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

Optimal. Leaf size=65 \[ \frac {\text {Si}\left (\cos ^{-1}(a+b x)\right )}{2 b}+\frac {a+b x}{2 b \cos ^{-1}(a+b x)}+\frac {\sqrt {1-(a+b x)^2}}{2 b \cos ^{-1}(a+b x)^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4804, 4622, 4720, 4624, 3299} \[ \frac {\text {Si}\left (\cos ^{-1}(a+b x)\right )}{2 b}+\frac {a+b x}{2 b \cos ^{-1}(a+b x)}+\frac {\sqrt {1-(a+b x)^2}}{2 b \cos ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(-3),x]

[Out]

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - 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 4624

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

Rule 4720

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp
[((f*x)^m*(a + b*ArcCos[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] + Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)
^(m - 1)*(a + b*ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && 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)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\cos ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=\frac {\sqrt {1-(a+b x)^2}}{2 b \cos ^{-1}(a+b x)^2}+\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \cos ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b}\\ &=\frac {\sqrt {1-(a+b x)^2}}{2 b \cos ^{-1}(a+b x)^2}+\frac {a+b x}{2 b \cos ^{-1}(a+b x)}-\frac {\operatorname {Subst}\left (\int \frac {1}{\cos ^{-1}(x)} \, dx,x,a+b x\right )}{2 b}\\ &=\frac {\sqrt {1-(a+b x)^2}}{2 b \cos ^{-1}(a+b x)^2}+\frac {a+b x}{2 b \cos ^{-1}(a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\cos ^{-1}(a+b x)\right )}{2 b}\\ &=\frac {\sqrt {1-(a+b x)^2}}{2 b \cos ^{-1}(a+b x)^2}+\frac {a+b x}{2 b \cos ^{-1}(a+b x)}+\frac {\text {Si}\left (\cos ^{-1}(a+b x)\right )}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 65, normalized size = 1.00 \[ \frac {\text {Si}\left (\cos ^{-1}(a+b x)\right )}{2 b}+\frac {a+b x}{2 b \cos ^{-1}(a+b x)}+\frac {\sqrt {1-(a+b x)^2}}{2 b \cos ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]^(-3),x]

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\arccos \left (b x + a\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^3,x, algorithm="fricas")

[Out]

integral(arccos(b*x + a)^(-3), x)

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giac [A]  time = 0.20, size = 57, normalized size = 0.88 \[ \frac {\operatorname {Si}\left (\arccos \left (b x + a\right )\right )}{2 \, b} + \frac {b x + a}{2 \, b \arccos \left (b x + a\right )} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{2 \, b \arccos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 53, normalized size = 0.82 \[ \frac {\frac {\sqrt {1-\left (b x +a \right )^{2}}}{2 \arccos \left (b x +a \right )^{2}}+\frac {b x +a}{2 \arccos \left (b x +a \right )}+\frac {\Si \left (\arccos \left (b x +a \right )\right )}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccos(b*x+a)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\mathrm {acos}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/acos(a + b*x)^3,x)

[Out]

int(1/acos(a + b*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {acos}^{3}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acos(b*x+a)**3,x)

[Out]

Integral(acos(a + b*x)**(-3), x)

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