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

Optimal. Leaf size=400 \[ \frac {2 b x^5 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 c \sqrt {d-c^2 d x^2}}-\frac {x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{5 c^2 d}-\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^6 d}+\frac {16 a b x \sqrt {1-c^2 x^2}}{15 c^5 \sqrt {d-c^2 d x^2}}-\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4 d}+\frac {8 b x^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c^3 \sqrt {d-c^2 d x^2}}+\frac {2 b^2 \left (1-c^2 x^2\right )^3}{125 c^6 \sqrt {d-c^2 d x^2}}-\frac {76 b^2 \left (1-c^2 x^2\right )^2}{675 c^6 \sqrt {d-c^2 d x^2}}+\frac {298 b^2 \left (1-c^2 x^2\right )}{225 c^6 \sqrt {d-c^2 d x^2}}+\frac {16 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{15 c^5 \sqrt {d-c^2 d x^2}} \]

[Out]

298/225*b^2*(-c^2*x^2+1)/c^6/(-c^2*d*x^2+d)^(1/2)-76/675*b^2*(-c^2*x^2+1)^2/c^6/(-c^2*d*x^2+d)^(1/2)+2/125*b^2
*(-c^2*x^2+1)^3/c^6/(-c^2*d*x^2+d)^(1/2)+16/15*a*b*x*(-c^2*x^2+1)^(1/2)/c^5/(-c^2*d*x^2+d)^(1/2)+16/15*b^2*x*a
rcsin(c*x)*(-c^2*x^2+1)^(1/2)/c^5/(-c^2*d*x^2+d)^(1/2)+8/45*b*x^3*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c^3/(-c
^2*d*x^2+d)^(1/2)+2/25*b*x^5*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c/(-c^2*d*x^2+d)^(1/2)-8/15*(a+b*arcsin(c*x)
)^2*(-c^2*d*x^2+d)^(1/2)/c^6/d-4/15*x^2*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/c^4/d-1/5*x^4*(a+b*arcsin(c*x
))^2*(-c^2*d*x^2+d)^(1/2)/c^2/d

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Rubi [A]  time = 0.58, antiderivative size = 400, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {4707, 4677, 4619, 261, 4627, 266, 43} \[ \frac {16 a b x \sqrt {1-c^2 x^2}}{15 c^5 \sqrt {d-c^2 d x^2}}+\frac {2 b x^5 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 c \sqrt {d-c^2 d x^2}}-\frac {x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{5 c^2 d}+\frac {8 b x^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c^3 \sqrt {d-c^2 d x^2}}-\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4 d}-\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^6 d}+\frac {2 b^2 \left (1-c^2 x^2\right )^3}{125 c^6 \sqrt {d-c^2 d x^2}}-\frac {76 b^2 \left (1-c^2 x^2\right )^2}{675 c^6 \sqrt {d-c^2 d x^2}}+\frac {298 b^2 \left (1-c^2 x^2\right )}{225 c^6 \sqrt {d-c^2 d x^2}}+\frac {16 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{15 c^5 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a*b*x*Sqrt[1 - c^2*x^2])/(15*c^5*Sqrt[d - c^2*d*x^2]) + (298*b^2*(1 - c^2*x^2))/(225*c^6*Sqrt[d - c^2*d*x^
2]) - (76*b^2*(1 - c^2*x^2)^2)/(675*c^6*Sqrt[d - c^2*d*x^2]) + (2*b^2*(1 - c^2*x^2)^3)/(125*c^6*Sqrt[d - c^2*d
*x^2]) + (16*b^2*x*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(15*c^5*Sqrt[d - c^2*d*x^2]) + (8*b*x^3*Sqrt[1 - c^2*x^2]*(a
+ b*ArcSin[c*x]))/(45*c^3*Sqrt[d - c^2*d*x^2]) + (2*b*x^5*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(25*c*Sqrt[d
- c^2*d*x^2]) - (8*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(15*c^6*d) - (4*x^2*Sqrt[d - c^2*d*x^2]*(a + b*
ArcSin[c*x])^2)/(15*c^4*d) - (x^4*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(5*c^2*d)

