3.154 \(\int x (a+b \cos ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=125 \[ -\frac{3}{4} b^2 x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac{3 b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^3+\frac{3 b^3 x \sqrt{1-c^2 x^2}}{8 c}-\frac{3 b^3 \sin ^{-1}(c x)}{8 c^2} \]

[Out]

(3*b^3*x*Sqrt[1 - c^2*x^2])/(8*c) - (3*b^2*x^2*(a + b*ArcCos[c*x]))/4 - (3*b*x*Sqrt[1 - c^2*x^2]*(a + b*ArcCos
[c*x])^2)/(4*c) - (a + b*ArcCos[c*x])^3/(4*c^2) + (x^2*(a + b*ArcCos[c*x])^3)/2 - (3*b^3*ArcSin[c*x])/(8*c^2)

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Rubi [A]  time = 0.207512, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {4628, 4708, 4642, 321, 216} \[ -\frac{3}{4} b^2 x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac{3 b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^3+\frac{3 b^3 x \sqrt{1-c^2 x^2}}{8 c}-\frac{3 b^3 \sin ^{-1}(c x)}{8 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcCos[c*x])^3,x]

[Out]

(3*b^3*x*Sqrt[1 - c^2*x^2])/(8*c) - (3*b^2*x^2*(a + b*ArcCos[c*x]))/4 - (3*b*x*Sqrt[1 - c^2*x^2]*(a + b*ArcCos
[c*x])^2)/(4*c) - (a + b*ArcCos[c*x])^3/(4*c^2) + (x^2*(a + b*ArcCos[c*x])^3)/2 - (3*b^3*ArcSin[c*x])/(8*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 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 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 )^3 \, dx &=\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^3+\frac{1}{2} (3 b c) \int \frac{x^2 \left (a+b \cos ^{-1}(c x)\right )^2}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{3 b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{4 c}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^3-\frac{1}{2} \left (3 b^2\right ) \int x \left (a+b \cos ^{-1}(c x)\right ) \, dx+\frac{(3 b) \int \frac{\left (a+b \cos ^{-1}(c x)\right )^2}{\sqrt{1-c^2 x^2}} \, dx}{4 c}\\ &=-\frac{3}{4} b^2 x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac{3 b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^3-\frac{1}{4} \left (3 b^3 c\right ) \int \frac{x^2}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{3 b^3 x \sqrt{1-c^2 x^2}}{8 c}-\frac{3}{4} b^2 x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac{3 b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^3-\frac{\left (3 b^3\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{8 c}\\ &=\frac{3 b^3 x \sqrt{1-c^2 x^2}}{8 c}-\frac{3}{4} b^2 x^2 \left (a+b \cos ^{-1}(c x)\right )-\frac{3 b x \sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{4 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^3}{4 c^2}+\frac{1}{2} x^2 \left (a+b \cos ^{-1}(c x)\right )^3-\frac{3 b^3 \sin ^{-1}(c x)}{8 c^2}\\ \end{align*}

Mathematica [A]  time = 0.167832, size = 185, normalized size = 1.48 \[ \frac{c x \left (-6 a^2 b \sqrt{1-c^2 x^2}+4 a^3 c x-6 a b^2 c x+3 b^3 \sqrt{1-c^2 x^2}\right )-6 b c x \cos ^{-1}(c x) \left (-2 a^2 c x+2 a b \sqrt{1-c^2 x^2}+b^2 c x\right )+\left (6 a^2 b-3 b^3\right ) \sin ^{-1}(c x)-6 b^2 \cos ^{-1}(c x)^2 \left (-2 a c^2 x^2+a+b c x \sqrt{1-c^2 x^2}\right )+2 b^3 \left (2 c^2 x^2-1\right ) \cos ^{-1}(c x)^3}{8 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcCos[c*x])^3,x]

[Out]

(c*x*(4*a^3*c*x - 6*a*b^2*c*x - 6*a^2*b*Sqrt[1 - c^2*x^2] + 3*b^3*Sqrt[1 - c^2*x^2]) - 6*b*c*x*(-2*a^2*c*x + b
^2*c*x + 2*a*b*Sqrt[1 - c^2*x^2])*ArcCos[c*x] - 6*b^2*(a - 2*a*c^2*x^2 + b*c*x*Sqrt[1 - c^2*x^2])*ArcCos[c*x]^
2 + 2*b^3*(-1 + 2*c^2*x^2)*ArcCos[c*x]^3 + (6*a^2*b - 3*b^3)*ArcSin[c*x])/(8*c^2)

