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

Optimal. Leaf size=40 \[ \frac{\sqrt{1-(a+b x)^2}}{b \cos ^{-1}(a+b x)}-\frac{\text{CosIntegral}\left (\cos ^{-1}(a+b x)\right )}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0784772, antiderivative size = 40, 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, 4622, 4724, 3302} \[ \frac{\sqrt{1-(a+b x)^2}}{b \cos ^{-1}(a+b x)}-\frac{\text{CosIntegral}\left (\cos ^{-1}(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(-2),x]

[Out]

Sqrt[1 - (a + b*x)^2]/(b*ArcCos[a + b*x]) - CosIntegral[ArcCos[a + b*x]]/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 4622

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

Rule 4724

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Cos[x]^m*Sin[x]^(2*p + 1), x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 0.0508103, size = 40, normalized size = 1. \[ \frac{\sqrt{1-(a+b x)^2}}{b \cos ^{-1}(a+b x)}-\frac{\text{CosIntegral}\left (\cos ^{-1}(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]^(-2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 37, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ({\frac{1}{\arccos \left ( bx+a \right ) }\sqrt{1- \left ( bx+a \right ) ^{2}}}-{\it Ci} \left ( \arccos \left ( bx+a \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\arccos \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(arccos(b*x + a)^(-2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{acos}^{2}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acos(b*x+a)**2,x)

[Out]

Integral(acos(a + b*x)**(-2), x)

________________________________________________________________________________________

Giac [A]  time = 1.32966, size = 51, normalized size = 1.27 \begin{align*} -\frac{\operatorname{Ci}\left (\arccos \left (b x + a\right )\right )}{b} + \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}}{b \arccos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^2,x, algorithm="giac")

[Out]

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