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

Optimal. Leaf size=47 \[ \frac {(a+b x) \cos ^{-1}(a+b x)^2}{b}-\frac {2 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)}{b}-2 x \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4804, 4620, 4678, 8} \[ \frac {(a+b x) \cos ^{-1}(a+b x)^2}{b}-\frac {2 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)}{b}-2 x \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 1.04 \[ \frac {-2 (a+b x)+(a+b x) \cos ^{-1}(a+b x)^2-2 \sqrt {1-(a+b x)^2} \cos ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]^2,x]

[Out]

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

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fricas [A]  time = 0.48, size = 53, normalized size = 1.13 \[ \frac {{\left (b x + a\right )} \arccos \left (b x + a\right )^{2} - 2 \, b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arccos \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.14, size = 52, normalized size = 1.11 \[ \frac {{\left (b x + a\right )} \arccos \left (b x + a\right )^{2}}{b} - \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} \arccos \left (b x + a\right )}{b} - \frac {2 \, {\left (b x + a\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 48, normalized size = 1.02 \[ \frac {\arccos \left (b x +a \right )^{2} \left (b x +a \right )-2 b x -2 a -2 \arccos \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.25, size = 44, normalized size = 0.94 \[ \frac {\left ({\mathrm {acos}\left (a+b\,x\right )}^2-2\right )\,\left (a+b\,x\right )}{b}-\frac {2\,\mathrm {acos}\left (a+b\,x\right )\,\sqrt {1-{\left (a+b\,x\right )}^2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a + b*x)^2,x)

[Out]

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

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sympy [A]  time = 0.24, size = 63, normalized size = 1.34 \[ \begin {cases} \frac {a \operatorname {acos}^{2}{\left (a + b x \right )}}{b} + x \operatorname {acos}^{2}{\left (a + b x \right )} - 2 x - \frac {2 \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {acos}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \operatorname {acos}^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(b*x+a)**2,x)

[Out]

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

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