3.229 \(\int (d-c^2 d x^2)^{5/2} (a+b \sin ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=438 \[ \frac {5 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{48 b c \sqrt {1-c^2 x^2}}+\frac {5}{16} d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{18 c}+\frac {5 b d^2 \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{48 c}-\frac {5 b c d^2 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 \sqrt {1-c^2 x^2}}+\frac {1}{6} x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5}{24} d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{108} b^2 d^2 x \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}-\frac {245 b^2 d^2 x \sqrt {d-c^2 d x^2}}{1152}-\frac {65 b^2 d^2 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{1728}+\frac {115 b^2 d^2 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{1152 c \sqrt {1-c^2 x^2}} \]

[Out]

5/24*d*x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^2+1/6*x*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))^2-245/1152*b^2*
d^2*x*(-c^2*d*x^2+d)^(1/2)-65/1728*b^2*d^2*x*(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/2)-1/108*b^2*d^2*x*(-c^2*x^2+1)^2*
(-c^2*d*x^2+d)^(1/2)+5/48*b*d^2*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c+1/18*b*d^2*(-c^2*x
^2+1)^(5/2)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c+5/16*d^2*x*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)+115/1
152*b^2*d^2*arcsin(c*x)*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-5/16*b*c*d^2*x^2*(a+b*arcsin(c*x))*(-c^2*d*x
^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+5/48*d^2*(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 = 0.39, antiderivative size = 438, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4649, 4647, 4641, 4627, 321, 216, 4677, 195} \[ \frac {5 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{48 b c \sqrt {1-c^2 x^2}}+\frac {5}{16} d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{18 c}+\frac {5 b d^2 \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{48 c}-\frac {5 b c d^2 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 \sqrt {1-c^2 x^2}}+\frac {1}{6} x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5}{24} d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{108} b^2 d^2 x \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}-\frac {245 b^2 d^2 x \sqrt {d-c^2 d x^2}}{1152}-\frac {65 b^2 d^2 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{1728}+\frac {115 b^2 d^2 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{1152 c \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x])^2,x]

[Out]

(-245*b^2*d^2*x*Sqrt[d - c^2*d*x^2])/1152 - (65*b^2*d^2*x*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2])/1728 - (b^2*d^2*x
*(1 - c^2*x^2)^2*Sqrt[d - c^2*d*x^2])/108 + (115*b^2*d^2*Sqrt[d - c^2*d*x^2]*ArcSin[c*x])/(1152*c*Sqrt[1 - c^2
*x^2]) - (5*b*c*d^2*x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(16*Sqrt[1 - c^2*x^2]) + (5*b*d^2*(1 - c^2*x^
2)^(3/2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(48*c) + (b*d^2*(1 - c^2*x^2)^(5/2)*Sqrt[d - c^2*d*x^2]*(a +
b*ArcSin[c*x]))/(18*c) + (5*d^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/16 + (5*d*x*(d - c^2*d*x^2)^(3/2
)*(a + b*ArcSin[c*x])^2)/24 + (x*(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x])^2)/6 + (5*d^2*Sqrt[d - c^2*d*x^2]*(
a + b*ArcSin[c*x])^3)/(48*b*c*Sqrt[1 - c^2*x^2])

Rule 195

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 216

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 321

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 4627

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 4641

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

Rule 4647

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[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(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 4649

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c
^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && Gt
Q[n, 0] && GtQ[p, 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]

