3.149 \(\int x (a+b \cos ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=76 \[ -\frac{b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )}{2 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^2-\frac{1}{4} b^2 x^2 \]

[Out]

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

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Rubi [A]  time = 0.117589, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4628, 4708, 4642, 30} \[ -\frac{b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )}{2 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^2-\frac{1}{4} b^2 x^2 \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcCos[c*x])^2,x]

[Out]

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

Mathematica [A]  time = 0.111045, size = 104, normalized size = 1.37 \[ \frac{c x \left (2 a^2 c x-2 a b \sqrt{1-c^2 x^2}-b^2 c x\right )+2 b c x \cos ^{-1}(c x) \left (2 a c x-b \sqrt{1-c^2 x^2}\right )+2 a b \sin ^{-1}(c x)+b^2 \left (2 c^2 x^2-1\right ) \cos ^{-1}(c x)^2}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcCos[c*x])^2,x]

[Out]

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

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Maple [A]  time = 0.051, size = 118, normalized size = 1.6 \begin{align*}{\frac{1}{{c}^{2}} \left ({\frac{{a}^{2}{c}^{2}{x}^{2}}{2}}+{b}^{2} \left ({\frac{{c}^{2}{x}^{2} \left ( \arccos \left ( cx \right ) \right ) ^{2}}{2}}-{\frac{\arccos \left ( cx \right ) }{2} \left ( cx\sqrt{-{c}^{2}{x}^{2}+1}+\arccos \left ( cx \right ) \right ) }+{\frac{ \left ( \arccos \left ( cx \right ) \right ) ^{2}}{4}}-{\frac{{c}^{2}{x}^{2}}{4}}+{\frac{1}{4}} \right ) +2\,ab \left ( 1/2\,{c}^{2}{x}^{2}\arccos \left ( cx \right ) -1/4\,cx\sqrt{-{c}^{2}{x}^{2}+1}+1/4\,\arcsin \left ( cx \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arccos(c*x))^2,x)

[Out]

1/c^2*(1/2*a^2*c^2*x^2+b^2*(1/2*c^2*x^2*arccos(c*x)^2-1/2*arccos(c*x)*(c*x*(-c^2*x^2+1)^(1/2)+arccos(c*x))+1/4
*arccos(c*x)^2-1/4*c^2*x^2+1/4)+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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{2} x^{2} + \frac{1}{2} \,{\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 )} a b + \frac{1}{2} \,{\left (x^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )^{2} - 2 \, c \int \frac{\sqrt{c x + 1} \sqrt{-c x + 1} x^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )}{c^{2} x^{2} - 1}\,{d x}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.64273, size = 221, normalized size = 2.91 \begin{align*} \frac{{\left (2 \, a^{2} - b^{2}\right )} c^{2} x^{2} +{\left (2 \, b^{2} c^{2} x^{2} - b^{2}\right )} \arccos \left (c x\right )^{2} + 2 \,{\left (2 \, a b c^{2} x^{2} - a b\right )} \arccos \left (c x\right ) - 2 \,{\left (b^{2} c x \arccos \left (c x\right ) + a b c x\right )} \sqrt{-c^{2} x^{2} + 1}}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.756254, size = 131, normalized size = 1.72 \begin{align*} \begin{cases} \frac{a^{2} x^{2}}{2} + a b x^{2} \operatorname{acos}{\left (c x \right )} - \frac{a b x \sqrt{- c^{2} x^{2} + 1}}{2 c} - \frac{a b \operatorname{acos}{\left (c x \right )}}{2 c^{2}} + \frac{b^{2} x^{2} \operatorname{acos}^{2}{\left (c x \right )}}{2} - \frac{b^{2} x^{2}}{4} - \frac{b^{2} x \sqrt{- c^{2} x^{2} + 1} \operatorname{acos}{\left (c x \right )}}{2 c} - \frac{b^{2} \operatorname{acos}^{2}{\left (c x \right )}}{4 c^{2}} & \text{for}\: c \neq 0 \\\frac{x^{2} \left (a + \frac{\pi b}{2}\right )^{2}}{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))**2,x)

[Out]

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

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Giac [A]  time = 1.17495, size = 161, normalized size = 2.12 \begin{align*} \frac{1}{2} \, b^{2} x^{2} \arccos \left (c x\right )^{2} + a b x^{2} \arccos \left (c x\right ) + \frac{1}{2} \, a^{2} x^{2} - \frac{1}{4} \, b^{2} x^{2} - \frac{\sqrt{-c^{2} x^{2} + 1} b^{2} x \arccos \left (c x\right )}{2 \, c} - \frac{\sqrt{-c^{2} x^{2} + 1} a b x}{2 \, c} - \frac{b^{2} \arccos \left (c x\right )^{2}}{4 \, c^{2}} - \frac{a b \arccos \left (c x\right )}{2 \, c^{2}} + \frac{b^{2}}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^2*arccos(c*x)^2 + a*b*x^2*arccos(c*x) + 1/2*a^2*x^2 - 1/4*b^2*x^2 - 1/2*sqrt(-c^2*x^2 + 1)*b^2*x*arc
cos(c*x)/c - 1/2*sqrt(-c^2*x^2 + 1)*a*b*x/c - 1/4*b^2*arccos(c*x)^2/c^2 - 1/2*a*b*arccos(c*x)/c^2 + 1/8*b^2/c^
2