3.305 \(\int \frac {x^2 \sin ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=107 \[ \frac {\sin ^{-1}(a x)^4}{8 a^3}-\frac {3 \sin ^{-1}(a x)^2}{8 a^3}-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 a^2}+\frac {3 x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{4 a^2}-\frac {3 x^2}{8 a}+\frac {3 x^2 \sin ^{-1}(a x)^2}{4 a} \]

[Out]

-3/8*x^2/a-3/8*arcsin(a*x)^2/a^3+3/4*x^2*arcsin(a*x)^2/a+1/8*arcsin(a*x)^4/a^3+3/4*x*arcsin(a*x)*(-a^2*x^2+1)^
(1/2)/a^2-1/2*x*arcsin(a*x)^3*(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.21, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4707, 4641, 4627, 30} \[ -\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 a^2}+\frac {3 x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{4 a^2}+\frac {\sin ^{-1}(a x)^4}{8 a^3}-\frac {3 \sin ^{-1}(a x)^2}{8 a^3}-\frac {3 x^2}{8 a}+\frac {3 x^2 \sin ^{-1}(a x)^2}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcSin[a*x]^3)/Sqrt[1 - a^2*x^2],x]

[Out]

(-3*x^2)/(8*a) + (3*x*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(4*a^2) - (3*ArcSin[a*x]^2)/(8*a^3) + (3*x^2*ArcSin[a*x]^
2)/(4*a) - (x*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^3)/(2*a^2) + ArcSin[a*x]^4/(8*a^3)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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^2 \sin ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 a^2}+\frac {\int \frac {\sin ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx}{2 a^2}+\frac {3 \int x \sin ^{-1}(a x)^2 \, dx}{2 a}\\ &=\frac {3 x^2 \sin ^{-1}(a x)^2}{4 a}-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 a^2}+\frac {\sin ^{-1}(a x)^4}{8 a^3}-\frac {3}{2} \int \frac {x^2 \sin ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {3 x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{4 a^2}+\frac {3 x^2 \sin ^{-1}(a x)^2}{4 a}-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 a^2}+\frac {\sin ^{-1}(a x)^4}{8 a^3}-\frac {3 \int \frac {\sin ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{4 a^2}-\frac {3 \int x \, dx}{4 a}\\ &=-\frac {3 x^2}{8 a}+\frac {3 x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{4 a^2}-\frac {3 \sin ^{-1}(a x)^2}{8 a^3}+\frac {3 x^2 \sin ^{-1}(a x)^2}{4 a}-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 a^2}+\frac {\sin ^{-1}(a x)^4}{8 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 85, normalized size = 0.79 \[ \frac {-3 a^2 x^2-4 a x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3+\left (6 a^2 x^2-3\right ) \sin ^{-1}(a x)^2+6 a x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)+\sin ^{-1}(a x)^4}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*ArcSin[a*x]^3)/Sqrt[1 - a^2*x^2],x]

[Out]

(-3*a^2*x^2 + 6*a*x*Sqrt[1 - a^2*x^2]*ArcSin[a*x] + (-3 + 6*a^2*x^2)*ArcSin[a*x]^2 - 4*a*x*Sqrt[1 - a^2*x^2]*A
rcSin[a*x]^3 + ArcSin[a*x]^4)/(8*a^3)

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fricas [A]  time = 0.46, size = 73, normalized size = 0.68 \[ -\frac {3 \, a^{2} x^{2} - \arcsin \left (a x\right )^{4} - 3 \, {\left (2 \, a^{2} x^{2} - 1\right )} \arcsin \left (a x\right )^{2} + 2 \, {\left (2 \, a x \arcsin \left (a x\right )^{3} - 3 \, a x \arcsin \left (a x\right )\right )} \sqrt {-a^{2} x^{2} + 1}}{8 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*a^2*x^2 - arcsin(a*x)^4 - 3*(2*a^2*x^2 - 1)*arcsin(a*x)^2 + 2*(2*a*x*arcsin(a*x)^3 - 3*a*x*arcsin(a*x)
)*sqrt(-a^2*x^2 + 1))/a^3

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giac [A]  time = 0.43, size = 108, normalized size = 1.01 \[ -\frac {\sqrt {-a^{2} x^{2} + 1} x \arcsin \left (a x\right )^{3}}{2 \, a^{2}} + \frac {\arcsin \left (a x\right )^{4}}{8 \, a^{3}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x \arcsin \left (a x\right )}{4 \, a^{2}} + \frac {3 \, {\left (a^{2} x^{2} - 1\right )} \arcsin \left (a x\right )^{2}}{4 \, a^{3}} + \frac {3 \, \arcsin \left (a x\right )^{2}}{8 \, a^{3}} - \frac {3 \, {\left (a^{2} x^{2} - 1\right )}}{8 \, a^{3}} - \frac {3}{16 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsin(a*x)^3/(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

-1/2*sqrt(-a^2*x^2 + 1)*x*arcsin(a*x)^3/a^2 + 1/8*arcsin(a*x)^4/a^3 + 3/4*sqrt(-a^2*x^2 + 1)*x*arcsin(a*x)/a^2
+ 3/4*(a^2*x^2 - 1)*arcsin(a*x)^2/a^3 + 3/8*arcsin(a*x)^2/a^3 - 3/8*(a^2*x^2 - 1)/a^3 - 3/16/a^3

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maple [A]  time = 0.14, size = 85, normalized size = 0.79 \[ \frac {-4 \arcsin \left (a x \right )^{3} \sqrt {-a^{2} x^{2}+1}\, x a +6 \arcsin \left (a x \right )^{2} x^{2} a^{2}+\arcsin \left (a x \right )^{4}+6 \arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}\, x a -3 a^{2} x^{2}-3 \arcsin \left (a x \right )^{2}}{8 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(-4*arcsin(a*x)^3*(-a^2*x^2+1)^(1/2)*x*a+6*arcsin(a*x)^2*x^2*a^2+arcsin(a*x)^4+6*arcsin(a*x)*(-a^2*x^2+1)^
(1/2)*x*a-3*a^2*x^2-3*arcsin(a*x)^2)/a^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \arcsin \left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsin(a*x)^3/sqrt(-a^2*x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\mathrm {asin}\left (a\,x\right )}^3}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*asin(a*x)^3)/(1 - a^2*x^2)^(1/2),x)

[Out]

int((x^2*asin(a*x)^3)/(1 - a^2*x^2)^(1/2), x)

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sympy [A]  time = 2.04, size = 100, normalized size = 0.93 \[ \begin {cases} \frac {3 x^{2} \operatorname {asin}^{2}{\left (a x \right )}}{4 a} - \frac {3 x^{2}}{8 a} - \frac {x \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}^{3}{\left (a x \right )}}{2 a^{2}} + \frac {3 x \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}{\left (a x \right )}}{4 a^{2}} + \frac {\operatorname {asin}^{4}{\left (a x \right )}}{8 a^{3}} - \frac {3 \operatorname {asin}^{2}{\left (a x \right )}}{8 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*x**2*asin(a*x)**2/(4*a) - 3*x**2/(8*a) - x*sqrt(-a**2*x**2 + 1)*asin(a*x)**3/(2*a**2) + 3*x*sqrt(
-a**2*x**2 + 1)*asin(a*x)/(4*a**2) + asin(a*x)**4/(8*a**3) - 3*asin(a*x)**2/(8*a**3), Ne(a, 0)), (0, True))

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