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

Optimal. Leaf size=98 \[ -\frac {3 \sin ^{-1}(a x)^2}{32 a^4}-\frac {3 x^2}{32 a^2}+\frac {x^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{8 a}+\frac {3 x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{16 a^3}+\frac {1}{4} x^4 \sin ^{-1}(a x)^2-\frac {x^4}{32} \]

[Out]

-3/32*x^2/a^2-1/32*x^4-3/32*arcsin(a*x)^2/a^4+1/4*x^4*arcsin(a*x)^2+3/16*x*arcsin(a*x)*(-a^2*x^2+1)^(1/2)/a^3+
1/8*x^3*arcsin(a*x)*(-a^2*x^2+1)^(1/2)/a

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Rubi [A]  time = 0.16, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4627, 4707, 4641, 30} \[ -\frac {3 x^2}{32 a^2}+\frac {x^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{8 a}+\frac {3 x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{16 a^3}-\frac {3 \sin ^{-1}(a x)^2}{32 a^4}+\frac {1}{4} x^4 \sin ^{-1}(a x)^2-\frac {x^4}{32} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSin[a*x]^2,x]

[Out]

(-3*x^2)/(32*a^2) - x^4/32 + (3*x*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(16*a^3) + (x^3*Sqrt[1 - a^2*x^2]*ArcSin[a*x]
)/(8*a) - (3*ArcSin[a*x]^2)/(32*a^4) + (x^4*ArcSin[a*x]^2)/4

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 x^3 \sin ^{-1}(a x)^2 \, dx &=\frac {1}{4} x^4 \sin ^{-1}(a x)^2-\frac {1}{2} a \int \frac {x^4 \sin ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {x^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{8 a}+\frac {1}{4} x^4 \sin ^{-1}(a x)^2-\frac {\int x^3 \, dx}{8}-\frac {3 \int \frac {x^2 \sin ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{8 a}\\ &=-\frac {x^4}{32}+\frac {3 x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{16 a^3}+\frac {x^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{8 a}+\frac {1}{4} x^4 \sin ^{-1}(a x)^2-\frac {3 \int \frac {\sin ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{16 a^3}-\frac {3 \int x \, dx}{16 a^2}\\ &=-\frac {3 x^2}{32 a^2}-\frac {x^4}{32}+\frac {3 x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{16 a^3}+\frac {x^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{8 a}-\frac {3 \sin ^{-1}(a x)^2}{32 a^4}+\frac {1}{4} x^4 \sin ^{-1}(a x)^2\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 74, normalized size = 0.76 \[ \frac {\left (8 a^4 x^4-3\right ) \sin ^{-1}(a x)^2-a^2 x^2 \left (a^2 x^2+3\right )+2 a x \sqrt {1-a^2 x^2} \left (2 a^2 x^2+3\right ) \sin ^{-1}(a x)}{32 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSin[a*x]^2,x]

[Out]

(-(a^2*x^2*(3 + a^2*x^2)) + 2*a*x*Sqrt[1 - a^2*x^2]*(3 + 2*a^2*x^2)*ArcSin[a*x] + (-3 + 8*a^4*x^4)*ArcSin[a*x]
^2)/(32*a^4)

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fricas [A]  time = 0.75, size = 70, normalized size = 0.71 \[ -\frac {a^{4} x^{4} + 3 \, a^{2} x^{2} - {\left (8 \, a^{4} x^{4} - 3\right )} \arcsin \left (a x\right )^{2} - 2 \, {\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )}{32 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(a*x)^2,x, algorithm="fricas")

[Out]

-1/32*(a^4*x^4 + 3*a^2*x^2 - (8*a^4*x^4 - 3)*arcsin(a*x)^2 - 2*(2*a^3*x^3 + 3*a*x)*sqrt(-a^2*x^2 + 1)*arcsin(a
*x))/a^4

