\(\int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx\) [200]

Optimal result

Integrand size = 12, antiderivative size = 163 \[ \int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx=\frac {2 \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}}+\frac {4 x}{3 b^2 \sqrt {a+b \arccos (c x)}}+\frac {4 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c}-\frac {4 \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{3 b^{5/2} c} \]

[Out]

4/3*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(5/2)/c-4/3*Fresnel
C(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*sin(a/b)*2^(1/2)*Pi^(1/2)/b^(5/2)/c+2/3*(-c^2*x^2+1)^(1/2)
/b/c/(a+b*arccos(c*x))^(3/2)+4/3*x/b^2/(a+b*arccos(c*x))^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4718, 4808, 4720, 3387, 3386, 3432, 3385, 3433} \[ \int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx=-\frac {4 \sqrt {2 \pi } \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c}+\frac {4 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c}+\frac {4 x}{3 b^2 \sqrt {a+b \arccos (c x)}}+\frac {2 \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}} \]

[In]

Int[(a + b*ArcCos[c*x])^(-5/2),x]

[Out]

(2*Sqrt[1 - c^2*x^2])/(3*b*c*(a + b*ArcCos[c*x])^(3/2)) + (4*x)/(3*b^2*Sqrt[a + b*ArcCos[c*x]]) + (4*Sqrt[2*Pi
]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/(3*b^(5/2)*c) - (4*Sqrt[2*Pi]*FresnelC[(Sqr
t[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b])/(3*b^(5/2)*c)

Rule 3385

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[f*(x^2/d)],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3386

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[f*(x^2/d)], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3387

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

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*} \text {integral}& = \frac {2 \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}}+\frac {(2 c) \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))^{3/2}} \, dx}{3 b} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}}+\frac {4 x}{3 b^2 \sqrt {a+b \arccos (c x)}}-\frac {4 \int \frac {1}{\sqrt {a+b \arccos (c x)}} \, dx}{3 b^2} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}}+\frac {4 x}{3 b^2 \sqrt {a+b \arccos (c x)}}-\frac {4 \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arccos (c x)\right )}{3 b^3 c} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}}+\frac {4 x}{3 b^2 \sqrt {a+b \arccos (c x)}}+\frac {\left (4 \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arccos (c x)\right )}{3 b^3 c}-\frac {\left (4 \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arccos (c x)\right )}{3 b^3 c} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}}+\frac {4 x}{3 b^2 \sqrt {a+b \arccos (c x)}}+\frac {\left (8 \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \arccos (c x)}\right )}{3 b^3 c}-\frac {\left (8 \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \arccos (c x)}\right )}{3 b^3 c} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{3 b c (a+b \arccos (c x))^{3/2}}+\frac {4 x}{3 b^2 \sqrt {a+b \arccos (c x)}}+\frac {4 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c}-\frac {4 \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{3 b^{5/2} c} \\ \end{align*}

Mathematica [F]

\[ \int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx=\int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx \]

[In]

Integrate[(a + b*ArcCos[c*x])^(-5/2),x]

[Out]

Integrate[(a + b*ArcCos[c*x])^(-5/2), x]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(340\) vs. \(2(129)=258\).

Time = 2.09 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.09

method	result	size
default	\(\frac {-\frac {4 \arccos \left (c x \right ) \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}{3}-\frac {4 \arccos \left (c x \right ) \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}{3}-\frac {4 \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, a}{3}-\frac {4 \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, a}{3}+\frac {4 \arccos \left (c x \right ) \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) b}{3}-\frac {2 \sin \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) b}{3}+\frac {4 \cos \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right ) a}{3}}{c \,b^{2} \left (a +b \arccos \left (c x \right )\right )^{\frac {3}{2}}}\)	\(341\)

[In]

int(1/(a+b*arccos(c*x))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3/c/b^2*(-2*arccos(c*x)*(a+b*arccos(c*x))^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(
c*x))^(1/2)/b)*2^(1/2)*Pi^(1/2)*(-1/b)^(1/2)*b-2*arccos(c*x)*(a+b*arccos(c*x))^(1/2)*sin(a/b)*FresnelC(2^(1/2)
/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*2^(1/2)*Pi^(1/2)*(-1/b)^(1/2)*b-2*(a+b*arccos(c*x))^(1/2)*co
s(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*2^(1/2)*Pi^(1/2)*(-1/b)^(1/2)*a-2*(a+
b*arccos(c*x))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*2^(1/2)*Pi^(1/
2)*(-1/b)^(1/2)*a+2*arccos(c*x)*cos(-(a+b*arccos(c*x))/b+a/b)*b-sin(-(a+b*arccos(c*x))/b+a/b)*b+2*cos(-(a+b*ar
ccos(c*x))/b+a/b)*a)/(a+b*arccos(c*x))^(3/2)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*arccos(c*x))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(a+b*acos(c*x))**(5/2),x)

[Out]

Integral((a + b*acos(c*x))**(-5/2), x)

Maxima [F]

\[ \int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arccos(c*x))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arccos(c*x) + a)^(-5/2), x)

Giac [F]

\[ \int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arccos(c*x))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccos(c*x) + a)^(-5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \arccos (c x))^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(1/(a + b*acos(c*x))^(5/2),x)

[Out]

int(1/(a + b*acos(c*x))^(5/2), x)