\(\int \frac {1}{(a+b \arccos (c x))^2} \, dx\) [165]

Optimal result

Integrand size = 10, antiderivative size = 86 \[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=\frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{b^2 c}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{b^2 c} \]

[Out]

-Ci((a+b*arccos(c*x))/b)*cos(a/b)/b^2/c-Si((a+b*arccos(c*x))/b)*sin(a/b)/b^2/c+(-c^2*x^2+1)^(1/2)/b/c/(a+b*arc
cos(c*x))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4718, 4810, 3384, 3380, 3383} \[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{b^2 c}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{b^2 c}+\frac {\sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))} \]

[In]

Int[(a + b*ArcCos[c*x])^(-2),x]

[Out]

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

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 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 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[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 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]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=\frac {\frac {b \sqrt {1-c^2 x^2}}{a+b \arccos (c x)}-\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right )-\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )}{b^2 c} \]

[In]

Integrate[(a + b*ArcCos[c*x])^(-2),x]

[Out]

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

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {-c^{2} x^{2}+1}}{\left (a +b \arccos \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{b^{2}}}{c}\)	\(74\)
default	\(\frac {\frac {\sqrt {-c^{2} x^{2}+1}}{\left (a +b \arccos \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{b^{2}}}{c}\)	\(74\)

[In]

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

[Out]

1/c*((-c^2*x^2+1)^(1/2)/(a+b*arccos(c*x))/b-(Si(arccos(c*x)+a/b)*sin(a/b)+Ci(arccos(c*x)+a/b)*cos(a/b))/b^2)

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*acos(c*x))**(-2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (84) = 168\).

Time = 0.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.24 \[ \int \frac {1}{(a+b \arccos (c x))^2} \, dx=-\frac {b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} - \frac {b \arccos \left (c x\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} - \frac {a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} - \frac {a \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} + \frac {\sqrt {-c^{2} x^{2} + 1} b}{b^{3} c \arccos \left (c x\right ) + a b^{2} c} \]

[In]

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

[Out]

-b*arccos(c*x)*cos(a/b)*cos_integral(a/b + arccos(c*x))/(b^3*c*arccos(c*x) + a*b^2*c) - b*arccos(c*x)*sin(a/b)
*sin_integral(a/b + arccos(c*x))/(b^3*c*arccos(c*x) + a*b^2*c) - a*cos(a/b)*cos_integral(a/b + arccos(c*x))/(b
^3*c*arccos(c*x) + a*b^2*c) - a*sin(a/b)*sin_integral(a/b + arccos(c*x))/(b^3*c*arccos(c*x) + a*b^2*c) + sqrt(
-c^2*x^2 + 1)*b/(b^3*c*arccos(c*x) + a*b^2*c)

Mupad [F(-1)]

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

[In]

int(1/(a + b*acos(c*x))^2,x)

[Out]

int(1/(a + b*acos(c*x))^2, x)