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

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Rubi [A]  time = 0.0544189, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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]

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 8

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

Mathematica [A]  time = 0.0228701, 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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Maple [A]  time = 0.049, size = 48, normalized size = 1. \begin{align*}{\frac{1}{b} \left ( \left ( \arccos \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) -2\,bx-2\,a-2\,\arccos \left ( bx+a \right ) \sqrt{1- \left ( bx+a \right ) ^{2}} \right ) } \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.33554, size = 130, normalized size = 2.77 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 0.282811, size = 63, normalized size = 1.34 \begin{align*} \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}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.34207, size = 70, normalized size = 1.49 \begin{align*} \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} \end{align*}

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