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

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

[Out]

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

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Rubi [A]  time = 0.0966761, 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 = {4628, 4708, 4642, 30} \[ -\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{2 a}-\frac{\cos ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^2-\frac{x^2}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4628

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
s[c*x])^n)/(d*(m + 1)), x] + Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCos[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 4708

Int[(((a_.) + ArcCos[(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*ArcCos[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcCos[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*ArcCos[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 4642

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp[(a + b*ArcCos[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 \cos ^{-1}(a x)^2 \, dx &=\frac{1}{2} x^2 \cos ^{-1}(a x)^2+a \int \frac{x^2 \cos ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{2 a}+\frac{1}{2} x^2 \cos ^{-1}(a x)^2-\frac{\int x \, dx}{2}+\frac{\int \frac{\cos ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{2 a}\\ &=-\frac{x^2}{4}-\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{2 a}-\frac{\cos ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^2\\ \end{align*}

Mathematica [A]  time = 0.0249349, size = 57, normalized size = 0.95 \[ -\frac{x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)}{2 a}+\frac{\left (2 a^2 x^2-1\right ) \cos ^{-1}(a x)^2}{4 a^2}-\frac{x^2}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.048, size = 63, normalized size = 1.1 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{{a}^{2}{x}^{2} \left ( \arccos \left ( ax \right ) \right ) ^{2}}{2}}-{\frac{\arccos \left ( ax \right ) }{2} \left ( ax\sqrt{-{a}^{2}{x}^{2}+1}+\arccos \left ( ax \right ) \right ) }+{\frac{ \left ( \arccos \left ( ax \right ) \right ) ^{2}}{4}}-{\frac{{a}^{2}{x}^{2}}{4}}+{\frac{1}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.92655, size = 123, normalized size = 2.05 \begin{align*} -\frac{a^{2} x^{2} + 2 \, \sqrt{-a^{2} x^{2} + 1} a x \arccos \left (a x\right ) -{\left (2 \, a^{2} x^{2} - 1\right )} \arccos \left (a x\right )^{2}}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.589721, size = 58, normalized size = 0.97 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{acos}^{2}{\left (a x \right )}}{2} - \frac{x^{2}}{4} - \frac{x \sqrt{- a^{2} x^{2} + 1} \operatorname{acos}{\left (a x \right )}}{2 a} - \frac{\operatorname{acos}^{2}{\left (a x \right )}}{4 a^{2}} & \text{for}\: a \neq 0 \\\frac{\pi ^{2} x^{2}}{8} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13698, size = 74, normalized size = 1.23 \begin{align*} \frac{1}{2} \, x^{2} \arccos \left (a x\right )^{2} - \frac{1}{4} \, x^{2} - \frac{\sqrt{-a^{2} x^{2} + 1} x \arccos \left (a x\right )}{2 \, a} - \frac{\arccos \left (a x\right )^{2}}{4 \, a^{2}} + \frac{1}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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