Rubi steps

\begin {align*} \int \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac {1}{6} x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{6} (5 d) \int \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx-\frac {\left (b c d^2 \sqrt {d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx}{3 \sqrt {1-c^2 x^2}}\\ &=\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{18 c}+\frac {5}{24} d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{6} x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{8} \left (5 d^2\right ) \int \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx-\frac {\left (b^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^{5/2} \, dx}{18 \sqrt {1-c^2 x^2}}-\frac {\left (5 b c d^2 \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}{12 \sqrt {1-c^2 x^2}}\\ &=-\frac {1}{108} b^2 d^2 x \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}+\frac {5 b d^2 \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{48 c}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{18 c}+\frac {5}{16} d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5}{24} d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{6} x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\left (5 d^2 \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}{16 \sqrt {1-c^2 x^2}}-\frac {\left (5 b^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \, dx}{108 \sqrt {1-c^2 x^2}}-\frac {\left (5 b^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \, dx}{48 \sqrt {1-c^2 x^2}}-\frac {\left (5 b c d^2 \sqrt {d-c^2 d x^2}\right ) \int x \left (a+b \sin ^{-1}(c x)\right ) \, dx}{8 \sqrt {1-c^2 x^2}}\\ &=-\frac {65 b^2 d^2 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{1728}-\frac {1}{108} b^2 d^2 x \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}-\frac {5 b c d^2 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 \sqrt {1-c^2 x^2}}+\frac {5 b d^2 \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{48 c}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{18 c}+\frac {5}{16} d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5}{24} d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{6} x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{48 b c \sqrt {1-c^2 x^2}}-\frac {\left (5 b^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \sqrt {1-c^2 x^2} \, dx}{144 \sqrt {1-c^2 x^2}}-\frac {\left (5 b^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \sqrt {1-c^2 x^2} \, dx}{64 \sqrt {1-c^2 x^2}}+\frac {\left (5 b^2 c^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{16 \sqrt {1-c^2 x^2}}\\ &=-\frac {245 b^2 d^2 x \sqrt {d-c^2 d x^2}}{1152}-\frac {65 b^2 d^2 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{1728}-\frac {1}{108} b^2 d^2 x \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}-\frac {5 b c d^2 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 \sqrt {1-c^2 x^2}}+\frac {5 b d^2 \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{48 c}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{18 c}+\frac {5}{16} d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5}{24} d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{6} x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{48 b c \sqrt {1-c^2 x^2}}-\frac {\left (5 b^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{288 \sqrt {1-c^2 x^2}}-\frac {\left (5 b^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{128 \sqrt {1-c^2 x^2}}+\frac {\left (5 b^2 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{32 \sqrt {1-c^2 x^2}}\\ &=-\frac {245 b^2 d^2 x \sqrt {d-c^2 d x^2}}{1152}-\frac {65 b^2 d^2 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{1728}-\frac {1}{108} b^2 d^2 x \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}+\frac {115 b^2 d^2 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{1152 c \sqrt {1-c^2 x^2}}-\frac {5 b c d^2 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 \sqrt {1-c^2 x^2}}+\frac {5 b d^2 \left (1-c^2 x^2\right )^{3/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{48 c}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{18 c}+\frac {5}{16} d^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5}{24} d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{6} x \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {5 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{48 b c \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 2.09, size = 407, normalized size = 0.93 \[ \frac {d^2 \left (\sqrt {d-c^2 d x^2} \left (9504 a^2 c x \sqrt {1-c^2 x^2}+2304 a^2 c^5 x^5 \sqrt {1-c^2 x^2}-7488 a^2 c^3 x^3 \sqrt {1-c^2 x^2}+3240 a b \cos \left (2 \sin ^{-1}(c x)\right )+324 a b \cos \left (4 \sin ^{-1}(c x)\right )+24 a b \cos \left (6 \sin ^{-1}(c x)\right )-1620 b^2 \sin \left (2 \sin ^{-1}(c x)\right )-81 b^2 \sin \left (4 \sin ^{-1}(c x)\right )-4 b^2 \sin \left (6 \sin ^{-1}(c x)\right )\right )-4320 a^2 \sqrt {d} \sqrt {1-c^2 x^2} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )+72 b \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)^2 \left (60 a+45 b \sin \left (2 \sin ^{-1}(c x)\right )+9 b \sin \left (4 \sin ^{-1}(c x)\right )+b \sin \left (6 \sin ^{-1}(c x)\right )\right )+12 b \sqrt {d-c^2 d x^2} \sin ^{-1}(c x) \left (540 a \sin \left (2 \sin ^{-1}(c x)\right )+108 a \sin \left (4 \sin ^{-1}(c x)\right )+12 a \sin \left (6 \sin ^{-1}(c x)\right )+270 b \cos \left (2 \sin ^{-1}(c x)\right )+27 b \cos \left (4 \sin ^{-1}(c x)\right )+2 b \cos \left (6 \sin ^{-1}(c x)\right )\right )+1440 b^2 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)^3\right )}{13824 c \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x])^2,x]

