3.3.65 \(\int \frac {x^3 \text {ArcSin}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\) [265]

Optimal. Leaf size=126 \[ \frac {14 \sqrt {1-a^2 x^2}}{9 a^4}-\frac {2 \left (1-a^2 x^2\right )^{3/2}}{27 a^4}+\frac {4 x \text {ArcSin}(a x)}{3 a^3}+\frac {2 x^3 \text {ArcSin}(a x)}{9 a}-\frac {2 \sqrt {1-a^2 x^2} \text {ArcSin}(a x)^2}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \text {ArcSin}(a x)^2}{3 a^2} \]

[Out]

-2/27*(-a^2*x^2+1)^(3/2)/a^4+4/3*x*arcsin(a*x)/a^3+2/9*x^3*arcsin(a*x)/a+14/9*(-a^2*x^2+1)^(1/2)/a^4-2/3*arcsi
n(a*x)^2*(-a^2*x^2+1)^(1/2)/a^4-1/3*x^2*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]
time = 0.14, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {4795, 4767, 4715, 267, 4723, 272, 45} \begin {gather*} \frac {4 x \text {ArcSin}(a x)}{3 a^3}-\frac {x^2 \sqrt {1-a^2 x^2} \text {ArcSin}(a x)^2}{3 a^2}-\frac {2 \sqrt {1-a^2 x^2} \text {ArcSin}(a x)^2}{3 a^4}-\frac {2 \left (1-a^2 x^2\right )^{3/2}}{27 a^4}+\frac {14 \sqrt {1-a^2 x^2}}{9 a^4}+\frac {2 x^3 \text {ArcSin}(a x)}{9 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(14*Sqrt[1 - a^2*x^2])/(9*a^4) - (2*(1 - a^2*x^2)^(3/2))/(27*a^4) + (4*x*ArcSin[a*x])/(3*a^3) + (2*x^3*ArcSin[
a*x])/(9*a) - (2*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^2)/(3*a^4) - (x^2*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^2)/(3*a^2)

Rule 45

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 267

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 272

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 4715

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 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 4767

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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 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^3 \sin ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^2}+\frac {2 \int \frac {x \sin ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{3 a^2}+\frac {2 \int x^2 \sin ^{-1}(a x) \, dx}{3 a}\\ &=\frac {2 x^3 \sin ^{-1}(a x)}{9 a}-\frac {2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^2}-\frac {2}{9} \int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx+\frac {4 \int \sin ^{-1}(a x) \, dx}{3 a^3}\\ &=\frac {4 x \sin ^{-1}(a x)}{3 a^3}+\frac {2 x^3 \sin ^{-1}(a x)}{9 a}-\frac {2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^2}-\frac {1}{9} \text {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )-\frac {4 \int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{3 a^2}\\ &=\frac {4 \sqrt {1-a^2 x^2}}{3 a^4}+\frac {4 x \sin ^{-1}(a x)}{3 a^3}+\frac {2 x^3 \sin ^{-1}(a x)}{9 a}-\frac {2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^2}-\frac {1}{9} \text {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 )\\ &=\frac {14 \sqrt {1-a^2 x^2}}{9 a^4}-\frac {2 \left (1-a^2 x^2\right )^{3/2}}{27 a^4}+\frac {4 x \sin ^{-1}(a x)}{3 a^3}+\frac {2 x^3 \sin ^{-1}(a x)}{9 a}-\frac {2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{3 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 81, normalized size = 0.64 \begin {gather*} \frac {2 \sqrt {1-a^2 x^2} \left (20+a^2 x^2\right )+6 a x \left (6+a^2 x^2\right ) \text {ArcSin}(a x)-9 \sqrt {1-a^2 x^2} \left (2+a^2 x^2\right ) \text {ArcSin}(a x)^2}{27 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 - a^2*x^2]*(20 + a^2*x^2) + 6*a*x*(6 + a^2*x^2)*ArcSin[a*x] - 9*Sqrt[1 - a^2*x^2]*(2 + a^2*x^2)*ArcS
in[a*x]^2)/(27*a^4)