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 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 \frac {x^5 \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d-c^2 d x^2}} \, dx &=-\frac {x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{5 c^2 d}+\frac {4 \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d-c^2 d x^2}} \, dx}{5 c^2}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int x^4 \left (a+b \sin ^{-1}(c x)\right ) \, dx}{5 c \sqrt {d-c^2 d x^2}}\\ &=\frac {2 b x^5 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 c \sqrt {d-c^2 d x^2}}-\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4 d}-\frac {x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{5 c^2 d}+\frac {8 \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d-c^2 d x^2}} \, dx}{15 c^4}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {x^5}{\sqrt {1-c^2 x^2}} \, dx}{25 \sqrt {d-c^2 d x^2}}+\frac {\left (8 b \sqrt {1-c^2 x^2}\right ) \int x^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx}{15 c^3 \sqrt {d-c^2 d x^2}}\\ &=\frac {8 b x^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c^3 \sqrt {d-c^2 d x^2}}+\frac {2 b x^5 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 c \sqrt {d-c^2 d x^2}}-\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^6 d}-\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4 d}-\frac {x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{5 c^2 d}-\frac {\left (b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )}{25 \sqrt {d-c^2 d x^2}}+\frac {\left (16 b \sqrt {1-c^2 x^2}\right ) \int \left (a+b \sin ^{-1}(c x)\right ) \, dx}{15 c^5 \sqrt {d-c^2 d x^2}}-\frac {\left (8 b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {x^3}{\sqrt {1-c^2 x^2}} \, dx}{45 c^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {16 a b x \sqrt {1-c^2 x^2}}{15 c^5 \sqrt {d-c^2 d x^2}}+\frac {8 b x^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c^3 \sqrt {d-c^2 d x^2}}+\frac {2 b x^5 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 c \sqrt {d-c^2 d x^2}}-\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^6 d}-\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4 d}-\frac {x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{5 c^2 d}-\frac {\left (b^2 \sqrt {1-c^2 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 {d-c^2 d x^2}}+\frac {\left (16 b^2 \sqrt {1-c^2 x^2}\right ) \int \sin ^{-1}(c x) \, dx}{15 c^5 \sqrt {d-c^2 d x^2}}-\frac {\left (4 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )}{45 c^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {16 a b x \sqrt {1-c^2 x^2}}{15 c^5 \sqrt {d-c^2 d x^2}}+\frac {2 b^2 \left (1-c^2 x^2\right )}{25 c^6 \sqrt {d-c^2 d x^2}}-\frac {4 b^2 \left (1-c^2 x^2\right )^2}{75 c^6 \sqrt {d-c^2 d x^2}}+\frac {2 b^2 \left (1-c^2 x^2\right )^3}{125 c^6 \sqrt {d-c^2 d x^2}}+\frac {16 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{15 c^5 \sqrt {d-c^2 d x^2}}+\frac {8 b x^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c^3 \sqrt {d-c^2 d x^2}}+\frac {2 b x^5 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 c \sqrt {d-c^2 d x^2}}-\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^6 d}-\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4 d}-\frac {x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{5 c^2 d}-\frac {\left (16 b^2 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{15 c^4 \sqrt {d-c^2 d x^2}}-\frac {\left (4 b^2 \sqrt {1-c^2 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 c^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {16 a b x \sqrt {1-c^2 x^2}}{15 c^5 \sqrt {d-c^2 d x^2}}+\frac {298 b^2 \left (1-c^2 x^2\right )}{225 c^6 \sqrt {d-c^2 d x^2}}-\frac {76 b^2 \left (1-c^2 x^2\right )^2}{675 c^6 \sqrt {d-c^2 d x^2}}+\frac {2 b^2 \left (1-c^2 x^2\right )^3}{125 c^6 \sqrt {d-c^2 d x^2}}+\frac {16 b^2 x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{15 c^5 \sqrt {d-c^2 d x^2}}+\frac {8 b x^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{45 c^3 \sqrt {d-c^2 d x^2}}+\frac {2 b x^5 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{25 c \sqrt {d-c^2 d x^2}}-\frac {8 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^6 d}-\frac {4 x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{15 c^4 d}-\frac {x^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{5 c^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 230, normalized size = 0.58 \[ \frac {225 a^2 \left (3 c^6 x^6+c^4 x^4+4 c^2 x^2-8\right )+30 a b c x \sqrt {1-c^2 x^2} \left (9 c^4 x^4+20 c^2 x^2+120\right )+30 b \sin ^{-1}(c x) \left (15 a \left (3 c^6 x^6+c^4 x^4+4 c^2 x^2-8\right )+b c x \sqrt {1-c^2 x^2} \left (9 c^4 x^4+20 c^2 x^2+120\right )\right )-2 b^2 \left (27 c^6 x^6+109 c^4 x^4+1936 c^2 x^2-2072\right )+225 b^2 \left (3 c^6 x^6+c^4 x^4+4 c^2 x^2-8\right ) \sin ^{-1}(c x)^2}{3375 c^6 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*a*b*c*x*Sqrt[1 - c^2*x^2]*(120 + 20*c^2*x^2 + 9*c^4*x^4) + 225*a^2*(-8 + 4*c^2*x^2 + c^4*x^4 + 3*c^6*x^6)
- 2*b^2*(-2072 + 1936*c^2*x^2 + 109*c^4*x^4 + 27*c^6*x^6) + 30*b*(b*c*x*Sqrt[1 - c^2*x^2]*(120 + 20*c^2*x^2 +
9*c^4*x^4) + 15*a*(-8 + 4*c^2*x^2 + c^4*x^4 + 3*c^6*x^6))*ArcSin[c*x] + 225*b^2*(-8 + 4*c^2*x^2 + c^4*x^4 + 3*
c^6*x^6)*ArcSin[c*x]^2)/(3375*c^6*Sqrt[d - c^2*d*x^2])