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Maple [A]  time = 0.052, size = 211, normalized size = 1.7 \begin{align*}{\frac{1}{{c}^{2}} \left ({\frac{{c}^{2}{x}^{2}{a}^{3}}{2}}+{b}^{3} \left ({\frac{{c}^{2}{x}^{2} \left ( \arccos \left ( cx \right ) \right ) ^{3}}{2}}-{\frac{3\, \left ( \arccos \left ( cx \right ) \right ) ^{2}}{4} \left ( cx\sqrt{-{c}^{2}{x}^{2}+1}+\arccos \left ( cx \right ) \right ) }-{\frac{3\,{c}^{2}{x}^{2}\arccos \left ( cx \right ) }{4}}+{\frac{3\,cx}{8}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{3\,\arccos \left ( cx \right ) }{8}}+{\frac{ \left ( \arccos \left ( cx \right ) \right ) ^{3}}{2}} \right ) +3\,a{b}^{2} \left ( 1/2\,{c}^{2}{x}^{2} \left ( \arccos \left ( cx \right ) \right ) ^{2}-1/2\,\arccos \left ( cx \right ) \left ( cx\sqrt{-{c}^{2}{x}^{2}+1}+\arccos \left ( cx \right ) \right ) +1/4\, \left ( \arccos \left ( cx \right ) \right ) ^{2}-1/4\,{c}^{2}{x}^{2}+1/4 \right ) +3\,{a}^{2}b \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))^3,x)

[Out]

1/c^2*(1/2*c^2*x^2*a^3+b^3*(1/2*c^2*x^2*arccos(c*x)^3-3/4*arccos(c*x)^2*(c*x*(-c^2*x^2+1)^(1/2)+arccos(c*x))-3
/4*c^2*x^2*arccos(c*x)+3/8*c*x*(-c^2*x^2+1)^(1/2)+3/8*arccos(c*x)+1/2*arccos(c*x)^3)+3*a*b^2*(1/2*c^2*x^2*arcc
os(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)+3*a^2*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} \, b^{3} x^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )^{3} + \frac{1}{2} \, a^{3} x^{2} + \frac{3}{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 )} a^{2} b - \int \frac{3 \,{\left (\sqrt{c x + 1} \sqrt{-c x + 1} b^{3} c x^{2} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )^{2} - 2 \,{\left (a b^{2} c^{2} x^{3} - a b^{2} x\right )} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )^{2}\right )}}{2 \,{\left (c^{2} x^{2} - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccos(c*x))^3,x, algorithm="maxima")

[Out]

1/2*b^3*x^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^3 + 1/2*a^3*x^2 + 3/4*(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)))*a^2*b - integrate(3/2*(sqrt(c*x + 1)*sqrt(-c*x + 1
)*b^3*c*x^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2 - 2*(a*b^2*c^2*x^3 - a*b^2*x)*arctan2(sqrt(c*x + 1)*s
qrt(-c*x + 1), c*x)^2)/(c^2*x^2 - 1), x)

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Fricas [A]  time = 2.29324, size = 378, normalized size = 3.02 \begin{align*} \frac{2 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} c^{2} x^{2} + 2 \,{\left (2 \, b^{3} c^{2} x^{2} - b^{3}\right )} \arccos \left (c x\right )^{3} + 6 \,{\left (2 \, a b^{2} c^{2} x^{2} - a b^{2}\right )} \arccos \left (c x\right )^{2} + 3 \,{\left (2 \,{\left (2 \, a^{2} b - b^{3}\right )} c^{2} x^{2} - 2 \, a^{2} b + b^{3}\right )} \arccos \left (c x\right ) - 3 \,{\left (2 \, b^{3} c x \arccos \left (c x\right )^{2} + 4 \, a b^{2} c x \arccos \left (c x\right ) +{\left (2 \, a^{2} b - b^{3}\right )} c x\right )} \sqrt{-c^{2} x^{2} + 1}}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccos(c*x))^3,x, algorithm="fricas")

[Out]

1/8*(2*(2*a^3 - 3*a*b^2)*c^2*x^2 + 2*(2*b^3*c^2*x^2 - b^3)*arccos(c*x)^3 + 6*(2*a*b^2*c^2*x^2 - a*b^2)*arccos(
c*x)^2 + 3*(2*(2*a^2*b - b^3)*c^2*x^2 - 2*a^2*b + b^3)*arccos(c*x) - 3*(2*b^3*c*x*arccos(c*x)^2 + 4*a*b^2*c*x*
arccos(c*x) + (2*a^2*b - b^3)*c*x)*sqrt(-c^2*x^2 + 1))/c^2