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giac [A]  time = 0.16, size = 133, normalized size = 1.36 \[ -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x \arcsin \left (a x\right )}{8 \, a^{3}} + \frac {{\left (a^{2} x^{2} - 1\right )}^{2} \arcsin \left (a x\right )^{2}}{4 \, a^{4}} + \frac {5 \, \sqrt {-a^{2} x^{2} + 1} x \arcsin \left (a x\right )}{16 \, a^{3}} + \frac {{\left (a^{2} x^{2} - 1\right )} \arcsin \left (a x\right )^{2}}{2 \, a^{4}} - \frac {{\left (a^{2} x^{2} - 1\right )}^{2}}{32 \, a^{4}} + \frac {5 \, \arcsin \left (a x\right )^{2}}{32 \, a^{4}} - \frac {5 \, {\left (a^{2} x^{2} - 1\right )}}{32 \, a^{4}} - \frac {17}{256 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(a*x)^2,x, algorithm="giac")

[Out]

-1/8*(-a^2*x^2 + 1)^(3/2)*x*arcsin(a*x)/a^3 + 1/4*(a^2*x^2 - 1)^2*arcsin(a*x)^2/a^4 + 5/16*sqrt(-a^2*x^2 + 1)*
x*arcsin(a*x)/a^3 + 1/2*(a^2*x^2 - 1)*arcsin(a*x)^2/a^4 - 1/32*(a^2*x^2 - 1)^2/a^4 + 5/32*arcsin(a*x)^2/a^4 -
5/32*(a^2*x^2 - 1)/a^4 - 17/256/a^4

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maple [A]  time = 0.14, size = 93, normalized size = 0.95 \[ \frac {\frac {a^{4} x^{4} \arcsin \left (a x \right )^{2}}{4}-\frac {\arcsin \left (a x \right ) \left (-2 a^{3} x^{3} \sqrt {-a^{2} x^{2}+1}-3 a x \sqrt {-a^{2} x^{2}+1}+3 \arcsin \left (a x \right )\right )}{16}+\frac {3 \arcsin \left (a x \right )^{2}}{32}-\frac {a^{4} x^{4}}{32}-\frac {3 a^{2} x^{2}}{32}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsin(a*x)^2,x)

[Out]

1/a^4*(1/4*a^4*x^4*arcsin(a*x)^2-1/16*arcsin(a*x)*(-2*a^3*x^3*(-a^2*x^2+1)^(1/2)-3*a*x*(-a^2*x^2+1)^(1/2)+3*ar
csin(a*x))+3/32*arcsin(a*x)^2-1/32*a^4*x^4-3/32*a^2*x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, x^{4} \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )^{2} + a \int \frac {\sqrt {a x + 1} \sqrt {-a x + 1} x^{4} \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )}{2 \, {\left (a^{2} x^{2} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(a*x)^2,x, algorithm="maxima")

[Out]

1/4*x^4*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2 + a*integrate(1/2*sqrt(a*x + 1)*sqrt(-a*x + 1)*x^4*arctan
2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))/(a^2*x^2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*asin(a*x)^2,x)

[Out]

int(x^3*asin(a*x)^2, x)

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sympy [A]  time = 1.85, size = 90, normalized size = 0.92 \[ \begin {cases} \frac {x^{4} \operatorname {asin}^{2}{\left (a x \right )}}{4} - \frac {x^{4}}{32} + \frac {x^{3} \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}{\left (a x \right )}}{8 a} - \frac {3 x^{2}}{32 a^{2}} + \frac {3 x \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}{\left (a x \right )}}{16 a^{3}} - \frac {3 \operatorname {asin}^{2}{\left (a x \right )}}{32 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asin(a*x)**2,x)

[Out]

Piecewise((x**4*asin(a*x)**2/4 - x**4/32 + x**3*sqrt(-a**2*x**2 + 1)*asin(a*x)/(8*a) - 3*x**2/(32*a**2) + 3*x*
sqrt(-a**2*x**2 + 1)*asin(a*x)/(16*a**3) - 3*asin(a*x)**2/(32*a**4), Ne(a, 0)), (0, True))

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