[Out]

(d^2*(1440*b^2*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^3 - 4320*a^2*Sqrt[d]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d - c^2
*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + 12*b*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]*(270*b*Cos[2*ArcSin[c*x]] + 27*b*Cos
[4*ArcSin[c*x]] + 2*b*Cos[6*ArcSin[c*x]] + 540*a*Sin[2*ArcSin[c*x]] + 108*a*Sin[4*ArcSin[c*x]] + 12*a*Sin[6*Ar
cSin[c*x]]) + 72*b*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^2*(60*a + 45*b*Sin[2*ArcSin[c*x]] + 9*b*Sin[4*ArcSin[c*x]]
+ b*Sin[6*ArcSin[c*x]]) + Sqrt[d - c^2*d*x^2]*(9504*a^2*c*x*Sqrt[1 - c^2*x^2] - 7488*a^2*c^3*x^3*Sqrt[1 - c^2*
x^2] + 2304*a^2*c^5*x^5*Sqrt[1 - c^2*x^2] + 3240*a*b*Cos[2*ArcSin[c*x]] + 324*a*b*Cos[4*ArcSin[c*x]] + 24*a*b*
Cos[6*ArcSin[c*x]] - 1620*b^2*Sin[2*ArcSin[c*x]] - 81*b^2*Sin[4*ArcSin[c*x]] - 4*b^2*Sin[6*ArcSin[c*x]])))/(13
824*c*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c^{4} d^{2} x^{4} - 2 \, a^{2} c^{2} d^{2} x^{2} + a^{2} d^{2} + {\left (b^{2} c^{4} d^{2} x^{4} - 2 \, b^{2} c^{2} d^{2} x^{2} + b^{2} d^{2}\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (a b c^{4} d^{2} x^{4} - 2 \, a b c^{2} d^{2} x^{2} + a b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.34, size = 5048, normalized size = 11.53 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{48} \, {\left (8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x + 10 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x + 15 \, \sqrt {-c^{2} d x^{2} + d} d^{2} x + \frac {15 \, d^{\frac {5}{2}} \arcsin \left (c x\right )}{c}\right )} a^{2} + \sqrt {d} \int {\left ({\left (b^{2} c^{4} d^{2} x^{4} - 2 \, b^{2} c^{2} d^{2} x^{2} + b^{2} d^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, {\left (a b c^{4} d^{2} x^{4} - 2 \, a b c^{2} d^{2} x^{2} + a b d^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

1/48*(8*(-c^2*d*x^2 + d)^(5/2)*x + 10*(-c^2*d*x^2 + d)^(3/2)*d*x + 15*sqrt(-c^2*d*x^2 + d)*d^2*x + 15*d^(5/2)*
arcsin(c*x)/c)*a^2 + sqrt(d)*integrate(((b^2*c^4*d^2*x^4 - 2*b^2*c^2*d^2*x^2 + b^2*d^2)*arctan2(c*x, sqrt(c*x
+ 1)*sqrt(-c*x + 1))^2 + 2*(a*b*c^4*d^2*x^4 - 2*a*b*c^2*d^2*x^2 + a*b*d^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*
x + 1)))*sqrt(c*x + 1)*sqrt(-c*x + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d-c^2\,d\,x^2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*d*x**2+d)**(5/2)*(a+b*asin(c*x))**2,x)

[Out]

Integral((-d*(c*x - 1)*(c*x + 1))**(5/2)*(a + b*asin(c*x))**2, x)

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