3.1.4 \(\int x \text {ArcCos}(a x) \, dx\) [4]

Optimal. Leaf size=45 \[ -\frac {x \sqrt {1-a^2 x^2}}{4 a}+\frac {1}{2} x^2 \text {ArcCos}(a x)+\frac {\text {ArcSin}(a x)}{4 a^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4724, 327, 222} \begin {gather*} \frac {\text {ArcSin}(a x)}{4 a^2}-\frac {x \sqrt {1-a^2 x^2}}{4 a}+\frac {1}{2} x^2 \text {ArcCos}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4724

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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 42, normalized size = 0.93 \begin {gather*} \frac {-a x \sqrt {1-a^2 x^2}+2 a^2 x^2 \text {ArcCos}(a x)+\text {ArcSin}(a x)}{4 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.01, size = 40, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\frac {a^{2} x^{2} \arccos \left (a x \right )}{2}-\frac {a x \sqrt {-a^{2} x^{2}+1}}{4}+\frac {\arcsin \left (a x \right )}{4}}{a^{2}}\)	\(40\)
default	\(\frac {\frac {a^{2} x^{2} \arccos \left (a x \right )}{2}-\frac {a x \sqrt {-a^{2} x^{2}+1}}{4}+\frac {\arcsin \left (a x \right )}{4}}{a^{2}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccos(a*x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.46, size = 40, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, x^{2} \arccos \left (a x\right ) - \frac {1}{4} \, a {\left (\frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{2}} - \frac {\arcsin \left (a x\right )}{a^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.64, size = 37, normalized size = 0.82 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + 1} a x - {\left (2 \, a^{2} x^{2} - 1\right )} \arccos \left (a x\right )}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.09, size = 42, normalized size = 0.93 \begin {gather*} \begin {cases} \frac {x^{2} \operatorname {acos}{\left (a x \right )}}{2} - \frac {x \sqrt {- a^{2} x^{2} + 1}}{4 a} - \frac {\operatorname {acos}{\left (a x \right )}}{4 a^{2}} & \text {for}\: a \neq 0 \\\frac {\pi x^{2}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 37, normalized size = 0.82 \begin {gather*} \frac {1}{2} \, x^{2} \arccos \left (a x\right ) - \frac {\sqrt {-a^{2} x^{2} + 1} x}{4 \, a} - \frac {\arccos \left (a x\right )}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 38, normalized size = 0.84 \begin {gather*} \frac {\mathrm {acos}\left (a\,x\right )\,\left (2\,a^2\,x^2-1\right )}{4\,a^2}-\frac {x\,\sqrt {1-a^2\,x^2}}{4\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acos(a*x),x)

[Out]

(acos(a*x)*(2*a^2*x^2 - 1))/(4*a^2) - (x*(1 - a^2*x^2)^(1/2))/(4*a)

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