3.210 \(\int x^3 \sqrt {d-c^2 d x^2} (a+b \sin ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=374 \[ -\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^2}-\frac {2 b c x^5 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 \sqrt {1-c^2 x^2}}+\frac {1}{5} x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {2 b x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c \sqrt {1-c^2 x^2}}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4}+\frac {4 a b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}-\frac {2 b^2 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}}{125 c^4}+\frac {26 b^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{675 c^4}+\frac {52 b^2 \sqrt {d-c^2 d x^2}}{225 c^4}+\frac {4 b^2 x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{15 c^3 \sqrt {1-c^2 x^2}} \]

[Out]

52/225*b^2*(-c^2*d*x^2+d)^(1/2)/c^4+26/675*b^2*(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/2)/c^4-2/125*b^2*(-c^2*x^2+1)^2*
(-c^2*d*x^2+d)^(1/2)/c^4-2/15*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/c^4-1/15*x^2*(a+b*arcsin(c*x))^2*(-c^2*
d*x^2+d)^(1/2)/c^2+1/5*x^4*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)+4/15*a*b*x*(-c^2*d*x^2+d)^(1/2)/c^3/(-c^2*
x^2+1)^(1/2)+4/15*b^2*x*arcsin(c*x)*(-c^2*d*x^2+d)^(1/2)/c^3/(-c^2*x^2+1)^(1/2)+2/45*b*x^3*(a+b*arcsin(c*x))*(
-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-2/25*b*c*x^5*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2
)

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Rubi [A]  time = 0.47, antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {4697, 4707, 4677, 4619, 261, 4627, 266, 43} \[ \frac {4 a b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}-\frac {2 b c x^5 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 \sqrt {1-c^2 x^2}}+\frac {1}{5} x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {2 b x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c \sqrt {1-c^2 x^2}}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^2}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4}-\frac {2 b^2 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}}{125 c^4}+\frac {26 b^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{675 c^4}+\frac {52 b^2 \sqrt {d-c^2 d x^2}}{225 c^4}+\frac {4 b^2 x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{15 c^3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(52*b^2*Sqrt[d - c^2*d*x^2])/(225*c^4) + (4*a*b*x*Sqrt[d - c^2*d*x^2])/(15*c^3*Sqrt[1 - c^2*x^2]) + (26*b^2*(1
- c^2*x^2)*Sqrt[d - c^2*d*x^2])/(675*c^4) - (2*b^2*(1 - c^2*x^2)^2*Sqrt[d - c^2*d*x^2])/(125*c^4) + (4*b^2*x*
Sqrt[d - c^2*d*x^2]*ArcSin[c*x])/(15*c^3*Sqrt[1 - c^2*x^2]) + (2*b*x^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])
)/(45*c*Sqrt[1 - c^2*x^2]) - (2*b*c*x^5*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(25*Sqrt[1 - c^2*x^2]) - (2*S
qrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(15*c^4) - (x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(15*c^2)
+ (x^4*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/5

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 261

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 266

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

Rule 4619

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 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 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 4697

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

Rule 4707

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSin[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcSin[c*x])^n)/Sqrt[d + e*x^2], x], x] + Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(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] && GtQ[m, 1] && IntegerQ[m]

