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

Optimal. Leaf size=60 \[ \frac{x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)}{2 a}-\frac{\sin ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \sin ^{-1}(a x)^2-\frac{x^2}{4} \]

[Out]

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

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Rubi [A]  time = 0.0932273, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4627, 4707, 4641, 30} \[ \frac{x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)}{2 a}-\frac{\sin ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \sin ^{-1}(a x)^2-\frac{x^2}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

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 30

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

Rubi steps

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

Mathematica [A]  time = 0.016981, size = 55, normalized size = 0.92 \[ \frac{-a^2 x^2+2 a x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)+\left (2 a^2 x^2-1\right ) \sin ^{-1}(a x)^2}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.025, size = 65, normalized size = 1.1 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{ \left ({a}^{2}{x}^{2}-1 \right ) \left ( \arcsin \left ( ax \right ) \right ) ^{2}}{2}}+{\frac{\arcsin \left ( ax \right ) }{2} \left ( ax\sqrt{-{a}^{2}{x}^{2}+1}+\arcsin \left ( ax \right ) \right ) }-{\frac{ \left ( \arcsin \left ( ax \right ) \right ) ^{2}}{4}}-{\frac{{a}^{2}{x}^{2}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x^{2} \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^{2} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.33855, size = 123, normalized size = 2.05 \begin{align*} -\frac{a^{2} x^{2} - 2 \, \sqrt{-a^{2} x^{2} + 1} a x \arcsin \left (a x\right ) -{\left (2 \, a^{2} x^{2} - 1\right )} \arcsin \left (a x\right )^{2}}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.769838, size = 51, normalized size = 0.85 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{asin}^{2}{\left (a x \right )}}{2} - \frac{x^{2}}{4} + \frac{x \sqrt{- a^{2} x^{2} + 1} \operatorname{asin}{\left (a x \right )}}{2 a} - \frac{\operatorname{asin}^{2}{\left (a x \right )}}{4 a^{2}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21496, size = 99, normalized size = 1.65 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1} x \arcsin \left (a x\right )}{2 \, a} + \frac{{\left (a^{2} x^{2} - 1\right )} \arcsin \left (a x\right )^{2}}{2 \, a^{2}} + \frac{\arcsin \left (a x\right )^{2}}{4 \, a^{2}} - \frac{a^{2} x^{2} - 1}{4 \, a^{2}} - \frac{1}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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