3.3.66 \(\int \frac {x^2 \text {ArcSin}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\) [266]

Optimal. Leaf size=89 \[ \frac {x \sqrt {1-a^2 x^2}}{4 a^2}-\frac {\text {ArcSin}(a x)}{4 a^3}+\frac {x^2 \text {ArcSin}(a x)}{2 a}-\frac {x \sqrt {1-a^2 x^2} \text {ArcSin}(a x)^2}{2 a^2}+\frac {\text {ArcSin}(a x)^3}{6 a^3} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4795, 4737, 4723, 327, 222} \begin {gather*} \frac {\text {ArcSin}(a x)^3}{6 a^3}-\frac {\text {ArcSin}(a x)}{4 a^3}-\frac {x \sqrt {1-a^2 x^2} \text {ArcSin}(a x)^2}{2 a^2}+\frac {x \sqrt {1-a^2 x^2}}{4 a^2}+\frac {x^2 \text {ArcSin}(a x)}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4723

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 4737

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*Si
mp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c
^2*d + e, 0] && NeQ[n, -1]

Rule 4795

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

Rubi steps

\begin {align*} \int \frac {x^2 \sin ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 a^2}+\frac {\int \frac {\sin ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{2 a^2}+\frac {\int x \sin ^{-1}(a x) \, dx}{a}\\ &=\frac {x^2 \sin ^{-1}(a x)}{2 a}-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 a^2}+\frac {\sin ^{-1}(a x)^3}{6 a^3}-\frac {1}{2} \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {x \sqrt {1-a^2 x^2}}{4 a^2}+\frac {x^2 \sin ^{-1}(a x)}{2 a}-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 a^2}+\frac {\sin ^{-1}(a x)^3}{6 a^3}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{4 a^2}\\ &=\frac {x \sqrt {1-a^2 x^2}}{4 a^2}-\frac {\sin ^{-1}(a x)}{4 a^3}+\frac {x^2 \sin ^{-1}(a x)}{2 a}-\frac {x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 a^2}+\frac {\sin ^{-1}(a x)^3}{6 a^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 73, normalized size = 0.82 \begin {gather*} \frac {3 a x \sqrt {1-a^2 x^2}+\left (-3+6 a^2 x^2\right ) \text {ArcSin}(a x)-6 a x \sqrt {1-a^2 x^2} \text {ArcSin}(a x)^2+2 \text {ArcSin}(a x)^3}{12 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x*Sqrt[1 - a^2*x^2] + (-3 + 6*a^2*x^2)*ArcSin[a*x] - 6*a*x*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^2 + 2*ArcSin[a*x
]^3)/(12*a^3)

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Maple [A]
time = 0.12, size = 71, normalized size = 0.80

method	result	size
default	\(\frac {-6 \arcsin \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}\, a x +6 a^{2} x^{2} \arcsin \left (a x \right )+2 \arcsin \left (a x \right )^{3}+3 a x \sqrt {-a^{2} x^{2}+1}-3 \arcsin \left (a x \right )}{12 a^{3}}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsin(a*x)^2/(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/12*(-6*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)*a*x+6*a^2*x^2*arcsin(a*x)+2*arcsin(a*x)^3+3*a*x*(-a^2*x^2+1)^(1/2)-3
*arcsin(a*x))/a^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.24, size = 59, normalized size = 0.66 \begin {gather*} \frac {2 \, \arcsin \left (a x\right )^{3} + 3 \, {\left (2 \, a^{2} x^{2} - 1\right )} \arcsin \left (a x\right ) - 3 \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a x \arcsin \left (a x\right )^{2} - a x\right )}}{12 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*arcsin(a*x)^3 + 3*(2*a^2*x^2 - 1)*arcsin(a*x) - 3*sqrt(-a^2*x^2 + 1)*(2*a*x*arcsin(a*x)^2 - a*x))/a^3

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Sympy [A]
time = 0.36, size = 78, normalized size = 0.88 \begin {gather*} \begin {cases} \frac {x^{2} \operatorname {asin}{\left (a x \right )}}{2 a} - \frac {x \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a x \right )}}{2 a^{2}} + \frac {x \sqrt {- a^{2} x^{2} + 1}}{4 a^{2}} + \frac {\operatorname {asin}^{3}{\left (a x \right )}}{6 a^{3}} - \frac {\operatorname {asin}{\left (a x \right )}}{4 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*asin(a*x)/(2*a) - x*sqrt(-a**2*x**2 + 1)*asin(a*x)**2/(2*a**2) + x*sqrt(-a**2*x**2 + 1)/(4*a**
2) + asin(a*x)**3/(6*a**3) - asin(a*x)/(4*a**3), Ne(a, 0)), (0, True))

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Giac [A]
time = 0.45, size = 81, normalized size = 0.91 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + 1} x \arcsin \left (a x\right )^{2}}{2 \, a^{2}} + \frac {\arcsin \left (a x\right )^{3}}{6 \, a^{3}} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{4 \, a^{2}} + \frac {{\left (a^{2} x^{2} - 1\right )} \arcsin \left (a x\right )}{2 \, a^{3}} + \frac {\arcsin \left (a x\right )}{4 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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