\(\int \frac {1}{\arccos (a+b x)^2} \, dx\) [35]

Optimal result

Integrand size = 8, antiderivative size = 40 \[ \int \frac {1}{\arccos (a+b x)^2} \, dx=\frac {\sqrt {1-(a+b x)^2}}{b \arccos (a+b x)}-\frac {\operatorname {CosIntegral}(\arccos (a+b x))}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4888, 4718, 4810, 3383} \[ \int \frac {1}{\arccos (a+b x)^2} \, dx=\frac {\sqrt {1-(a+b x)^2}}{b \arccos (a+b x)}-\frac {\operatorname {CosIntegral}(\arccos (a+b x))}{b} \]

[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 3383

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 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 4810

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\arccos (a+b x)^2} \, dx=\frac {\sqrt {1-(a+b x)^2}}{b \arccos (a+b x)}-\frac {\operatorname {CosIntegral}(\arccos (a+b x))}{b} \]

[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] (verified)

Time = 0.71 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {1-\left (b x +a \right )^{2}}}{\arccos \left (b x +a \right )}-\operatorname {Ci}\left (\arccos \left (b x +a \right )\right )}{b}\)	\(37\)
default	\(\frac {\frac {\sqrt {1-\left (b x +a \right )^{2}}}{\arccos \left (b x +a \right )}-\operatorname {Ci}\left (\arccos \left (b x +a \right )\right )}{b}\)	\(37\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {1}{\arccos (a+b x)^2} \, dx=\int { \frac {1}{\arccos \left (b x + a\right )^{2}} \,d x } \]

[In]

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

[Out]

integral(arccos(b*x + a)^(-2), x)

Sympy [F]

\[ \int \frac {1}{\arccos (a+b x)^2} \, dx=\int \frac {1}{\operatorname {acos}^{2}{\left (a + b x \right )}}\, dx \]

[In]

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

[Out]

Integral(acos(a + b*x)**(-2), x)

Maxima [F]

\[ \int \frac {1}{\arccos (a+b x)^2} \, dx=\int { \frac {1}{\arccos \left (b x + a\right )^{2}} \,d x } \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\arccos (a+b x)^2} \, dx=-\frac {\operatorname {Ci}\left (\arccos \left (b x + a\right )\right )}{b} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{b \arccos \left (b x + a\right )} \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\arccos (a+b x)^2} \, dx=\int \frac {1}{{\mathrm {acos}\left (a+b\,x\right )}^2} \,d x \]

[In]

int(1/acos(a + b*x)^2,x)

[Out]

int(1/acos(a + b*x)^2, x)