3.22 \(\int x^4 \sin ^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=201 \[ -\frac {16 x \sin ^{-1}(a x)}{25 a^4}-\frac {8 x^3 \sin ^{-1}(a x)}{75 a^2}+\frac {3 x^4 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a}-\frac {6 \left (1-a^2 x^2\right )^{5/2}}{625 a^5}+\frac {76 \left (1-a^2 x^2\right )^{3/2}}{1125 a^5}-\frac {298 \sqrt {1-a^2 x^2}}{375 a^5}+\frac {8 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^3}+\frac {1}{5} x^5 \sin ^{-1}(a x)^3-\frac {6}{125} x^5 \sin ^{-1}(a x) \]

[Out]

76/1125*(-a^2*x^2+1)^(3/2)/a^5-6/625*(-a^2*x^2+1)^(5/2)/a^5-16/25*x*arcsin(a*x)/a^4-8/75*x^3*arcsin(a*x)/a^2-6
/125*x^5*arcsin(a*x)+1/5*x^5*arcsin(a*x)^3-298/375*(-a^2*x^2+1)^(1/2)/a^5+8/25*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2
)/a^5+4/25*x^2*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)/a^3+3/25*x^4*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)/a

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Rubi [A]  time = 0.38, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4627, 4707, 4677, 4619, 261, 266, 43} \[ -\frac {6 \left (1-a^2 x^2\right )^{5/2}}{625 a^5}+\frac {76 \left (1-a^2 x^2\right )^{3/2}}{1125 a^5}-\frac {298 \sqrt {1-a^2 x^2}}{375 a^5}+\frac {3 x^4 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a}-\frac {8 x^3 \sin ^{-1}(a x)}{75 a^2}+\frac {4 x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^3}+\frac {8 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^5}-\frac {16 x \sin ^{-1}(a x)}{25 a^4}+\frac {1}{5} x^5 \sin ^{-1}(a x)^3-\frac {6}{125} x^5 \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcSin[a*x]^3,x]

[Out]

