Integrand size = 10, antiderivative size = 111 \[ \int \frac {1}{(a+b \arccos (c x))^3} \, dx=\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}+\frac {x}{2 b^2 (a+b \arccos (c x))}-\frac {\operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{2 b^3 c}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{2 b^3 c} \] Output:
1/2*(-c^2*x^2+1)^(1/2)/b/c/(a+b*arccos(c*x))^2+1/2*x/b^2/(a+b*arccos(c*x)) -1/2*Ci((a+b*arccos(c*x))/b)*sin(a/b)/b^3/c+1/2*cos(a/b)*Si((a+b*arccos(c* x))/b)/b^3/c
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(a+b \arccos (c x))^3} \, dx=\frac {\frac {b \left (a c x+b \sqrt {1-c^2 x^2}+b c x \arccos (c x)\right )}{(a+b \arccos (c x))^2}-\operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right ) \sin \left (\frac {a}{b}\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )}{2 b^3 c} \] Input:
Integrate[(a + b*ArcCos[c*x])^(-3),x]
Output:
((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)
Time = 0.76 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5133, 5223, 5135, 25, 3042, 3784, 25, 3042, 3780, 3783}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(a+b \arccos (c x))^3} \, dx\) |
\(\Big \downarrow \) 5133 |
\(\displaystyle \frac {c \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))^2}dx}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 5223 |
\(\displaystyle \frac {c \left (\frac {x}{b c (a+b \arccos (c x))}-\frac {\int \frac {1}{a+b \arccos (c x)}dx}{b c}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 5135 |
\(\displaystyle \frac {c \left (\frac {\int -\frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^2}+\frac {x}{b c (a+b \arccos (c x))}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {c \left (\frac {x}{b c (a+b \arccos (c x))}-\frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^2}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {c \left (\frac {x}{b c (a+b \arccos (c x))}-\frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^2}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle \frac {c \left (\frac {-\sin \left (\frac {a}{b}\right ) \int \frac {\cos \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))-\cos \left (\frac {a}{b}\right ) \int -\frac {\sin \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^2}+\frac {x}{b c (a+b \arccos (c x))}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {c \left (\frac {\cos \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))-\sin \left (\frac {a}{b}\right ) \int \frac {\cos \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^2}+\frac {x}{b c (a+b \arccos (c x))}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {c \left (\frac {\cos \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))-\sin \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arccos (c x)}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^2}+\frac {x}{b c (a+b \arccos (c x))}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle \frac {c \left (\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )-\sin \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arccos (c x)}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^2}+\frac {x}{b c (a+b \arccos (c x))}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {c \left (\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )-\sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{b^2 c^2}+\frac {x}{b c (a+b \arccos (c x))}\right )}{2 b}+\frac {\sqrt {1-c^2 x^2}}{2 b c (a+b \arccos (c x))^2}\) |
Input:
Int[(a + b*ArcCos[c*x])^(-3),x]
Output:
Sqrt[1 - c^2*x^2]/(2*b*c*(a + b*ArcCos[c*x])^2) + (c*(x/(b*c*(a + b*ArcCos [c*x])) + (-(CosIntegral[(a + b*ArcCos[c*x])/b]*Sin[a/b]) + Cos[a/b]*SinIn tegral[(a + b*ArcCos[c*x])/b])/(b^2*c^2)))/(2*b)
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* e - c*f)/d] Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[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]
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] - Simp[c/(b*(n + 1 )) Int[x*((a + b*ArcCos[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ [{a, b, c}, x] && LtQ[n, -1]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[-(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]
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] + Simp[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]
Time = 0.00 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(\frac {\frac {\sqrt {-c^{2} x^{2}+1}}{2 \left (a +b \arccos \left (c x \right )\right )^{2} b}+\frac {\arccos \left (c x \right ) \operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -\arccos \left (c x \right ) \operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +c x b}{2 \left (a +b \arccos \left (c x \right )\right ) b^{3}}}{c}\) | \(139\) |
default | \(\frac {\frac {\sqrt {-c^{2} x^{2}+1}}{2 \left (a +b \arccos \left (c x \right )\right )^{2} b}+\frac {\arccos \left (c x \right ) \operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -\arccos \left (c x \right ) \operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +c x b}{2 \left (a +b \arccos \left (c x \right )\right ) b^{3}}}{c}\) | \(139\) |
Input:
int(1/(a+b*arccos(c*x))^3,x,method=_RETURNVERBOSE)
Output:
1/c*(1/2*(-c^2*x^2+1)^(1/2)/(a+b*arccos(c*x))^2/b+1/2*(arccos(c*x)*Si(arcc os(c*x)+a/b)*cos(a/b)*b-arccos(c*x)*Ci(arccos(c*x)+a/b)*sin(a/b)*b+Si(arcc os(c*x)+a/b)*cos(a/b)*a-Ci(arccos(c*x)+a/b)*sin(a/b)*a+c*x*b)/(a+b*arccos( c*x))/b^3)
\[ \int \frac {1}{(a+b \arccos (c x))^3} \, dx=\int { \frac {1}{{\left (b \arccos \left (c x\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(1/(a+b*arccos(c*x))^3,x, algorithm="fricas")
Output:
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)
\[ \int \frac {1}{(a+b \arccos (c x))^3} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{3}}\, dx \] Input:
integrate(1/(a+b*acos(c*x))**3,x)
Output:
Integral((a + b*acos(c*x))**(-3), x)
\[ \int \frac {1}{(a+b \arccos (c x))^3} \, dx=\int { \frac {1}{{\left (b \arccos \left (c x\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(1/(a+b*arccos(c*x))^3,x, algorithm="maxima")
Output:
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*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)*int egrate(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(sqr t(c*x + 1)*sqrt(-c*x + 1), c*x) + a^2*b^2*c)
Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (101) = 202\).
Time = 0.14 (sec) , antiderivative size = 481, normalized size of antiderivative = 4.33 \[ \int \frac {1}{(a+b \arccos (c x))^3} \, dx=-\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 )}} \] Input:
integrate(1/(a+b*arccos(c*x))^3,x, algorithm="giac")
Output:
-1/2*b^2*arccos(c*x)^2*cos_integral(a/b + arccos(c*x))*sin(a/b)/(b^5*c*arc cos(c*x)^2 + 2*a*b^4*c*arccos(c*x) + a^2*b^3*c) + 1/2*b^2*arccos(c*x)^2*co s(a/b)*sin_integral(a/b + arccos(c*x))/(b^5*c*arccos(c*x)^2 + 2*a*b^4*c*ar ccos(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 + arcc os(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 + ar ccos(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_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*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)
Timed out. \[ \int \frac {1}{(a+b \arccos (c x))^3} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^3} \,d x \] Input:
int(1/(a + b*acos(c*x))^3,x)
Output:
int(1/(a + b*acos(c*x))^3, x)
\[ \int \frac {1}{(a+b \arccos (c x))^3} \, dx=\int \frac {1}{\mathit {acos} \left (c x \right )^{3} b^{3}+3 \mathit {acos} \left (c x \right )^{2} a \,b^{2}+3 \mathit {acos} \left (c x \right ) a^{2} b +a^{3}}d x \] Input:
int(1/(a+b*acos(c*x))^3,x)
Output:
int(1/(acos(c*x)**3*b**3 + 3*acos(c*x)**2*a*b**2 + 3*acos(c*x)*a**2*b + a* *3),x)