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

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

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 40, 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, 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 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]

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 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 \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*}

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Mathematica [A]  time = 0.05, size = 40, normalized size = 1.00 \[ \frac {\sqrt {1-(a+b x)^2}}{b \cos ^{-1}(a+b x)}-\frac {\text {Ci}\left (\cos ^{-1}(a+b x)\right )}{b} \]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\arccos \left (b x + a\right )^{2}}, x\right ) \]

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)

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giac [A]  time = 2.78, size = 38, normalized size = 0.95 \[ -\frac {\operatorname {Ci}\left (\arccos \left (b x + a\right )\right )}{b} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{b \arccos \left (b x + a\right )} \]

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))

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maple [A]  time = 0.07, size = 37, normalized size = 0.92 \[ \frac {\frac {\sqrt {1-\left (b x +a \right )^{2}}}{\arccos \left (b x +a \right )}-\Ci \left (\arccos \left (b x +a \right )\right )}{b} \]

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)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\mathrm {acos}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/acos(a + b*x)^2,x)

[Out]

int(1/acos(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {acos}^{2}{\left (a + b x \right )}}\, dx \]

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)

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