3.142 \(\int x (a+b \cos ^{-1}(c x)) \, dx\)

Optimal. Leaf size=51 \[ \frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac{b x \sqrt{1-c^2 x^2}}{4 c}+\frac{b \sin ^{-1}(c x)}{4 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0191739, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4628, 321, 216} \[ \frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac{b x \sqrt{1-c^2 x^2}}{4 c}+\frac{b \sin ^{-1}(c x)}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcCos[c*x]),x]

[Out]

-(b*x*Sqrt[1 - c^2*x^2])/(4*c) + (x^2*(a + b*ArcCos[c*x]))/2 + (b*ArcSin[c*x])/(4*c^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 321

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 216

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]

Rubi steps

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

Mathematica [A]  time = 0.03564, size = 56, normalized size = 1.1 \[ \frac{a x^2}{2}-\frac{b x \sqrt{1-c^2 x^2}}{4 c}+\frac{b \sin ^{-1}(c x)}{4 c^2}+\frac{1}{2} b x^2 \cos ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcCos[c*x]),x]

[Out]

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

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Maple [A]  time = 0.005, size = 52, normalized size = 1. \begin{align*}{\frac{1}{{c}^{2}} \left ({\frac{{c}^{2}{x}^{2}a}{2}}+b \left ({\frac{{c}^{2}{x}^{2}\arccos \left ( cx \right ) }{2}}-{\frac{cx}{4}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{\arcsin \left ( cx \right ) }{4}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arccos(c*x)),x)

[Out]

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

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Maxima [A]  time = 1.43945, size = 84, normalized size = 1.65 \begin{align*} \frac{1}{2} \, a x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \arccos \left (c x\right ) - c{\left (\frac{\sqrt{-c^{2} x^{2} + 1} x}{c^{2}} - \frac{\arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.57809, size = 111, normalized size = 2.18 \begin{align*} \frac{2 \, a c^{2} x^{2} - \sqrt{-c^{2} x^{2} + 1} b c x +{\left (2 \, b c^{2} x^{2} - b\right )} \arccos \left (c x\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.331354, size = 60, normalized size = 1.18 \begin{align*} \begin{cases} \frac{a x^{2}}{2} + \frac{b x^{2} \operatorname{acos}{\left (c x \right )}}{2} - \frac{b x \sqrt{- c^{2} x^{2} + 1}}{4 c} - \frac{b \operatorname{acos}{\left (c x \right )}}{4 c^{2}} & \text{for}\: c \neq 0 \\\frac{x^{2} \left (a + \frac{\pi b}{2}\right )}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*acos(c*x)),x)

[Out]

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

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Giac [A]  time = 1.15468, size = 62, normalized size = 1.22 \begin{align*} \frac{1}{2} \, b x^{2} \arccos \left (c x\right ) + \frac{1}{2} \, a x^{2} - \frac{\sqrt{-c^{2} x^{2} + 1} b x}{4 \, c} - \frac{b \arccos \left (c x\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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