\(\int \frac {1}{\arccos (a+b x)^3} \, dx\) [36]

Optimal result

Integrand size = 8, antiderivative size = 65 \[ \int \frac {1}{\arccos (a+b x)^3} \, dx=\frac {\sqrt {1-(a+b x)^2}}{2 b \arccos (a+b x)^2}+\frac {a+b x}{2 b \arccos (a+b x)}+\frac {\text {Si}(\arccos (a+b x))}{2 b} \]

[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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4888, 4718, 4808, 4720, 3380} \[ \int \frac {1}{\arccos (a+b x)^3} \, dx=\frac {\text {Si}(\arccos (a+b x))}{2 b}+\frac {a+b x}{2 b \arccos (a+b x)}+\frac {\sqrt {1-(a+b x)^2}}{2 b \arccos (a+b x)^2} \]

[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 3380

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 4718

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] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 4720

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

Rule 4808

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(-(f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcCos[c*x])^(n + 1), x] + Dist[f*(m/(
b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], 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]

Rule 4888

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\arccos (x)^3} \, dx,x,a+b x\right )}{b} \\ & = \frac {\sqrt {1-(a+b x)^2}}{2 b \arccos (a+b x)^2}+\frac {\text {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \arccos (x)^2} \, dx,x,a+b x\right )}{2 b} \\ & = \frac {\sqrt {1-(a+b x)^2}}{2 b \arccos (a+b x)^2}+\frac {a+b x}{2 b \arccos (a+b x)}-\frac {\text {Subst}\left (\int \frac {1}{\arccos (x)} \, dx,x,a+b x\right )}{2 b} \\ & = \frac {\sqrt {1-(a+b x)^2}}{2 b \arccos (a+b x)^2}+\frac {a+b x}{2 b \arccos (a+b x)}+\frac {\text {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\arccos (a+b x)\right )}{2 b} \\ & = \frac {\sqrt {1-(a+b x)^2}}{2 b \arccos (a+b x)^2}+\frac {a+b x}{2 b \arccos (a+b x)}+\frac {\text {Si}(\arccos (a+b x))}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\arccos (a+b x)^3} \, dx=\frac {\sqrt {1-(a+b x)^2}}{2 b \arccos (a+b x)^2}+\frac {a+b x}{2 b \arccos (a+b x)}+\frac {\text {Si}(\arccos (a+b x))}{2 b} \]

[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)

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {1-\left (b x +a \right )^{2}}}{2 \arccos \left (b x +a \right )^{2}}+\frac {b x +a}{2 \arccos \left (b x +a \right )}+\frac {\operatorname {Si}\left (\arccos \left (b x +a \right )\right )}{2}}{b}\)	\(53\)
default	\(\frac {\frac {\sqrt {1-\left (b x +a \right )^{2}}}{2 \arccos \left (b x +a \right )^{2}}+\frac {b x +a}{2 \arccos \left (b x +a \right )}+\frac {\operatorname {Si}\left (\arccos \left (b x +a \right )\right )}{2}}{b}\)	\(53\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {1}{\arccos (a+b x)^3} \, dx=\int { \frac {1}{\arccos \left (b x + a\right )^{3}} \,d x } \]

[In]

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

[Out]

integral(arccos(b*x + a)^(-3), x)

Sympy [F]

\[ \int \frac {1}{\arccos (a+b x)^3} \, dx=\int \frac {1}{\operatorname {acos}^{3}{\left (a + b x \right )}}\, dx \]

[In]

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

[Out]

Integral(acos(a + b*x)**(-3), x)

Maxima [F]

\[ \int \frac {1}{\arccos (a+b x)^3} \, dx=\int { \frac {1}{\arccos \left (b x + a\right )^{3}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\arccos (a+b x)^3} \, dx=\frac {\operatorname {Si}\left (\arccos \left (b x + a\right )\right )}{2 \, b} + \frac {b x + a}{2 \, b \arccos \left (b x + a\right )} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{2 \, b \arccos \left (b x + a\right )^{2}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\arccos (a+b x)^3} \, dx=\int \frac {1}{{\mathrm {acos}\left (a+b\,x\right )}^3} \,d x \]

[In]

int(1/acos(a + b*x)^3,x)

[Out]

int(1/acos(a + b*x)^3, x)