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

Optimal. Leaf size=82 \[ \frac {6 \sqrt {1-(a+b x)^2}}{b}+\frac {(a+b x) \cos ^{-1}(a+b x)^3}{b}-\frac {3 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^2}{b}-\frac {6 (a+b x) \cos ^{-1}(a+b x)}{b} \]

[Out]

-6*(b*x+a)*arccos(b*x+a)/b+(b*x+a)*arccos(b*x+a)^3/b+6*(1-(b*x+a)^2)^(1/2)/b-3*arccos(b*x+a)^2*(1-(b*x+a)^2)^(
1/2)/b

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Rubi [A]  time = 0.08, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4804, 4620, 4678, 261} \[ \frac {6 \sqrt {1-(a+b x)^2}}{b}+\frac {(a+b x) \cos ^{-1}(a+b x)^3}{b}-\frac {3 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^2}{b}-\frac {6 (a+b x) \cos ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^3,x]

[Out]

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4620

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcCos[c*x])^n, x] + Dist[b*c*n, Int[
(x*(a + b*ArcCos[c*x])^(n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4678

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcCos[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4804

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCos[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

\begin {align*} \int \cos ^{-1}(a+b x)^3 \, dx &=\frac {\operatorname {Subst}\left (\int \cos ^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \cos ^{-1}(a+b x)^3}{b}+\frac {3 \operatorname {Subst}\left (\int \frac {x \cos ^{-1}(x)^2}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{b}\\ &=-\frac {3 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \cos ^{-1}(a+b x)^3}{b}-\frac {6 \operatorname {Subst}\left (\int \cos ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=-\frac {6 (a+b x) \cos ^{-1}(a+b x)}{b}-\frac {3 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \cos ^{-1}(a+b x)^3}{b}-\frac {6 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {6 \sqrt {1-(a+b x)^2}}{b}-\frac {6 (a+b x) \cos ^{-1}(a+b x)}{b}-\frac {3 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \cos ^{-1}(a+b x)^3}{b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 74, normalized size = 0.90 \[ \frac {6 \sqrt {1-(a+b x)^2}+(a+b x) \cos ^{-1}(a+b x)^3-3 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)^2-6 (a+b x) \cos ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]^3,x]

[Out]

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

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fricas [A]  time = 0.47, size = 66, normalized size = 0.80 \[ \frac {{\left (b x + a\right )} \arccos \left (b x + a\right )^{3} - 6 \, {\left (b x + a\right )} \arccos \left (b x + a\right ) - 3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (\arccos \left (b x + a\right )^{2} - 2\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 78, normalized size = 0.95 \[ \frac {{\left (b x + a\right )} \arccos \left (b x + a\right )^{3}}{b} - \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} \arccos \left (b x + a\right )^{2}}{b} - \frac {6 \, {\left (b x + a\right )} \arccos \left (b x + a\right )}{b} + \frac {6 \, \sqrt {-{\left (b x + a\right )}^{2} + 1}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 71, normalized size = 0.87 \[ \frac {\arccos \left (b x +a \right )^{3} \left (b x +a \right )-3 \arccos \left (b x +a \right )^{2} \sqrt {1-\left (b x +a \right )^{2}}+6 \sqrt {1-\left (b x +a \right )^{2}}-6 \left (b x +a \right ) \arccos \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(b*x+a)^3,x)

[Out]

1/b*(arccos(b*x+a)^3*(b*x+a)-3*arccos(b*x+a)^2*(1-(b*x+a)^2)^(1/2)+6*(1-(b*x+a)^2)^(1/2)-6*(b*x+a)*arccos(b*x+
a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ x \arctan \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1}, b x + a\right )^{3} - 3 \, b \int \frac {\sqrt {b x + a + 1} \sqrt {-b x - a + 1} x \arctan \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1}, b x + a\right )^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.28, size = 60, normalized size = 0.73 \[ -\frac {\left (3\,{\mathrm {acos}\left (a+b\,x\right )}^2-6\right )\,\sqrt {1-{\left (a+b\,x\right )}^2}}{b}-\frac {\left (6\,\mathrm {acos}\left (a+b\,x\right )-{\mathrm {acos}\left (a+b\,x\right )}^3\right )\,\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a + b*x)^3,x)

[Out]

- ((3*acos(a + b*x)^2 - 6)*(1 - (a + b*x)^2)^(1/2))/b - ((6*acos(a + b*x) - acos(a + b*x)^3)*(a + b*x))/b

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sympy [A]  time = 0.53, size = 109, normalized size = 1.33 \[ \begin {cases} \frac {a \operatorname {acos}^{3}{\left (a + b x \right )}}{b} - \frac {6 a \operatorname {acos}{\left (a + b x \right )}}{b} + x \operatorname {acos}^{3}{\left (a + b x \right )} - 6 x \operatorname {acos}{\left (a + b x \right )} - \frac {3 \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {acos}^{2}{\left (a + b x \right )}}{b} + \frac {6 \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{b} & \text {for}\: b \neq 0 \\x \operatorname {acos}^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(b*x+a)**3,x)

[Out]

Piecewise((a*acos(a + b*x)**3/b - 6*a*acos(a + b*x)/b + x*acos(a + b*x)**3 - 6*x*acos(a + b*x) - 3*sqrt(-a**2
- 2*a*b*x - b**2*x**2 + 1)*acos(a + b*x)**2/b + 6*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/b, Ne(b, 0)), (x*acos(
a)**3, True))

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