Rubi steps

\begin {align*} \int x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac {1}{5} x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\sqrt {d-c^2 d x^2} \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}} \, dx}{5 \sqrt {1-c^2 x^2}}-\frac {\left (2 b c \sqrt {d-c^2 d x^2}\right ) \int x^4 \left (a+b \sin ^{-1}(c x)\right ) \, dx}{5 \sqrt {1-c^2 x^2}}\\ &=-\frac {2 b c x^5 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 \sqrt {1-c^2 x^2}}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^2}+\frac {1}{5} x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\left (2 \sqrt {d-c^2 d x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}} \, dx}{15 c^2 \sqrt {1-c^2 x^2}}+\frac {\left (2 b \sqrt {d-c^2 d x^2}\right ) \int x^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx}{15 c \sqrt {1-c^2 x^2}}+\frac {\left (2 b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {x^5}{\sqrt {1-c^2 x^2}} \, dx}{25 \sqrt {1-c^2 x^2}}\\ &=\frac {2 b x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c \sqrt {1-c^2 x^2}}-\frac {2 b c x^5 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 \sqrt {1-c^2 x^2}}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^2}+\frac {1}{5} x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {\left (2 b^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {x^3}{\sqrt {1-c^2 x^2}} \, dx}{45 \sqrt {1-c^2 x^2}}+\frac {\left (4 b \sqrt {d-c^2 d x^2}\right ) \int \left (a+b \sin ^{-1}(c x)\right ) \, dx}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {\left (b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )}{25 \sqrt {1-c^2 x^2}}\\ &=\frac {4 a b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {2 b x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c \sqrt {1-c^2 x^2}}-\frac {2 b c x^5 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 \sqrt {1-c^2 x^2}}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^2}+\frac {1}{5} x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {\left (b^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )}{45 \sqrt {1-c^2 x^2}}+\frac {\left (4 b^2 \sqrt {d-c^2 d x^2}\right ) \int \sin ^{-1}(c x) \, dx}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {\left (b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{c^4 \sqrt {1-c^2 x}}-\frac {2 \sqrt {1-c^2 x}}{c^4}+\frac {\left (1-c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right )}{25 \sqrt {1-c^2 x^2}}\\ &=-\frac {2 b^2 \sqrt {d-c^2 d x^2}}{25 c^4}+\frac {4 a b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {4 b^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{75 c^4}-\frac {2 b^2 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}}{125 c^4}+\frac {4 b^2 x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {2 b x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c \sqrt {1-c^2 x^2}}-\frac {2 b c x^5 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 \sqrt {1-c^2 x^2}}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^2}+\frac {1}{5} x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {\left (b^2 \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{c^2 \sqrt {1-c^2 x}}-\frac {\sqrt {1-c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{45 \sqrt {1-c^2 x^2}}-\frac {\left (4 b^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{15 c^2 \sqrt {1-c^2 x^2}}\\ &=\frac {52 b^2 \sqrt {d-c^2 d x^2}}{225 c^4}+\frac {4 a b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {26 b^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{675 c^4}-\frac {2 b^2 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}}{125 c^4}+\frac {4 b^2 x \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {2 b x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c \sqrt {1-c^2 x^2}}-\frac {2 b c x^5 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 \sqrt {1-c^2 x^2}}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^2}+\frac {1}{5} x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 242, normalized size = 0.65 \[ \frac {\sqrt {d-c^2 d x^2} \left (225 a^2 \sqrt {1-c^2 x^2} \left (3 c^4 x^4-c^2 x^2-2\right )-30 a b c x \left (9 c^4 x^4-5 c^2 x^2-30\right )-30 b \sin ^{-1}(c x) \left (15 a \sqrt {1-c^2 x^2} \left (-3 c^4 x^4+c^2 x^2+2\right )+b c x \left (9 c^4 x^4-5 c^2 x^2-30\right )\right )-2 b^2 \sqrt {1-c^2 x^2} \left (27 c^4 x^4+11 c^2 x^2-428\right )+225 b^2 \sqrt {1-c^2 x^2} \left (3 c^4 x^4-c^2 x^2-2\right ) \sin ^{-1}(c x)^2\right )}{3375 c^4 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d - c^2*d*x^2]*(225*a^2*Sqrt[1 - c^2*x^2]*(-2 - c^2*x^2 + 3*c^4*x^4) - 30*a*b*c*x*(-30 - 5*c^2*x^2 + 9*c
^4*x^4) - 2*b^2*Sqrt[1 - c^2*x^2]*(-428 + 11*c^2*x^2 + 27*c^4*x^4) - 30*b*(15*a*Sqrt[1 - c^2*x^2]*(2 + c^2*x^2
- 3*c^4*x^4) + b*c*x*(-30 - 5*c^2*x^2 + 9*c^4*x^4))*ArcSin[c*x] + 225*b^2*Sqrt[1 - c^2*x^2]*(-2 - c^2*x^2 + 3
*c^4*x^4)*ArcSin[c*x]^2))/(3375*c^4*Sqrt[1 - c^2*x^2])

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fricas [A]  time = 0.85, size = 277, normalized size = 0.74 \[ \frac {30 \, {\left (9 \, a b c^{5} x^{5} - 5 \, a b c^{3} x^{3} - 30 \, a b c x + {\left (9 \, b^{2} c^{5} x^{5} - 5 \, b^{2} c^{3} x^{3} - 30 \, b^{2} c x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + {\left (27 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{6} x^{6} - 4 \, {\left (225 \, a^{2} - 8 \, b^{2}\right )} c^{4} x^{4} - {\left (225 \, a^{2} - 878 \, b^{2}\right )} c^{2} x^{2} + 225 \, {\left (3 \, b^{2} c^{6} x^{6} - 4 \, b^{2} c^{4} x^{4} - b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} \arcsin \left (c x\right )^{2} + 450 \, a^{2} - 856 \, b^{2} + 450 \, {\left (3 \, a b c^{6} x^{6} - 4 \, a b c^{4} x^{4} - a b c^{2} x^{2} + 2 \, a b\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{3375 \, {\left (c^{6} x^{2} - c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

1/3375*(30*(9*a*b*c^5*x^5 - 5*a*b*c^3*x^3 - 30*a*b*c*x + (9*b^2*c^5*x^5 - 5*b^2*c^3*x^3 - 30*b^2*c*x)*arcsin(c
*x))*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + (27*(25*a^2 - 2*b^2)*c^6*x^6 - 4*(225*a^2 - 8*b^2)*c^4*x^4 - (2
25*a^2 - 878*b^2)*c^2*x^2 + 225*(3*b^2*c^6*x^6 - 4*b^2*c^4*x^4 - b^2*c^2*x^2 + 2*b^2)*arcsin(c*x)^2 + 450*a^2
- 856*b^2 + 450*(3*a*b*c^6*x^6 - 4*a*b*c^4*x^4 - a*b*c^2*x^2 + 2*a*b)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^6*
x^2 - c^4)