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Maple [A]
time = 0.10, size = 127, normalized size = 1.01

method	result	size
default	\(-\frac {\left (9 a^{4} x^{4} \arcsin \left (a x \right )^{2}+9 \arcsin \left (a x \right )^{2} a^{2} x^{2}+6 \arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-2 a^{4} x^{4}-38 a^{2} x^{2}-18 \arcsin \left (a x \right )^{2}+36 a x \arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}+40\right ) \sqrt {-a^{2} x^{2}+1}}{27 a^{4} \left (a^{2} x^{2}-1\right )}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27/a^4*(9*a^4*x^4*arcsin(a*x)^2+9*arcsin(a*x)^2*a^2*x^2+6*arcsin(a*x)*(-a^2*x^2+1)^(1/2)*a^3*x^3-2*a^4*x^4-
38*a^2*x^2-18*arcsin(a*x)^2+36*a*x*arcsin(a*x)*(-a^2*x^2+1)^(1/2)+40)*(-a^2*x^2+1)^(1/2)/(a^2*x^2-1)

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Maxima [A]
time = 0.48, size = 105, normalized size = 0.83 \begin {gather*} -\frac {1}{3} \, {\left (\frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{a^{2}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{4}}\right )} \arcsin \left (a x\right )^{2} + \frac {2 \, {\left (\sqrt {-a^{2} x^{2} + 1} x^{2} + \frac {20 \, \sqrt {-a^{2} x^{2} + 1}}{a^{2}}\right )}}{27 \, a^{2}} + \frac {2 \, {\left (a^{2} x^{3} + 6 \, x\right )} \arcsin \left (a x\right )}{9 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(sqrt(-a^2*x^2 + 1)*x^2/a^2 + 2*sqrt(-a^2*x^2 + 1)/a^4)*arcsin(a*x)^2 + 2/27*(sqrt(-a^2*x^2 + 1)*x^2 + 20
*sqrt(-a^2*x^2 + 1)/a^2)/a^2 + 2/9*(a^2*x^3 + 6*x)*arcsin(a*x)/a^3

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Fricas [A]
time = 4.03, size = 64, normalized size = 0.51 \begin {gather*} \frac {6 \, {\left (a^{3} x^{3} + 6 \, a x\right )} \arcsin \left (a x\right ) + {\left (2 \, a^{2} x^{2} - 9 \, {\left (a^{2} x^{2} + 2\right )} \arcsin \left (a x\right )^{2} + 40\right )} \sqrt {-a^{2} x^{2} + 1}}{27 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*(6*(a^3*x^3 + 6*a*x)*arcsin(a*x) + (2*a^2*x^2 - 9*(a^2*x^2 + 2)*arcsin(a*x)^2 + 40)*sqrt(-a^2*x^2 + 1))/a
^4

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Sympy [A]
time = 0.47, size = 121, normalized size = 0.96 \begin {gather*} \begin {cases} \frac {2 x^{3} \operatorname {asin}{\left (a x \right )}}{9 a} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a x \right )}}{3 a^{2}} + \frac {2 x^{2} \sqrt {- a^{2} x^{2} + 1}}{27 a^{2}} + \frac {4 x \operatorname {asin}{\left (a x \right )}}{3 a^{3}} - \frac {2 \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a x \right )}}{3 a^{4}} + \frac {40 \sqrt {- a^{2} x^{2} + 1}}{27 a^{4}} & \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**3*asin(a*x)**2/(-a**2*x**2+1)**(1/2),x)

[Out]

Piecewise((2*x**3*asin(a*x)/(9*a) - x**2*sqrt(-a**2*x**2 + 1)*asin(a*x)**2/(3*a**2) + 2*x**2*sqrt(-a**2*x**2 +
1)/(27*a**2) + 4*x*asin(a*x)/(3*a**3) - 2*sqrt(-a**2*x**2 + 1)*asin(a*x)**2/(3*a**4) + 40*sqrt(-a**2*x**2 + 1
)/(27*a**4), Ne(a, 0)), (0, True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,{\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^3*asin(a*x)^2)/(1 - a^2*x^2)^(1/2),x)

[Out]

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

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