\(\int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx\) [195]

Optimal result

Integrand size = 12, antiderivative size = 137 \[ \int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx=\frac {2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{b^{3/2} c}-\frac {2 \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{b^{3/2} c} \]

[Out]

-2*cos(a/b)*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arccos(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/c-2*FresnelS(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^(3/2)/c+2*(-c^2*x^2+1)^(1/2)/b/c/
(a+b*arccos(c*x))^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {4718, 4810, 3387, 3386, 3432, 3385, 3433} \[ \int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx=-\frac {2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{b^{3/2} c}-\frac {2 \sqrt {2 \pi } \sin \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{b^{3/2} c}+\frac {2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arccos (c x)}} \]

[In]

Int[(a + b*ArcCos[c*x])^(-3/2),x]

[Out]

(2*Sqrt[1 - c^2*x^2])/(b*c*Sqrt[a + b*ArcCos[c*x]]) - (2*Sqrt[2*Pi]*Cos[a/b]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*A
rcCos[c*x]])/Sqrt[b]])/(b^(3/2)*c) - (2*Sqrt[2*Pi]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[
a/b])/(b^(3/2)*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 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 {2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arccos (c x)}}+\frac {(2 c) \int \frac {x}{\sqrt {1-c^2 x^2} \sqrt {a+b \arccos (c x)}} \, dx}{b} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {2 \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \arccos (c x)\right )}{b^2 c} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {\left (2 \cos \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 )}{b^2 c}-\frac {\left (2 \sin \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 )}{b^2 c} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {\left (4 \cos \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 )}{b^2 c}-\frac {\left (4 \sin \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 )}{b^2 c} \\ & = \frac {2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arccos (c x)}}-\frac {2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right )}{b^{3/2} c}-\frac {2 \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{b^{3/2} c} \\ \end{align*}

Mathematica [F]

\[ \int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx=\int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx \]

[In]

Integrate[(a + b*ArcCos[c*x])^(-3/2),x]

[Out]

Integrate[(a + b*ArcCos[c*x])^(-3/2), x]

Maple [A] (verified)

Time = 1.96 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15

method	result	size
default	\(-\frac {2 \left (\sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}-\sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}+\sin \left (-\frac {a +b \arccos \left (c x \right )}{b}+\frac {a}{b}\right )\right )}{c b \sqrt {a +b \arccos \left (c x \right )}}\)	\(157\)

[In]

int(1/(a+b*arccos(c*x))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/c/b/(a+b*arccos(c*x))^(1/2)*(Pi^(1/2)*2^(1/2)*(a+b*arccos(c*x))^(1/2)*cos(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(-
1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*(-1/b)^(1/2)-Pi^(1/2)*2^(1/2)*(a+b*arccos(c*x))^(1/2)*sin(a/b)*FresnelS(
2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*(-1/b)^(1/2)+sin(-(a+b*arccos(c*x))/b+a/b))

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*arccos(c*x))^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(a+b*acos(c*x))**(3/2),x)

[Out]

Integral((a + b*acos(c*x))**(-3/2), x)

Maxima [F]

\[ \int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arccos(c*x))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*arccos(c*x) + a)^(-3/2), x)

Giac [F]

\[ \int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arccos(c*x))^(3/2),x, algorithm="giac")

[Out]

integrate((b*arccos(c*x) + a)^(-3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \arccos (c x))^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(a + b*acos(c*x))^(3/2),x)

[Out]

int(1/(a + b*acos(c*x))^(3/2), x)