3.2.70 \(\int \frac {1}{(a+b \text {ArcCos}(c x))^3} \, dx\) [170]

Optimal. Leaf size=111 \[ \frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \text {ArcCos}(c x))^2}+\frac {x}{2 b^2 (a+b \text {ArcCos}(c x))}-\frac {\text {CosIntegral}\left (\frac {a+b \text {ArcCos}(c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{2 b^3 c}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcCos}(c x)}{b}\right )}{2 b^3 c} \]

[Out]

1/2*x/b^2/(a+b*arccos(c*x))+1/2*cos(a/b)*Si((a+b*arccos(c*x))/b)/b^3/c-1/2*Ci((a+b*arccos(c*x))/b)*sin(a/b)/b^
3/c+1/2*(-c^2*x^2+1)^(1/2)/b/c/(a+b*arccos(c*x))^2

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Rubi [A]
time = 0.13, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4718, 4808, 4720, 3384, 3380, 3383} \begin {gather*} -\frac {\sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcCos}(c x)}{b}\right )}{2 b^3 c}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcCos}(c x)}{b}\right )}{2 b^3 c}+\frac {x}{2 b^2 (a+b \text {ArcCos}(c x))}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \text {ArcCos}(c x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])^(-3),x]

[Out]

Sqrt[1 - c^2*x^2]/(2*b*c*(a + b*ArcCos[c*x])^2) + x/(2*b^2*(a + b*ArcCos[c*x])) - (CosIntegral[(a + b*ArcCos[c
*x])/b]*Sin[a/b])/(2*b^3*c) + (Cos[a/b]*SinIntegral[(a + b*ArcCos[c*x])/b])/(2*b^3*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 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]

Rubi steps

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

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Mathematica [A]
time = 0.18, size = 89, normalized size = 0.80 \begin {gather*} \frac {\frac {b \left (a c x+b \sqrt {1-c^2 x^2}+b c x \text {ArcCos}(c x)\right )}{(a+b \text {ArcCos}(c x))^2}-\text {CosIntegral}\left (\frac {a}{b}+\text {ArcCos}(c x)\right ) \sin \left (\frac {a}{b}\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\text {ArcCos}(c x)\right )}{2 b^3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCos[c*x])^(-3),x]

[Out]

((b*(a*c*x + b*Sqrt[1 - c^2*x^2] + b*c*x*ArcCos[c*x]))/(a + b*ArcCos[c*x])^2 - CosIntegral[a/b + ArcCos[c*x]]*
Sin[a/b] + Cos[a/b]*SinIntegral[a/b + ArcCos[c*x]])/(2*b^3*c)

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Maple [A]
time = 0.08, size = 139, normalized size = 1.25

method result size
derivativedivides \(\frac {\frac {\sqrt {-c^{2} x^{2}+1}}{2 \left (a +b \arccos \left (c x \right )\right )^{2} b}+\frac {\arccos \left (c x \right ) \sinIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -\arccos \left (c x \right ) \cosineIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +\sinIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -\cosineIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +x b c}{2 \left (a +b \arccos \left (c x \right )\right ) b^{3}}}{c}\) \(139\)
default \(\frac {\frac {\sqrt {-c^{2} x^{2}+1}}{2 \left (a +b \arccos \left (c x \right )\right )^{2} b}+\frac {\arccos \left (c x \right ) \sinIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -\arccos \left (c x \right ) \cosineIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +\sinIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -\cosineIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +x b c}{2 \left (a +b \arccos \left (c x \right )\right ) b^{3}}}{c}\) \(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(1/2/(a+b*arccos(c*x))^2/b*(-c^2*x^2+1)^(1/2)+1/2*(arccos(c*x)*Si(arccos(c*x)+a/b)*cos(a/b)*b-arccos(c*x)*
Ci(arccos(c*x)+a/b)*sin(a/b)*b+Si(arccos(c*x)+a/b)*cos(a/b)*a-Ci(arccos(c*x)+a/b)*sin(a/b)*a+x*b*c)/(a+b*arcco
s(c*x))/b^3)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*c*x*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*c*x + sqrt(c*x + 1)*sqrt(-c*x + 1)*b - 2*(b^4*c*arct
an2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2 + 2*a*b^3*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a^2*b^2*c)*i
ntegrate(1/2/(b^3*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b^2), x))/(b^4*c*arctan2(sqrt(c*x + 1)*sqrt(-
c*x + 1), c*x)^2 + 2*a*b^3*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a^2*b^2*c)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/(b^3*arccos(c*x)^3 + 3*a*b^2*arccos(c*x)^2 + 3*a^2*b*arccos(c*x) + a^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acos(c*x))**(-3), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (101) = 202\).
time = 0.42, size = 481, normalized size = 4.33 \begin {gather*} -\frac {b^{2} \arccos \left (c x\right )^{2} \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{2 \, {\left (b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c\right )}} + \frac {b^{2} \arccos \left (c x\right )^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{2 \, {\left (b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c\right )}} + \frac {b^{2} c x \arccos \left (c x\right )}{2 \, {\left (b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c\right )}} - \frac {a b \arccos \left (c x\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c} + \frac {a b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c} + \frac {a b c x}{2 \, {\left (b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c\right )}} - \frac {a^{2} \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{2 \, {\left (b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c\right )}} + \frac {a^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{2 \, {\left (b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c\right )}} + \frac {\sqrt {-c^{2} x^{2} + 1} b^{2}}{2 \, {\left (b^{5} c \arccos \left (c x\right )^{2} + 2 \, a b^{4} c \arccos \left (c x\right ) + a^{2} b^{3} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^2*arccos(c*x)^2*cos_integral(a/b + arccos(c*x))*sin(a/b)/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*arccos(c*x) +
 a^2*b^3*c) + 1/2*b^2*arccos(c*x)^2*cos(a/b)*sin_integral(a/b + arccos(c*x))/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*
arccos(c*x) + a^2*b^3*c) + 1/2*b^2*c*x*arccos(c*x)/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*arccos(c*x) + a^2*b^3*c) -
 a*b*arccos(c*x)*cos_integral(a/b + arccos(c*x))*sin(a/b)/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*arccos(c*x) + a^2*b
^3*c) + a*b*arccos(c*x)*cos(a/b)*sin_integral(a/b + arccos(c*x))/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*arccos(c*x)
+ a^2*b^3*c) + 1/2*a*b*c*x/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*arccos(c*x) + a^2*b^3*c) - 1/2*a^2*cos_integral(a/
b + arccos(c*x))*sin(a/b)/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*arccos(c*x) + a^2*b^3*c) + 1/2*a^2*cos(a/b)*sin_int
egral(a/b + arccos(c*x))/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*arccos(c*x) + a^2*b^3*c) + 1/2*sqrt(-c^2*x^2 + 1)*b^
2/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*arccos(c*x) + a^2*b^3*c)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acos(c*x))^3,x)

[Out]

int(1/(a + b*acos(c*x))^3, x)

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