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Sympy [A]  time = 1.62723, size = 269, normalized size = 2.15 \begin{align*} \begin{cases} \frac{a^{3} x^{2}}{2} + \frac{3 a^{2} b x^{2} \operatorname{acos}{\left (c x \right )}}{2} - \frac{3 a^{2} b x \sqrt{- c^{2} x^{2} + 1}}{4 c} - \frac{3 a^{2} b \operatorname{acos}{\left (c x \right )}}{4 c^{2}} + \frac{3 a b^{2} x^{2} \operatorname{acos}^{2}{\left (c x \right )}}{2} - \frac{3 a b^{2} x^{2}}{4} - \frac{3 a b^{2} x \sqrt{- c^{2} x^{2} + 1} \operatorname{acos}{\left (c x \right )}}{2 c} - \frac{3 a b^{2} \operatorname{acos}^{2}{\left (c x \right )}}{4 c^{2}} + \frac{b^{3} x^{2} \operatorname{acos}^{3}{\left (c x \right )}}{2} - \frac{3 b^{3} x^{2} \operatorname{acos}{\left (c x \right )}}{4} - \frac{3 b^{3} x \sqrt{- c^{2} x^{2} + 1} \operatorname{acos}^{2}{\left (c x \right )}}{4 c} + \frac{3 b^{3} x \sqrt{- c^{2} x^{2} + 1}}{8 c} - \frac{b^{3} \operatorname{acos}^{3}{\left (c x \right )}}{4 c^{2}} + \frac{3 b^{3} \operatorname{acos}{\left (c x \right )}}{8 c^{2}} & \text{for}\: c \neq 0 \\\frac{x^{2} \left (a + \frac{\pi b}{2}\right )^{3}}{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))**3,x)

[Out]

Piecewise((a**3*x**2/2 + 3*a**2*b*x**2*acos(c*x)/2 - 3*a**2*b*x*sqrt(-c**2*x**2 + 1)/(4*c) - 3*a**2*b*acos(c*x
)/(4*c**2) + 3*a*b**2*x**2*acos(c*x)**2/2 - 3*a*b**2*x**2/4 - 3*a*b**2*x*sqrt(-c**2*x**2 + 1)*acos(c*x)/(2*c)
- 3*a*b**2*acos(c*x)**2/(4*c**2) + b**3*x**2*acos(c*x)**3/2 - 3*b**3*x**2*acos(c*x)/4 - 3*b**3*x*sqrt(-c**2*x*
*2 + 1)*acos(c*x)**2/(4*c) + 3*b**3*x*sqrt(-c**2*x**2 + 1)/(8*c) - b**3*acos(c*x)**3/(4*c**2) + 3*b**3*acos(c*
x)/(8*c**2), Ne(c, 0)), (x**2*(a + pi*b/2)**3/2, True))

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Giac [B]  time = 1.20801, size = 312, normalized size = 2.5 \begin{align*} \frac{1}{2} \, b^{3} x^{2} \arccos \left (c x\right )^{3} + \frac{3}{2} \, a b^{2} x^{2} \arccos \left (c x\right )^{2} + \frac{3}{2} \, a^{2} b x^{2} \arccos \left (c x\right ) - \frac{3}{4} \, b^{3} x^{2} \arccos \left (c x\right ) - \frac{3 \, \sqrt{-c^{2} x^{2} + 1} b^{3} x \arccos \left (c x\right )^{2}}{4 \, c} + \frac{1}{2} \, a^{3} x^{2} - \frac{3}{4} \, a b^{2} x^{2} - \frac{3 \, \sqrt{-c^{2} x^{2} + 1} a b^{2} x \arccos \left (c x\right )}{2 \, c} - \frac{b^{3} \arccos \left (c x\right )^{3}}{4 \, c^{2}} - \frac{3 \, \sqrt{-c^{2} x^{2} + 1} a^{2} b x}{4 \, c} + \frac{3 \, \sqrt{-c^{2} x^{2} + 1} b^{3} x}{8 \, c} - \frac{3 \, a b^{2} \arccos \left (c x\right )^{2}}{4 \, c^{2}} - \frac{3 \, a^{2} b \arccos \left (c x\right )}{4 \, c^{2}} + \frac{3 \, b^{3} \arccos \left (c x\right )}{8 \, c^{2}} + \frac{3 \, a 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))^3,x, algorithm="giac")

[Out]

1/2*b^3*x^2*arccos(c*x)^3 + 3/2*a*b^2*x^2*arccos(c*x)^2 + 3/2*a^2*b*x^2*arccos(c*x) - 3/4*b^3*x^2*arccos(c*x)
- 3/4*sqrt(-c^2*x^2 + 1)*b^3*x*arccos(c*x)^2/c + 1/2*a^3*x^2 - 3/4*a*b^2*x^2 - 3/2*sqrt(-c^2*x^2 + 1)*a*b^2*x*
arccos(c*x)/c - 1/4*b^3*arccos(c*x)^3/c^2 - 3/4*sqrt(-c^2*x^2 + 1)*a^2*b*x/c + 3/8*sqrt(-c^2*x^2 + 1)*b^3*x/c
- 3/4*a*b^2*arccos(c*x)^2/c^2 - 3/4*a^2*b*arccos(c*x)/c^2 + 3/8*b^3*arccos(c*x)/c^2 + 3/8*a*b^2/c^2