - 64*a^2*c^3*x^3*Sqrt[1 - c^2*x^2] + 64*
a*b*Cos[2*ArcSin[c*x]] + 4*a*b*Cos[4*ArcSin[c*x]] - 32*b^2*Sin[2*ArcSin[c*x]] - b^2*Sin[4*ArcSin[c*x]]) + 4*b*
d*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]*(16*b*Cos[2*ArcSin[c*x]] + b*Cos[4*ArcSin[c*x]] + 4*a*(8*Sin[2*ArcSin[c*x]]
+ Sin[4*ArcSin[c*x]])))/(256*c*Sqrt[1 - c^2*x^2])

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Maple [C] Result contains complex when optimal does not.
time = 0.18, size = 929, normalized size = 3.05

method	result	size
default	\(\frac {x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} a^{2}}{4}+\frac {3 a^{2} d x \sqrt {-c^{2} d \,x^{2}+d}}{8}+\frac {3 a^{2} d^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 \sqrt {c^{2} d}}+b^{2} \left (-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{3} d}{8 c \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-8 i \sqrt {-c^{2} x^{2}+1}\, x^{4} c^{4}+8 c^{5} x^{5}+8 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}-12 c^{3} x^{3}-i \sqrt {-c^{2} x^{2}+1}+4 c x \right ) \left (4 i \arcsin \left (c x \right )+8 \arcsin \left (c x \right )^{2}-1\right ) d}{512 c \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (2 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}+2 c^{3} x^{3}-i \sqrt {-c^{2} x^{2}+1}-2 c x \right ) \left (-2 i \arcsin \left (c x \right )+2 \arcsin \left (c x \right )^{2}-1\right ) d}{16 c \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i x^{2} c^{2}-c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (68 i \arcsin \left (c x \right )+56 \arcsin \left (c x \right )^{2}-31\right ) \cos \left (3 \arcsin \left (c x \right )\right ) d}{512 c \left (c^{2} x^{2}-1\right )}+\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (20 i \arcsin \left (c x \right )+24 \arcsin \left (c x \right )^{2}-11\right ) \sin \left (3 \arcsin \left (c x \right )\right ) d}{512 c \left (c^{2} x^{2}-1\right )}\right )+2 a b \left (-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2} d}{16 c \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-8 i \sqrt {-c^{2} x^{2}+1}\, x^{4} c^{4}+8 c^{5} x^{5}+8 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}-12 c^{3} x^{3}-i \sqrt {-c^{2} x^{2}+1}+4 c x \right ) \left (i+4 \arcsin \left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (2 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}+2 c^{3} x^{3}-i \sqrt {-c^{2} x^{2}+1}-2 c x \right ) \left (-i+2 \arcsin \left (c x \right )\right ) d}{16 c \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i x^{2} c^{2}-c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (17 i+28 \arcsin \left (c x \right )\right ) \cos \left (3 \arcsin \left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}-1\right )}+\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (5 i+12 \arcsin \left (c x \right )\right ) \sin \left (3 \arcsin \left (c x \right )\right ) d}{256 c \left (c^{2} x^{2}-1\right )}\right )\)	\(929\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(-c^2*d*x^2+d)^(3/2)*a^2+3/8*a^2*d*x*(-c^2*d*x^2+d)^(1/2)+3/8*a^2*d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)
*x/(-c^2*d*x^2+d)^(1/2))+b^2*(-1/8*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/(c^2*x^2-1)*arcsin(c*x)^3*d-1/5
12*(-d*(c^2*x^2-1))^(1/2)*(-8*I*(-c^2*x^2+1)^(1/2)*x^4*c^4+8*c^5*x^5+8*I*(-c^2*x^2+1)^(1/2)*x^2*c^2-12*c^3*x^3
-I*(-c^2*x^2+1)^(1/2)+4*c*x)*(4*I*arcsin(c*x)+8*arcsin(c*x)^2-1)*d/c/(c^2*x^2-1)+1/16*(-d*(c^2*x^2-1))^(1/2)*(
2*I*(-c^2*x^2+1)^(1/2)*x^2*c^2+2*c^3*x^3-I*(-c^2*x^2+1)^(1/2)-2*c*x)*(-2*I*arcsin(c*x)+2*arcsin(c*x)^2-1)*d/c/
(c^2*x^2-1)-1/512*(-d*(c^2*x^2-1))^(1/2)*(I*x^2*c^2-c*x*(-c^2*x^2+1)^(1/2)-I)*(68*I*arcsin(c*x)+56*arcsin(c*x)
^2-31)*cos(3*arcsin(c*x))*d/c/(c^2*x^2-1)+3/512*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(2
0*I*arcsin(c*x)+24*arcsin(c*x)^2-11)*sin(3*arcsin(c*x))*d/c/(c^2*x^2-1))+2*a*b*(-3/16*(-d*(c^2*x^2-1))^(1/2)*(
-c^2*x^2+1)^(1/2)/c/(c^2*x^2-1)*arcsin(c*x)^2*d-1/256*(-d*(c^2*x^2-1))^(1/2)*(-8*I*(-c^2*x^2+1)^(1/2)*x^4*c^4+
8*c^5*x^5+8*I*(-c^2*x^2+1)^(1/2)*x^2*c^2-12*c^3*x^3-I*(-c^2*x^2+1)^(1/2)+4*c*x)*(I+4*arcsin(c*x))*d/c/(c^2*x^2
-1)+1/16*(-d*(c^2*x^2-1))^(1/2)*(2*I*(-c^2*x^2+1)^(1/2)*x^2*c^2+2*c^3*x^3-I*(-c^2*x^2+1)^(1/2)-2*c*x)*(-I+2*ar
csin(c*x))*d/c/(c^2*x^2-1)-1/256*(-d*(c^2*x^2-1))^(1/2)*(I*x^2*c^2-c*x*(-c^2*x^2+1)^(1/2)-I)*(17*I+28*arcsin(c
*x))*cos(3*arcsin(c*x))*d/c/(c^2*x^2-1)+3/256*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(5*I
+12*arcsin(c*x))*sin(3*arcsin(c*x))*d/c/(c^2*x^2-1))

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

[Out]

1/8*(2*(-c^2*d*x^2 + d)^(3/2)*x + 3*sqrt(-c^2*d*x^2 + d)*d*x + 3*d^(3/2)*arcsin(c*x)/c)*a^2 + sqrt(d)*integrat
e(-((b^2*c^2*d*x^2 - b^2*d)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*(a*b*c^2*d*x^2 - a*b*d)*arctan2(c
*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sqrt(-c*x + 1), x)

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-d*(c*x - 1)*(c*x + 1))**(3/2)*(a + b*asin(c*x))**2, x)

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

[Out]

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

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