(-298*Sqrt[1 - a^2*x^2])/(375*a^5) + (76*(1 - a^2*x^2)^(3/2))/(1125*a^5) - (6*(1 - a^2*x^2)^(5/2))/(625*a^5) -
(16*x*ArcSin[a*x])/(25*a^4) - (8*x^3*ArcSin[a*x])/(75*a^2) - (6*x^5*ArcSin[a*x])/125 + (8*Sqrt[1 - a^2*x^2]*A
rcSin[a*x]^2)/(25*a^5) + (4*x^2*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^2)/(25*a^3) + (3*x^4*Sqrt[1 - a^2*x^2]*ArcSin[a*
x]^2)/(25*a) + (x^5*ArcSin[a*x]^3)/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 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^4 \sin ^{-1}(a x)^3 \, dx &=\frac {1}{5} x^5 \sin ^{-1}(a x)^3-\frac {1}{5} (3 a) \int \frac {x^5 \sin ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {3 x^4 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sin ^{-1}(a x)^3-\frac {6}{25} \int x^4 \sin ^{-1}(a x) \, dx-\frac {12 \int \frac {x^3 \sin ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{25 a}\\ &=-\frac {6}{125} x^5 \sin ^{-1}(a x)+\frac {4 x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^3}+\frac {3 x^4 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sin ^{-1}(a x)^3-\frac {8 \int \frac {x \sin ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{25 a^3}-\frac {8 \int x^2 \sin ^{-1}(a x) \, dx}{25 a^2}+\frac {1}{125} (6 a) \int \frac {x^5}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {8 x^3 \sin ^{-1}(a x)}{75 a^2}-\frac {6}{125} x^5 \sin ^{-1}(a x)+\frac {8 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^3}+\frac {3 x^4 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sin ^{-1}(a x)^3-\frac {16 \int \sin ^{-1}(a x) \, dx}{25 a^4}+\frac {8 \int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx}{75 a}+\frac {1}{125} (3 a) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {16 x \sin ^{-1}(a x)}{25 a^4}-\frac {8 x^3 \sin ^{-1}(a x)}{75 a^2}-\frac {6}{125} x^5 \sin ^{-1}(a x)+\frac {8 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^3}+\frac {3 x^4 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sin ^{-1}(a x)^3+\frac {16 \int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{25 a^3}+\frac {4 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )}{75 a}+\frac {1}{125} (3 a) \operatorname {Subst}\left (\int \left (\frac {1}{a^4 \sqrt {1-a^2 x}}-\frac {2 \sqrt {1-a^2 x}}{a^4}+\frac {\left (1-a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )\\ &=-\frac {86 \sqrt {1-a^2 x^2}}{125 a^5}+\frac {4 \left (1-a^2 x^2\right )^{3/2}}{125 a^5}-\frac {6 \left (1-a^2 x^2\right )^{5/2}}{625 a^5}-\frac {16 x \sin ^{-1}(a x)}{25 a^4}-\frac {8 x^3 \sin ^{-1}(a x)}{75 a^2}-\frac {6}{125} x^5 \sin ^{-1}(a x)+\frac {8 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^3}+\frac {3 x^4 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sin ^{-1}(a x)^3+\frac {4 \operatorname {Subst}\left (\int \left (\frac {1}{a^2 \sqrt {1-a^2 x}}-\frac {\sqrt {1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )}{75 a}\\ &=-\frac {298 \sqrt {1-a^2 x^2}}{375 a^5}+\frac {76 \left (1-a^2 x^2\right )^{3/2}}{1125 a^5}-\frac {6 \left (1-a^2 x^2\right )^{5/2}}{625 a^5}-\frac {16 x \sin ^{-1}(a x)}{25 a^4}-\frac {8 x^3 \sin ^{-1}(a x)}{75 a^2}-\frac {6}{125} x^5 \sin ^{-1}(a x)+\frac {8 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a^3}+\frac {3 x^4 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sin ^{-1}(a x)^3\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 122, normalized size = 0.61 \[ \frac {1125 a^5 x^5 \sin ^{-1}(a x)^3-2 \sqrt {1-a^2 x^2} \left (27 a^4 x^4+136 a^2 x^2+2072\right )-30 a x \left (9 a^4 x^4+20 a^2 x^2+120\right ) \sin ^{-1}(a x)+225 \sqrt {1-a^2 x^2} \left (3 a^4 x^4+4 a^2 x^2+8\right ) \sin ^{-1}(a x)^2}{5625 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcSin[a*x]^3,x]

[Out]

(-2*Sqrt[1 - a^2*x^2]*(2072 + 136*a^2*x^2 + 27*a^4*x^4) - 30*a*x*(120 + 20*a^2*x^2 + 9*a^4*x^4)*ArcSin[a*x] +
225*Sqrt[1 - a^2*x^2]*(8 + 4*a^2*x^2 + 3*a^4*x^4)*ArcSin[a*x]^2 + 1125*a^5*x^5*ArcSin[a*x]^3)/(5625*a^5)

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fricas [A]  time = 0.70, size = 105, normalized size = 0.52 \[ \frac {1125 \, a^{5} x^{5} \arcsin \left (a x\right )^{3} - 30 \, {\left (9 \, a^{5} x^{5} + 20 \, a^{3} x^{3} + 120 \, a x\right )} \arcsin \left (a x\right ) - {\left (54 \, a^{4} x^{4} + 272 \, a^{2} x^{2} - 225 \, {\left (3 \, a^{4} x^{4} + 4 \, a^{2} x^{2} + 8\right )} \arcsin \left (a x\right )^{2} + 4144\right )} \sqrt {-a^{2} x^{2} + 1}}{5625 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsin(a*x)^3,x, algorithm="fricas")

[Out]

1/5625*(1125*a^5*x^5*arcsin(a*x)^3 - 30*(9*a^5*x^5 + 20*a^3*x^3 + 120*a*x)*arcsin(a*x) - (54*a^4*x^4 + 272*a^2
*x^2 - 225*(3*a^4*x^4 + 4*a^2*x^2 + 8)*arcsin(a*x)^2 + 4144)*sqrt(-a^2*x^2 + 1))/a^5