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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(x^3*(-c^2*d*x^2+d)^(1/2)*(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 = 0.58, size = 1165, normalized size = 3.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(-1/5*x^2*(-c^2*d*x^2+d)^(3/2)/c^2/d-2/15/d/c^4*(-c^2*d*x^2+d)^(3/2))+b^2*(1/4000*(-d*(c^2*x^2-1))^(1/2)*(
16*c^6*x^6-28*c^4*x^4-16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-5*I*(-c^2*x^2
+1)^(1/2)*x*c-1)*(10*I*arcsin(c*x)+25*arcsin(c*x)^2-2)/c^4/(c^2*x^2-1)+1/864*(-d*(c^2*x^2-1))^(1/2)*(4*c^4*x^4
-5*c^2*x^2-4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+3*I*(-c^2*x^2+1)^(1/2)*x*c+1)*(6*I*arcsin(c*x)+9*arcsin(c*x)^2-2)/c^
4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)^2-2+2*I*arcsin(c*x
))/c^4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(arcsin(c*x)^2-2-2*I*arcsi
n(c*x))/c^4/(c^2*x^2-1)+1/864*(-d*(c^2*x^2-1))^(1/2)*(4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+4*c^4*x^4-3*I*(-c^2*x^2+1
)^(1/2)*x*c-5*c^2*x^2+1)*(-6*I*arcsin(c*x)+9*arcsin(c*x)^2-2)/c^4/(c^2*x^2-1)+1/4000*(-d*(c^2*x^2-1))^(1/2)*(1
6*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+16*c^6*x^6-20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-28*c^4*x^4+5*I*(-c^2*x^2+1)^(1/2)*x*
c+13*c^2*x^2-1)*(-10*I*arcsin(c*x)+25*arcsin(c*x)^2-2)/c^4/(c^2*x^2-1))+2*a*b*(1/800*(-d*(c^2*x^2-1))^(1/2)*(1
6*c^6*x^6-28*c^4*x^4-16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-5*I*(-c^2*x^2+
1)^(1/2)*x*c-1)*(I+5*arcsin(c*x))/c^4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*
c-1)*(I+arcsin(c*x))/c^4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(arcsin(
c*x)-I)/c^4/(c^2*x^2-1)+1/288*(-d*(c^2*x^2-1))^(1/2)*(4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+4*c^4*x^4-3*I*(-c^2*x^2+1
)^(1/2)*x*c-5*c^2*x^2+1)*(-I+3*arcsin(c*x))/c^4/(c^2*x^2-1)-1/3600*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2
)*x*c+c^2*x^2-1)*(17*I+15*arcsin(c*x))*cos(4*arcsin(c*x))/c^4/(c^2*x^2-1)-1/900*(-d*(c^2*x^2-1))^(1/2)*(I*c^2*
x^2-c*x*(-c^2*x^2+1)^(1/2)-I)*(2*I+15*arcsin(c*x))*sin(4*arcsin(c*x))/c^4/(c^2*x^2-1))

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maxima [A]  time = 1.06, size = 311, normalized size = 0.83 \[ -\frac {1}{15} \, b^{2} {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} \arcsin \left (c x\right )^{2} - \frac {2}{15} \, a b {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} \arcsin \left (c x\right ) - \frac {1}{15} \, a^{2} {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} - \frac {2}{3375} \, b^{2} {\left (\frac {27 \, \sqrt {-c^{2} x^{2} + 1} c^{2} \sqrt {d} x^{4} + 11 \, \sqrt {-c^{2} x^{2} + 1} \sqrt {d} x^{2} - \frac {428 \, \sqrt {-c^{2} x^{2} + 1} \sqrt {d}}{c^{2}}}{c^{2}} + \frac {15 \, {\left (9 \, c^{4} \sqrt {d} x^{5} - 5 \, c^{2} \sqrt {d} x^{3} - 30 \, \sqrt {d} x\right )} \arcsin \left (c x\right )}{c^{3}}\right )} - \frac {2 \, {\left (9 \, c^{4} \sqrt {d} x^{5} - 5 \, c^{2} \sqrt {d} x^{3} - 30 \, \sqrt {d} x\right )} a b}{225 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

-1/15*b^2*(3*(-c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d))*arcsin(c*x)^2 - 2/15*a*b*(
3*(-c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d))*arcsin(c*x) - 1/15*a^2*(3*(-c^2*d*x^2
+ d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d)) - 2/3375*b^2*((27*sqrt(-c^2*x^2 + 1)*c^2*sqrt(d)*x
^4 + 11*sqrt(-c^2*x^2 + 1)*sqrt(d)*x^2 - 428*sqrt(-c^2*x^2 + 1)*sqrt(d)/c^2)/c^2 + 15*(9*c^4*sqrt(d)*x^5 - 5*c
^2*sqrt(d)*x^3 - 30*sqrt(d)*x)*arcsin(c*x)/c^3) - 2/225*(9*c^4*sqrt(d)*x^5 - 5*c^2*sqrt(d)*x^3 - 30*sqrt(d)*x)
*a*b/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d-c^2\,d\,x^2} \,d x \]

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)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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