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fricas [A]  time = 0.46, size = 276, normalized size = 0.69 \[ -\frac {30 \, {\left (9 \, a b c^{5} x^{5} + 20 \, a b c^{3} x^{3} + 120 \, a b c x + {\left (9 \, b^{2} c^{5} x^{5} + 20 \, b^{2} c^{3} x^{3} + 120 \, b^{2} c x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + {\left (27 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{6} x^{6} + {\left (225 \, a^{2} - 218 \, b^{2}\right )} c^{4} x^{4} + 4 \, {\left (225 \, a^{2} - 968 \, b^{2}\right )} c^{2} x^{2} + 225 \, {\left (3 \, b^{2} c^{6} x^{6} + b^{2} c^{4} x^{4} + 4 \, b^{2} c^{2} x^{2} - 8 \, b^{2}\right )} \arcsin \left (c x\right )^{2} - 1800 \, a^{2} + 4144 \, b^{2} + 450 \, {\left (3 \, a b c^{6} x^{6} + a b c^{4} x^{4} + 4 \, a b c^{2} x^{2} - 8 \, a b\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{3375 \, {\left (c^{8} d x^{2} - c^{6} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

-1/3375*(30*(9*a*b*c^5*x^5 + 20*a*b*c^3*x^3 + 120*a*b*c*x + (9*b^2*c^5*x^5 + 20*b^2*c^3*x^3 + 120*b^2*c*x)*arc
sin(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 + (225*a^2 - 218*b^2)*c^4*x^4
+ 4*(225*a^2 - 968*b^2)*c^2*x^2 + 225*(3*b^2*c^6*x^6 + b^2*c^4*x^4 + 4*b^2*c^2*x^2 - 8*b^2)*arcsin(c*x)^2 - 1
800*a^2 + 4144*b^2 + 450*(3*a*b*c^6*x^6 + a*b*c^4*x^4 + 4*a*b*c^2*x^2 - 8*a*b)*arcsin(c*x))*sqrt(-c^2*d*x^2 +
d))/(c^8*d*x^2 - c^6*d)

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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^5*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),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 = 0.79, size = 1020, normalized size = 2.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),x)

[Out]

a^2*(-1/5*x^4/c^2/d*(-c^2*d*x^2+d)^(1/2)+4/5/c^2*(-1/3*x^2/c^2/d*(-c^2*d*x^2+d)^(1/2)-2/3/d/c^4*(-c^2*d*x^2+d)
^(1/2)))+b^2*(5/1728*(-d*(c^2*x^2-1))^(1/2)*(2*c^2*x^2-2*I*(-c^2*x^2+1)^(1/2)*x*c-1)*(6*I*arcsin(c*x)+9*arcsin
(c*x)^2-2)/c^6/d/(c^2*x^2-1)-5/16*(-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^6/d/(c^2*x^2-1)-5/16*(-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*arcsin(c*x))/c^6/d/(c^2*x^2-1)+5/1728*(-d*(c^2*x^2-1))^(1/2)*(2*I*(-c^2*x^2+1)^(1/2)*x*c+2*c^2*x^2
-1)*(-6*I*arcsin(c*x)+9*arcsin(c*x)^2-2)/c^6/d/(c^2*x^2-1)+1/4000*(-d*(c^2*x^2-1))^(1/2)/c^6/d/(c^2*x^2-1)*(25
*arcsin(c*x)^2-2)*cos(6*arcsin(c*x))-1/400*(-d*(c^2*x^2-1))^(1/2)/c^6/d/(c^2*x^2-1)*arcsin(c*x)*sin(6*arcsin(c
*x))-1/54000*(-d*(c^2*x^2-1))^(1/2)/c^6/d/(c^2*x^2-1)*(2475*arcsin(c*x)^2-598)*cos(4*arcsin(c*x))+29/900*(-d*(
c^2*x^2-1))^(1/2)/c^6/d/(c^2*x^2-1)*arcsin(c*x)*sin(4*arcsin(c*x)))+2*a*b*(5/576*(-d*(c^2*x^2-1))^(1/2)*(2*c^2
*x^2-2*I*(-c^2*x^2+1)^(1/2)*x*c-1)*(I+3*arcsin(c*x))/c^6/d/(c^2*x^2-1)-5/16*(-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^6/d/(c^2*x^2-1)-5/16*(-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^6/d/(c^2*x^2-1)+5/576*(-d*(c^2*x^2-1))^(1/2)*(2*I*(-c^2*x^2+1)^(1/2)*x*c+2*c^
2*x^2-1)*(-I+3*arcsin(c*x))/c^6/d/(c^2*x^2-1)+1/160*(-d*(c^2*x^2-1))^(1/2)/c^6/d/(c^2*x^2-1)*arcsin(c*x)*cos(6
*arcsin(c*x))-1/800*(-d*(c^2*x^2-1))^(1/2)/c^6/d/(c^2*x^2-1)*sin(6*arcsin(c*x))-11/240*(-d*(c^2*x^2-1))^(1/2)/
c^6/d/(c^2*x^2-1)*arcsin(c*x)*cos(4*arcsin(c*x))+29/1800*(-d*(c^2*x^2-1))^(1/2)/c^6/d/(c^2*x^2-1)*sin(4*arcsin
(c*x)))