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giac [A]  time = 0.18, size = 249, normalized size = 1.24 \[ \frac {{\left (a^{2} x^{2} - 1\right )}^{2} x \arcsin \left (a x\right )^{3}}{5 \, a^{4}} + \frac {2 \, {\left (a^{2} x^{2} - 1\right )} x \arcsin \left (a x\right )^{3}}{5 \, a^{4}} - \frac {6 \, {\left (a^{2} x^{2} - 1\right )}^{2} x \arcsin \left (a x\right )}{125 \, a^{4}} + \frac {x \arcsin \left (a x\right )^{3}}{5 \, a^{4}} + \frac {3 \, {\left (a^{2} x^{2} - 1\right )}^{2} \sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )^{2}}{25 \, a^{5}} - \frac {76 \, {\left (a^{2} x^{2} - 1\right )} x \arcsin \left (a x\right )}{375 \, a^{4}} - \frac {2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \arcsin \left (a x\right )^{2}}{5 \, a^{5}} - \frac {298 \, x \arcsin \left (a x\right )}{375 \, a^{4}} - \frac {6 \, {\left (a^{2} x^{2} - 1\right )}^{2} \sqrt {-a^{2} x^{2} + 1}}{625 \, a^{5}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )^{2}}{5 \, a^{5}} + \frac {76 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{1125 \, a^{5}} - \frac {298 \, \sqrt {-a^{2} x^{2} + 1}}{375 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsin(a*x)^3,x, algorithm="giac")

[Out]

1/5*(a^2*x^2 - 1)^2*x*arcsin(a*x)^3/a^4 + 2/5*(a^2*x^2 - 1)*x*arcsin(a*x)^3/a^4 - 6/125*(a^2*x^2 - 1)^2*x*arcs
in(a*x)/a^4 + 1/5*x*arcsin(a*x)^3/a^4 + 3/25*(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)*arcsin(a*x)^2/a^5 - 76/375*(a^
2*x^2 - 1)*x*arcsin(a*x)/a^4 - 2/5*(-a^2*x^2 + 1)^(3/2)*arcsin(a*x)^2/a^5 - 298/375*x*arcsin(a*x)/a^4 - 6/625*
(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)/a^5 + 3/5*sqrt(-a^2*x^2 + 1)*arcsin(a*x)^2/a^5 + 76/1125*(-a^2*x^2 + 1)^(3/
2)/a^5 - 298/375*sqrt(-a^2*x^2 + 1)/a^5

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maple [A]  time = 0.08, size = 159, normalized size = 0.79 \[ \frac {\frac {a^{5} x^{5} \arcsin \left (a x \right )^{3}}{5}+\frac {\arcsin \left (a x \right )^{2} \left (3 a^{4} x^{4}+4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{25}-\frac {16 \sqrt {-a^{2} x^{2}+1}}{25}-\frac {16 a x \arcsin \left (a x \right )}{25}-\frac {6 a^{5} x^{5} \arcsin \left (a x \right )}{125}-\frac {2 \left (3 a^{4} x^{4}+4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{625}-\frac {8 a^{3} x^{3} \arcsin \left (a x \right )}{75}-\frac {8 \left (a^{2} x^{2}+2\right ) \sqrt {-a^{2} x^{2}+1}}{225}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsin(a*x)^3,x)

[Out]

1/a^5*(1/5*a^5*x^5*arcsin(a*x)^3+1/25*arcsin(a*x)^2*(3*a^4*x^4+4*a^2*x^2+8)*(-a^2*x^2+1)^(1/2)-16/25*(-a^2*x^2
+1)^(1/2)-16/25*a*x*arcsin(a*x)-6/125*a^5*x^5*arcsin(a*x)-2/625*(3*a^4*x^4+4*a^2*x^2+8)*(-a^2*x^2+1)^(1/2)-8/7
5*a^3*x^3*arcsin(a*x)-8/225*(a^2*x^2+2)*(-a^2*x^2+1)^(1/2))