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maxima [A]  time = 1.47, size = 365, normalized size = 0.91 \[ -\frac {1}{15} \, {\left (\frac {3 \, \sqrt {-c^{2} d x^{2} + d} x^{4}}{c^{2} d} + \frac {4 \, \sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{4} d} + \frac {8 \, \sqrt {-c^{2} d x^{2} + d}}{c^{6} d}\right )} b^{2} \arcsin \left (c x\right )^{2} - \frac {2}{15} \, {\left (\frac {3 \, \sqrt {-c^{2} d x^{2} + d} x^{4}}{c^{2} d} + \frac {4 \, \sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{4} d} + \frac {8 \, \sqrt {-c^{2} d x^{2} + d}}{c^{6} d}\right )} a b \arcsin \left (c x\right ) - \frac {1}{15} \, {\left (\frac {3 \, \sqrt {-c^{2} d x^{2} + d} x^{4}}{c^{2} d} + \frac {4 \, \sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{4} d} + \frac {8 \, \sqrt {-c^{2} d x^{2} + d}}{c^{6} d}\right )} a^{2} + \frac {2}{3375} \, b^{2} {\left (\frac {27 \, \sqrt {-c^{2} x^{2} + 1} c^{2} x^{4} + 136 \, \sqrt {-c^{2} x^{2} + 1} x^{2} + \frac {2072 \, \sqrt {-c^{2} x^{2} + 1}}{c^{2}}}{c^{4} \sqrt {d}} + \frac {15 \, {\left (9 \, c^{4} x^{5} + 20 \, c^{2} x^{3} + 120 \, x\right )} \arcsin \left (c x\right )}{c^{5} \sqrt {d}}\right )} + \frac {2 \, {\left (9 \, c^{4} x^{5} + 20 \, c^{2} x^{3} + 120 \, x\right )} a b}{225 \, c^{5} \sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

-1/15*(3*sqrt(-c^2*d*x^2 + d)*x^4/(c^2*d) + 4*sqrt(-c^2*d*x^2 + d)*x^2/(c^4*d) + 8*sqrt(-c^2*d*x^2 + d)/(c^6*d
))*b^2*arcsin(c*x)^2 - 2/15*(3*sqrt(-c^2*d*x^2 + d)*x^4/(c^2*d) + 4*sqrt(-c^2*d*x^2 + d)*x^2/(c^4*d) + 8*sqrt(
-c^2*d*x^2 + d)/(c^6*d))*a*b*arcsin(c*x) - 1/15*(3*sqrt(-c^2*d*x^2 + d)*x^4/(c^2*d) + 4*sqrt(-c^2*d*x^2 + d)*x
^2/(c^4*d) + 8*sqrt(-c^2*d*x^2 + d)/(c^6*d))*a^2 + 2/3375*b^2*((27*sqrt(-c^2*x^2 + 1)*c^2*x^4 + 136*sqrt(-c^2*
x^2 + 1)*x^2 + 2072*sqrt(-c^2*x^2 + 1)/c^2)/(c^4*sqrt(d)) + 15*(9*c^4*x^5 + 20*c^2*x^3 + 120*x)*arcsin(c*x)/(c
^5*sqrt(d))) + 2/225*(9*c^4*x^5 + 20*c^2*x^3 + 120*x)*a*b/(c^5*sqrt(d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^5*(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 \[ \int \frac {x^{5} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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