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maxima [A]  time = 0.52, size = 171, normalized size = 0.85 \[ \frac {1}{5} \, x^{5} \arcsin \left (a x\right )^{3} + \frac {1}{25} \, {\left (\frac {3 \, \sqrt {-a^{2} x^{2} + 1} x^{4}}{a^{2}} + \frac {4 \, \sqrt {-a^{2} x^{2} + 1} x^{2}}{a^{4}} + \frac {8 \, \sqrt {-a^{2} x^{2} + 1}}{a^{6}}\right )} a \arcsin \left (a x\right )^{2} - \frac {2}{5625} \, a {\left (\frac {27 \, \sqrt {-a^{2} x^{2} + 1} a^{2} x^{4} + 136 \, \sqrt {-a^{2} x^{2} + 1} x^{2} + \frac {2072 \, \sqrt {-a^{2} x^{2} + 1}}{a^{2}}}{a^{4}} + \frac {15 \, {\left (9 \, a^{4} x^{5} + 20 \, a^{2} x^{3} + 120 \, x\right )} \arcsin \left (a x\right )}{a^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsin(a*x)^3,x, algorithm="maxima")

[Out]

1/5*x^5*arcsin(a*x)^3 + 1/25*(3*sqrt(-a^2*x^2 + 1)*x^4/a^2 + 4*sqrt(-a^2*x^2 + 1)*x^2/a^4 + 8*sqrt(-a^2*x^2 +
1)/a^6)*a*arcsin(a*x)^2 - 2/5625*a*((27*sqrt(-a^2*x^2 + 1)*a^2*x^4 + 136*sqrt(-a^2*x^2 + 1)*x^2 + 2072*sqrt(-a
^2*x^2 + 1)/a^2)/a^4 + 15*(9*a^4*x^5 + 20*a^2*x^3 + 120*x)*arcsin(a*x)/a^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,{\mathrm {asin}\left (a\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*asin(a*x)^3,x)

[Out]

int(x^4*asin(a*x)^3, x)

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sympy [A]  time = 5.68, size = 196, normalized size = 0.98 \[ \begin {cases} \frac {x^{5} \operatorname {asin}^{3}{\left (a x \right )}}{5} - \frac {6 x^{5} \operatorname {asin}{\left (a x \right )}}{125} + \frac {3 x^{4} \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a x \right )}}{25 a} - \frac {6 x^{4} \sqrt {- a^{2} x^{2} + 1}}{625 a} - \frac {8 x^{3} \operatorname {asin}{\left (a x \right )}}{75 a^{2}} + \frac {4 x^{2} \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a x \right )}}{25 a^{3}} - \frac {272 x^{2} \sqrt {- a^{2} x^{2} + 1}}{5625 a^{3}} - \frac {16 x \operatorname {asin}{\left (a x \right )}}{25 a^{4}} + \frac {8 \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a x \right )}}{25 a^{5}} - \frac {4144 \sqrt {- a^{2} x^{2} + 1}}{5625 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*asin(a*x)**3,x)

[Out]

Piecewise((x**5*asin(a*x)**3/5 - 6*x**5*asin(a*x)/125 + 3*x**4*sqrt(-a**2*x**2 + 1)*asin(a*x)**2/(25*a) - 6*x*
*4*sqrt(-a**2*x**2 + 1)/(625*a) - 8*x**3*asin(a*x)/(75*a**2) + 4*x**2*sqrt(-a**2*x**2 + 1)*asin(a*x)**2/(25*a*
*3) - 272*x**2*sqrt(-a**2*x**2 + 1)/(5625*a**3) - 16*x*asin(a*x)/(25*a**4) + 8*sqrt(-a**2*x**2 + 1)*asin(a*x)*
*2/(25*a**5) - 4144*sqrt(-a**2*x**2 + 1)/(5625*a**5), Ne(a, 0)), (0, True))

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